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General Physics I
Course Outline
Textbook: the class notes beside the following textbooks:
1. Physics for Scientists and Engineers, Raymond A. Serway, 6th Edition
2. University Physics, Sears, Zemansky and Young
Physics and Measurement: Standards of Length, Mass, and Time, Density and Atomic Mass, Dimensional Analysis, Conversion of Units.
Motion in One Dimension: Position, Velocity, and Speed, Instantaneous Velocity and Speed, Acceleration, Motion Diagrams, One-Dimensional Motion with Constant Acceleration, Freely Falling Objects, Kinematics Equations Derived from Calculus.
Vectors: Coordinate Systems, Vector and Scalar Quantities, Some Properties of Vectors, Components of a Vector and Unit Vectors, The Scalar Product of Two Vectors, The Vector Product of Two Vectors.
Motion in Two Dimensions: The Position, Velocity, and Acceleration Vectors, Two-Dimensional Motion with Constant Acceleration, Projectile Motion, Uniform Circular Motion, Tangential and Radial Acceleration, Relative Velocity and Relative Acceleration.
The Laws of Motion: The Concept of Force. Newton's First Law and Inertial Frames, Newton's Second Law, The Gravitational Force and Weight, Newton's Third Law, Some Applications of Newton's Laws, Forces of Friction.
Energy and Energy Transfer: Systems and Environments, Work Done by a Constant Force, Work Done by a Varying Force, Kinetic Energy and the Work-Kinetic Energy Theorem, The Non-Isolated System, Conservation of Energy,Situations Involving Kinetic Friction, Power.
Potential Energy: Potential Energy of a System, The Isolated System, Conservation of Mechanical Energy, Conservative and Nonconservative Forces, Changes in Mechanical Energy for Nonconservative Forces, Relationship Between Conservative Forces and Potential Energy, Energy Diagrams and Equilibrium of a System.
Linear Momentum and Collisions: Linear Momentum and Its Conservation,Impulse and Momentum, Collisions in One Dimension, Two-Dimensional Collisions, The Center of Mass, Motion of a System of Particles.
Universal Gravitation: Newton's Law of Universal Gravitation, Measuring the Gravitational Constant, Free-Fall Acceleration and the Gravitational Force,Kepler's Laws and the Motion of Planets, The Gravitational Field, Gravitational Potential Energy, Energy Considerations in Planetary and Satellite Motion.
GRADING POLICY
Your grade will be judged on your performance in Home work, Quizzes, tow tests and the Lab. Points will be allocated to each of these in the following manner :
GRADING SCALE:
Grade ComponentWeight
HW/Quizzes30
Labs10
Midterm Exam25
Final Exam35
Total100
Measurement & Units
Physical quantities (in mechanics)
Basic quantities : in mechanics the three fundamental quantities are
Length (L), mass (M), time (T)
Derived quantities : all other physical quantities in mechanics can be expressed in term of basic quantities
Area Volume VelocityAccelerationForce MomentumWork…..…..
MassThe SI unit of mass is the Kilogram, which is defined as the mass of a specific platinum-iridium alloy cylinder.
TimeThe SI unit of time is the Second, which is the time required for a cesium-133 atom to undergo 9192631770 vibrations.
LengthThe SI unit of length is Meter, which is the distance traveled by light is vacuum during a time of 1/2999792458 second.
Systems of Units
SI units (International System of Units):
length: meter (m), mass: kilogram (kg), time: second (s) *This system is also referred to as the mks system for meter-kilogram-second .
Gaussian units
length: centimeter (cm), mass: gram (g), time: second (s) *This system is also referred to as the cgs system for centimeter-gram-second.
British engineering system:
Length: inches, feet, miles, mass: slugs (pounds), time: seconds
We will use mostly SI units, but you may run across some problems using British units. You should know how to convert back & forth.
ConversionsConversionsWhen units are not consistent, you may need to convert When units are not consistent, you may need to convert toto
appropriate onesappropriate ones..
Units can be treated like algebraic quantities that can Units can be treated like algebraic quantities that can cancel each other outcancel each other out..1 mile = 1609 m = 1.609 km1 ft = 0.3048 m = 30.48 cm1m = 39.37 in = 3.281 ft1 in = 0.0254 m = 2.54 cm
1 mile = 5280 ft
Example: Convert miles per hour to meters per second:Example: Convert miles per hour to meters per second:
s
m
2
1
s
m447.0
s 3600
hr 1
ft28.3
m 1
mi
ft 5280
hr
mi 1
hr
mi1
Questions:1. Convert 500 millimeters into meters.2. Convert 4.2 liters into milliliters.3. Convert 1.45 meters into inches.4. Convert 65 miles per hour into
kilometers per second.
PrefixesPrefixesPrefixes correspond to powers of 10Prefixes correspond to powers of 10
Each prefix has a specific name/abbreviationEach prefix has a specific name/abbreviation
Power Prefix Abbrev.
1015 peta P109 giga G106 mega M103 kilo k10-2 centi P10-3 milli m10-6 micro 10-9 nano n10-12 pico p10-15 femto f
Distance from Earth to nearest star 40 PmMean radius of Earth 6 MmLength of a housefly 5 mmSize of living cells 10 mSize of an atom 0.1 nm
QuantityQuantityLengthLengthMassMasstimetime
DimensionDimension[[LL=]=]LL
[[MM=]=]MM[[TT=]=]TT
areaarea[[AA = ] = ]LL22
volumevolume[[VV=]=]LL33
velocityvelocity[ [ vv =] =]L/TL/T
AccelerationAccelerationforceforce
[[aa = ] = ]L/TL/T22
[[ff=]=]M L/TM L/T22
Dimensional Analysis
Definition: The Dimension is the qualitative nature of a physical quantity (length, mass, time) .
brackets ] [ denote the dimension or units of a physical quantity:
Idea: Dimensional analysis can be used to derive or check formulas by treating dimensions as algebraic quantities. Quantities can be added or subtracted only if they have the same dimensions, and quantities on two sides of an equation must have the same dimensions
Example: Using the dimensional analysis check that this equation
x = ½ at2 is correct, where x is the distance, a is the acceleration and t is the time.
L]x[ left hand side
LTT
L]at
2
1[ 2
22 right hand side
This equation is correct because the dimension of the left and right side of the equation have the same dimensions.
Solution
Example:Suppose that the acceleration of a particle moving in circle of radius r with uniform velocity v is proportional to the rn and vm.Use the dimensional analysis to determine the power n and m.
The left hand side
Therefore
or
SolutionLet us assume a is represented in this expression
a = k rn vm
Where k is the proportionality constant of dimensionless unit.
The right hand side [a] = L/T2
mnm2 TLLT
Hence
n+m=1 and m=2 Therefore
n =-1
and the acceleration a is a = k r -1 v2
Problem:
1 .Show that the expression x = vt +1/2 at2 is dimensionally correct , where x is coordinate and has unit of length, v is velocity, a is acceleration and t is the time.
2 .Show that the period T of a simple pendulum is measured in time unit given by
g
lT 2
DensityDensity
Every substance has a density, designated Every substance has a density, designated = M/V = M/V
Dimensions of density are, units (kg/mDimensions of density are, units (kg/m33))3L
M
• Some examples,
Substance (103 kg/m3)
Gold 19.3
Lead 11.3
Aluminum 2.70
Water 1.00
Atomic DensityAtomic Density In dealing with macroscopic numbers of atoms (and similarIn dealing with macroscopic numbers of atoms (and similar small particles) we often use a convenient quantity called small particles) we often use a convenient quantity called
AvogadroAvogadro’’s Number, Ns Number, NAA = 6.023 x 10 = 6.023 x 1023 23 atoms per moleatoms per mole
Commonly used mass units in regards to elements Commonly used mass units in regards to elements 1. Molar Mass = mass in grams of one mole of the 1. Molar Mass = mass in grams of one mole of the
substance (averaging over natural isotope occurrences)substance (averaging over natural isotope occurrences)2. Atomic Mass = mass in u (a.m.u.) of one atom of a 2. Atomic Mass = mass in u (a.m.u.) of one atom of a
substance. substance. It is approximately the total number of protons and It is approximately the total number of protons and
neutrons in one atom of that substance. neutrons in one atom of that substance. 1u = 1.660 538 7 x 101u = 1.660 538 7 x 10-27-27 kg kg
atom/mol10023.6
g/mol .12 (carbon)
23M
What is the mass of a single carbon (C12) atom ?
= 2 x 10-23 g/atom
Coordinate Systems & Vectors
Coordinate Systems and Frames of Reference
The location of a point on a line can be described by one coordinate; a point on a plane can be described by two coordinates; a point in a three dimensional volume can be described by three coordinates. In general, the number of coordinates equals the number of dimensions. A coordinate system consists of:
1 .a fixed reference point (origin)
2 .a set of axes with specified directions and scales
3 .instructions that specify how to label a point in space relative to the origin and axes
Coordinate SystemsCoordinate Systems
In 1 dimension, only 1 kind of system, In 1 dimension, only 1 kind of system, – Linear Coordinates Linear Coordinates ((xx) ) +/-+/-
In 2 dimensions there are two commonly used In 2 dimensions there are two commonly used systems,systems,– Cartesian CoordinatesCartesian Coordinates ((xx,,yy))– Polar CoordinatesPolar Coordinates ((rr,,))
In 3 dimensions there are three commonly used In 3 dimensions there are three commonly used systems,systems,– Cartesian CoordinatesCartesian Coordinates ((x,y,zx,y,z) ) – Cylindrical Coordinates Cylindrical Coordinates ( (rr,,,z,z))– Spherical CoordinatesSpherical Coordinates ((rr,,))
Cartesian coordinate systemCartesian coordinate system also called rectangular also called rectangular
coordinate systemcoordinate system x and y axesx and y axes points are labeled (x,y)points are labeled (x,y)
Plane polar coordinate systemPlane polar coordinate system
origin and origin and reference line are reference line are notednoted
point is distance r point is distance r from the origin in from the origin in the direction of the direction of angle angle
points are labeled points are labeled (r,(r,))
The relation between coordinates
x rcos sinry
22 yxr
x
ytan
Furthermore, it follows that
Problem: A point is located in polar coordinate system by the coordinate and.
Find the x and y coordinates of this point, assuming the two coordinate systems have the same origin .
5.2r 35
Example: The Cartesian coordinates of a point are given by
)x,y-) =(3.5,-2.5 (meter. Find the polar coordinate of this point .
Solution:
21636180
714.05.3
5.2
x
ytan
m3.4)5.2()5.3(yxr 2222
Note that you must use the signs of x and y to find that is in the third quadrant of coordinate system. That is not 36
216
Scalars and Vectors Scalars have magnitude only. Length, time, mass, speed and volume are examples of scalars.
Vectors have magnitude and direction. The magnitude of is written Position, displacement, velocity, acceleration and force are examples of vector quantities.
v
v
Properties of VectorsProperties of Vectors
Equality of Two VectorsEquality of Two Vectors
Two vectors are Two vectors are equalequal if they have the same if they have the same magnitude and the same directionmagnitude and the same direction
Movement of vectors in a diagramMovement of vectors in a diagram
Any vector can be moved parallel to itself without Any vector can be moved parallel to itself without being affectedbeing affected
Negative VectorsNegative Vectors
Two vectors are Two vectors are negativenegative if they have the same if they have the same magnitudemagnitude
but are 180° apart (opposite directions)but are 180° apart (opposite directions)
Multiplication or division of a vector by a scalar results in a vector for which
)a (only the magnitude changes if the scalar is positive )b (the magnitude changes and the direction
is reversed if the scalar is negative.
Adding VectorsAdding Vectors
When adding vectors, their directions When adding vectors, their directions must be taken into account and units must be taken into account and units must be the samemust be the same
First: Graphical MethodsFirst: Graphical Methods
Second: Algebraic MethodsSecond: Algebraic Methods
Adding Vectors Graphically Adding Vectors Graphically (Triangle Method)(Triangle Method)
Continue drawing the Continue drawing the vectors vectors ““tip-to-tailtip-to-tail””
The resultant is drawn The resultant is drawn from the origin of from the origin of AA to to the end of the last the end of the last vectorvector
Measure the length of Measure the length of RR and its angleand its angle
When you have many When you have many vectors, just keep vectors, just keep repeating the repeating the process until all are process until all are includedincluded
The resultant is still The resultant is still drawn from the drawn from the origin of the first origin of the first vector to the end of vector to the end of the last vectorthe last vector
Alternative Graphical Method Alternative Graphical Method (Parallelogram Method)(Parallelogram Method)
When you have only two When you have only two vectors, you may use the vectors, you may use the Parallelogram MethodParallelogram Method
All vectors, including the All vectors, including the resultant, are drawn resultant, are drawn from a common originfrom a common origin
The remaining sides of The remaining sides of the parallelogram are the parallelogram are sketched to determine sketched to determine the diagonal, the diagonal, RR
Vector SubtractionVector Subtraction
Special case of vector Special case of vector additionaddition
If If AA –– BB, then use , then use AA+(+(--BB))
Continue with Continue with standard vector standard vector addition procedureaddition procedure
Components of a VectorComponents of a Vector
These are the These are the projections of the projections of the vector along the x- vector along the x- and y-axesand y-axes
The x-component of a vector is the projection along the x-axisThe x-component of a vector is the projection along the x-axis
The y-component of a vector is the projection along the y-axisThe y-component of a vector is the projection along the y-axis
ThenThen,,
cosxA A
sinyA A
x yA A ADDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD
x
y12y
2x A
AtanandAAA
Adding Vectors AlgebraicallyAdding Vectors Algebraically
))11((Choose a coordinate system and sketch the Choose a coordinate system and sketch the vectorsvectors
))22((Find the x- and y-components of all the vectorFind the x- and y-components of all the vector
))33((Add all the x-componentsAdd all the x-components
This gives RThis gives Rxx:: xx vR
yy vR
)4(Add all the y-components
This gives Ry
))55((find the magnitude of the Resultantfind the magnitude of the Resultant
Use the inverse tangent function to find Use the inverse tangent function to find the direction of Rthe direction of R::
2y
2x RRR
x
y1
R
Rtan
Unit VectorsUnit Vectors
A A Unit Vector Unit Vector is a vector having is a vector having length 1 and no unitslength 1 and no units
It is used to specify a direction.It is used to specify a direction. Unit vector Unit vector uu points in the points in the
direction of direction of UU– Often denoted with a Often denoted with a ““hathat””: : u = u =
ûû
U = |U| U = |U| ûû
û û
x
y
z
ii
jj
kk
Useful examples are the cartesian unit vectors ] i, j, ki, j, k [
Point in the direction of the x, y and z axes.
R = rx i + ry j + rz k
Example: A particle undergoes three consecutive displacements given by
,cm)kj3i(d1 cm)k3ji2(d 2 cm)ji(d3
Find the resultant displacement of the particle
cm)k4j3i2(R
k)031(j)113(i)121(dddR 321
cm4R,cm3R,cm2R zyx
cm39.5RRRR z2
y2
x2
Solution:
The resultant displacement has component
The magnitude is
Product of a vector
1-The scalar product )dot product(
There are two different ways in which we can usefully define the multiplication of two vectors
Each of the lengths |A| and |B| is a number and is number, so A.B is not a vector but a number or scalar. This is why it's called the scalar product.
Special cases of the dot product
Since i and j and k are all one unit in length and they are all mutually perpendicular, we have
i.i = j.j = k.k = 1 and i.j = j.i = i.k = k.i = j.k = k.j = 0 .
cos. BABA
The angle between the two vectorIf A and B both have x,y and z components, we express them in
the formkAjAiAA zyx kBjBiBB zyx
2 -The vector product )cross product)
Special cases of the cross product
nABBA ˆsin.
Problem 1: Find the sum of two vectors A and B lying in the xy plane and given by
Problem 2: A particle undergoes three consecutive displacements:
Find the components of the resultant displacement and its magnitude.
mjiBmjiA )42(,)22(
cmjid
cmkjidcmkjid
)1513(
,)51423(,)123015(
3
21
Discussion
]tka[]s[ nm
m2nmnm2m TLTTLL
1m 2n0m2n
Solution
and
[1 ]The position of a particle moving under uniform acceleration is some function of time and the acceleration. Suppose we write this position s = kam tn where , k is a dimensionless constant. Show by dimensional analysis that this expression
is satisfied if if m = 1 and n = 2
Mm
FrG
r
MmGF
2
2
2
2
2
322
kg
m.N
s.kg
m
kgkg
ms/m.kgG
Solution:
[2 ]Newton’s law of universal gravitation is represented by
Here F is the magnitude of the gravitational force exerted by one small object on another , M and m are the masses of the objects, and r is a distance. Force has the SI units kg ·m/ s2. What are the SI units of the proportionality
constant G ?
2r
MmGF
336
3
m/kg104.111010.2
1094.23
V
m
Solution
One centimeter (cm) equals 0.01 m.
One kilometer (km) equals 1000 m.
One inch equals 2.54 cmOne foot equals 30 cm…
cmin
incm
in
cmininin 065.12)54.275.4(
1
54.2)75.4(1)75.4()75.4(
Example:
]3[ A solid piece of lead has a mass of 23.94 g and a volume of 2.10 cm3. From these data, calculate the density of lead in SI units (kg/ m3).
,r
xcoscosrx
3
34
3
4r
r
ysinsinry
3
32y
Solution:
then
then
212
212 )yy()xx(d
m74)43()23(d 22
a(Solution:
[4 ]If the rectangular coordinates of a point are given by (2, y) and its polar coordinates are ( r , 30°), determine y and r .
[4 ]Two points in the xy plane have Cartesian coordinates (2.00, -4.00) m and ( -3.00, 3.00) m. Determine (a) the distance between these points and
(b) their polar coordinates .
b For (2,-4) the polar coordinate is (2,2√5) since
m52164yxr 22
5.2692tanx
ytan 1
[5 ]Vector A has a magnitude of 8.00 units and makes an angle of 45.0 ° with the positive x axis. Vector B also has a magnitude of 8.00 units and is directed along the negative x axis. Using graphical methods, find (a) the
vector sum A + B and (b) the vector difference A - B .
Solution:
j4i5BAC
Solution:
j8i)B(AD
7.385
4tan 1
2.978tan 1
[6 ]Given the vectors A = 2.00 i +6.00 j and B = 3.00 i - 2.00 j, (a) draw the vector sum , C = A + B and the vector difference D = A - B. (b) Calculate C and D, first in terms of unit vectors and then in terms of polar coordinates,
with angles measured with respect to the , +x axis .
Solution:
j6i2j)42(i)13(BA j2i4j)42(i)13()B(ABA
10240364BA
5220416BA
2883tan 1
6.262
1tan 1
Direction of A+B
Direction of A-B
[7 ]Consider the two vectors A = 3 i - 2 j and B = i - 4 j. Calculate (a) A + B, (b) A - B, (c) │A + B│, (d) │A - B│, and (e) the directions of A + B and A - B .
k4j12i8A
k16j48i32A4B
k12j36i24A3C
Solution:
a(
b(
c(
[8 ]The vector A has x, y, and , z components of 8.00, 12.0, and -4.00 units, respectively. (a) Write a vector expression for A in unit vector notation. (b) Obtain a unit vector expression for a vector B four time the length of A pointing in the same direction as A. (c) Obtain a unit vector expression for a vector C
three times the length of A pointing in the direction opposite the direction of A .
unit5.4945cos3045cos40dx
unit2745sin3045sin4020dy
unit4.56)d()d(R 2y
2x
55.0d
dtan
x
y 8.28
Solution:
[9 ]Three displacement vectors of a croquet ball are shown in Figure, where A = 20.0 units, B = 40.0 units, and C = 30.0 units. Find the magnitude and
direction of the resultant displacement
[ 10 ]Find the magnitude and the direction of resultant force
N30240cos853cos1525Fx N15240sin853sin1510Fy
N5.33)F()F(R 2y
2x
5.0F
Ftan
x
y
5.26
Solution:
Additional Questions
[1 ]The position of a particle moving under uniform acceleration is some function of time and the acceleration. Suppose we write this position s = kam tn where , k is a dimensionless constant. Show by dimensional analysis that this expression is
satisfied if if m = 1 and n = 2
[2 ]Newton’s law of universal gravitation is represented by
Here F is the magnitude of the gravitational force exerted by one small object on another , M and m are the masses of the objects, and r is a distance. Force has
the SI units kg ·m/ s2. What are the SI units of the proportionality constant G ?
2r
MmGF
]3[ A solid piece of lead has a mass of 23.94 g and a volume of 2.10 cm3. From these data, calculate the density of lead in SI units (kg/ m3).
[4 ]If the rectangular coordinates of a point are given by (2, y) and its polar coordinates are (r,30°), determine y and r .
[4 ]Two points in the xy plane have Cartesian coordinates (2.00, -4.00) m and ( -3.00, 3.00) m. Determine (a) the distance between these points and (b)
their polar coordinates .
[5 ]Vector A has a magnitude of 8.00 units and makes an angle of 45.0 ° with the positive x axis. Vector B also has a magnitude of 8.00 units and is directed along the negative x axis. Using graphical methods, find (a) the
vector sum A + B and (b) the vector difference A - B .
[6 ]Given the vectors A = 2.00 i +6.00 j and B = 3.00 i - 2.00 j, (a) draw the vector sum , C = A + B and the vector difference D = A - B. (b) Calculate C and D, first in terms of unit vectors and then in terms of polar
coordinates, with angles measured with respect to the , +x axis .
[7 ]Consider the two vectors A = 3 i - 2 j and B = i - 4 j. Calculate (a) A + B, (b) A - B, (c) │A + B│, (d) │A - B│, and (e) the directions of A + B and A
- B .
[8 ]The vector A has x, y, and , z components of 8.00, 12.0, and -4.00 units, respectively. (a) Write a vector expression for A in unit vector notation. (b) Obtain a unit vector expression for a vector B four time the length of A pointing in the same direction as A. (c) Obtain a unit vector expression for a vector C three times the length of A pointing in the direction opposite the
direction of A .
[9 ]Three displacement vectors of a croquet ball are shown in Figure, where A = 20.0 units, B = 40.0 units, and C = 30.0 units. Find the magnitude and
direction of the resultant displacement
[10 ]A rectangular building lot is 100 ft by 150 ft. Determine the area of this lot in m2
[11 ]A solid piece of lead has a mass of 23.94 g and a volume of 2.10 cm3. From these data, calculate the density of lead in SI units (kg/m3).
[12 ]If an equation is dimensionally correct, does this mean that the equation must be true? If an equation is not dimensionally correct, does this mean that the equation cannot be true?
[13 ]Which of the following equations are dimensionally correct?
[14) ]a (A fundamental law of motion states that the acceleration of an object is directly proportional to the resultant force exerted on the object and inversely proportional to its mass. If the proportionality constant is defined to have no dimensions, determine the dimensions of force. (b) The Newton is the SI unit of force. According to the results for (a), how can you express a force having units of Newton's using the fundamentalunits of mass, length, and time?
[15 ]The volume of a wallet is 8.50 in.3 Convert this value to m3, using the definition ( 1 in= 2.54 cm.)
1
( )
( ) (2 )cos( ), 2
f ia v v ax
b y m kx where k m
[16 ]The polar coordinates of a point are r = 5.50 m and = 240°. What are the Cartesian coordinates of this point?
[5 ]If the polar coordinates of the point (x, y) are (r, θ), determine the polar coordinates for the points:
) a-) (x, y ) ,(b-) (2x, -2y ,(and (c) (3x, -3y).
[17 ]A force F1 of magnitude 6.00 units acts at the origin in a direction 30.0° above the positive x axis. A second force F2 of magnitude 5.00 units acts at the origin in the direction of the positive y axis. Find graphically the magnitude and direction of the resultant force F1 + F2.
[18 ]A vector has an x component of -25.0 units and a y component of 40.0 units. Find the magnitude and direction of this vector.
[19 ]Consider the three displacement vectors A =(3ˆi - 3ˆj) m, B = (ˆi -4ˆj) m, and C = (-2ˆi + 5ˆj) m. Use the component method to determine (a) the magnitude and direction of the vector D = A + B + C, (b) the magnitude and direction of E = -A - B + C.
[20 ]If A = (6.00ˆi - 8.00ˆj ) units, B = (-8.00ˆi + 3.00ˆj ) units, and C = (26.0ˆi +19.0ˆj ) units, determine a and b such that a A + b B + C =0.
[21 ]A vector is given by R =2ˆi + ˆj + 3ˆk. Find (a) the magnitudes of the x, y, and z components, (b) the magnitude of R, and (c) the angles between R and the x, y, and z axes.
[22]If v = 2i + 3j + k and w = 4i + j + 2kFind dot and cross product
Find the angle between V and W
One Dimensional Kinematics
In lecture this we discuss motion in one dimension. We introduce definitions for displacement, velocity and acceleration, and derive equations of motion for bodies moving in one dimension with constant acceleration. We apply these equations to the situation of a body moving
under the influence of gravity alone .
One Dimensional Kinematics**
**One dimensional kinematics refers to motion along a straight line.
Kinematics is that branch of physics which involves the description of motion, without examining the forces which produce the motion.
Dynamics, on the other hand, involves an examination of both a description of motion and the forces which produce it.
Distance and displacementDistance and displacement
DisplacementDisplacement is defined as the is defined as the change in position change in position of an objectof an object..
• xxff = final value of = final value of xx, , xxii = initial value of = initial value of xx• Change can be positive, negative or zero.Change can be positive, negative or zero.• Displacement is a vectorDisplacement is a vector
Distance is the total length of travel.
• It is always positive.
• It is measured by the meter.
if xxxΔ ‘’ (Delta)=change
Distance or DisplacementDistance or Displacement??
Distance may be, but is not necessarily, the Distance may be, but is not necessarily, the magnitude of the displacementmagnitude of the displacement
DistanceDisplacement
Example, starting with xi = 60 m and ending at xf = 150
m, the displacement is Δ x = xf - xi = 150 m - 60 m = 90 m
Δ x > 0 since xi < xf .
Example, starting with xi = 150 m and ending at xf = 60
m, the displacement is Δx = xf - xi = 60 m - 150 m = -90 m
Δx < 0 since xi > xf .
Average Speed and VelocityAverage Speed and Velocity
Speed and velocity are not the same in physics!
Speed is rate of change of distance:
time
distance speed average
if
if
tt
xx
time
ntdisplaceme velocityaverage
(always positive)
(positive, negative or zero)
velocity is a vector
Here we are just giving the ‘x-component’ of velocity, assuming the other components are either zero or irrelevant to our present discussion
Velocity is rate of change of displacement:
The Position - Time graphThe Position - Time graph
The average velocity between two times is the slope of the straight line connecting those two points.
average velocity from 0 to 3 sec is positive
average velocity from 2 to 3 sec is negative
Instantaneous VelocityInstantaneous Velocity
The velocity at one instant in time is known as the instantaneous velocity and is found by taking the average velocity for smaller and smaller time intervals:
dt
dx
Δt
Δxlimv
0tΔ
The speedometer indicates instantaneous velocity (t 1 s).
On an x vs t plot, the slope of the line tangent to the curve at a point in time is the instantaneous velocity at that time.
The average velocity of a particle is defined as theratio of the displacement to the time interval.
The instantaneous velocity of a particle is defined as the limit of the average velocity as the time interval approaches zero.
dt
dx
Δt
Δxlimv
0tΔ
if
if
tt
xx
Δt
Δxv
Example: A Particle movies along the x-axis. According to the expression
1- determine the displacement of the particle in the time interval t=0s to t=1s
x(0)=0 and x(1)=-2 m m202x
2 -Calculate the average velocity in the time interval t=0 to t=1
s/m21
2
t
xv
3 -Find the instantaneous velocity of the particle at t=2.5 second
t44dt
dxv
At t=2.5s, s/m6v
2t2t4)t(x
AccelerationAcceleration
Often, velocity is not constant, Often, velocity is not constant, rather it changes with timerather it changes with time..
The The raterate of change of velocity is of change of velocity is known as known as accelerationacceleration..
This is theThis is the average acceleration average acceleration.. Acceleration is a vectorAcceleration is a vector The unit of acceleration is: The unit of acceleration is: m/sm/s22
if
ifav tt
vv
ΔtΔv
a
positive, negative or zero
Instantaneous AccelerationInstantaneous Acceleration
If we wish to know the If we wish to know the instantaneous instantaneous accelerationacceleration, we once again let , we once again let t t 0 0::
dt
dv
Δt
Δvlima
0tΔ
The Velocity - Time graphThe Velocity - Time graph
Graphically, acceleration can be found from the slope of a velocity vs. time curve.
For these curves, the average acceleration and the instantaneous acceleration are the same, because the acceleration is constant.
ExampleExampleA car moves from a position of +4.0 m to a position of –1.0 m in 2.0 sec. The initial velocity of the car is –4.0 m/s and the final velocity is –1 m/s.
(a) What is the displacement of the car?(b) What is the average velocity of the car?(c) What is the average acceleration of the car?
Answer: (a) x = xf – xi = –1.0 m – (+4.0 m) = – 5 m (b) vav = x/t = (– 5.0 m)/(2.0 s) = – 2.5 m/s (c)
2sm
sm
m/s5.1s2
)4(1
if
ifav tt
vv
ΔtΔv
a
s/m)t540(v 2
s/m20)2(v s/m40)0(v
2s/m1002
4020
t
va
2s/tm10dt
dva
2s/m20a
Example: The velocity of a particle moving according to the expression
1 (Find the average acceleration in the time interval t=0 to t=2s
2 (Determine the acceleration at t=2s
At, t=2 then
Motion with Constant Acceleration
Motion with Constant AccelerationMotion with Constant Acceleration
If acceleration is constant, there are four useful formulae relating position x, velocity v, acceleration a at time t:
xavxxavv
attvxx
tvvxx
atvv
2)(2
)(
200
20
2
221
00
021
0
0 v0 = initial velocity (at t=0)
x0 = initial position (at t=0)
t0 = initial time – assumed here to be at 0 s.
If t0 0, replace t in these formulae with t – t0
(Instead of xf, xi, we are using x and x0)
Where do these formulae come fromWhere do these formulae come from??If acceleration is constant, then a = averageIf acceleration is constant, then a = average
accelerationacceleration..
if
if
tta
ta 0
0 , (1)at
tt,0t fi let ,
Also let,
f0i ,
then
If a = constant, then velocity If a = constant, then velocity vsvs time graph is a time graph is a straightstraight
line .line .For a straight line graphFor a straight line graph::v
t
v0
t
xx 0
20
But
2t
xx 00
00 ( ) , (2)
2x x t
then
Substitute eq.(1) into eq. (2)
t)at(2
1xx 000
20 0
1, (3)
2x x t at
then
Finally , substituting the value of t from equation (1) into equation (2)
from equation (1) a
vvt 0
then )a
)(2
(xx 000
)a2
(xx20
2
0
or
2 20 02 ( ), (4)a x x
SummaryThe equations of motion are valid only when …
• acceleration is constant • Motion is constrained to a straight line.
The One Dimensional Equations of Motion for Constant The One Dimensional Equations of Motion for Constant AccelerationAcceleration
traditional traditional namename
equationequationrelationshiprelationship
11stst equation equation22ndnd equation equation33rdrd equation equation
vv ==vv00 ++ atat
xx ==xx00 ++ vv00tt ++ ½ ½ atat22
vv22 ==vv0022 ++ 22aa((xx −− xx00
))
velocity-velocity- timetimedisplacement-displacement- timetime
velocity-velocity- displacementdisplacement
Example:Example: particle moves from a position of (A) particle moves from a position of (A) to a position of (B) in 2.0 sec. The initial to a position of (B) in 2.0 sec. The initial velocity of the car is 1.0 m/s and the final velocity of the car is 1.0 m/s and the final velocity is 4.0 m/svelocity is 4.0 m/s..
))11 ( (calculate the accelerationcalculate the acceleration..
RecallRecall
v = vv = v00 + at + at
44 m/s = 1 m/s + m/s = 1 m/s + aa(2 s)(2 s)
33 m/s = m/s = aa(2 s)(2 s)
aa = 1.5 m/s = 1.5 m/s22
2st m/s,4vm/s,1v0
200 at
2
1txx
m345.12
121x
)2 (find the distance the particle travels in the first 2 sec
Let the origin be at the initial position (( 0x 0
Example: A car decelerates at 2.0 m/s 2 and comes to a stop after traveling 25 m. Finda) The speed of the car at the start of the deceleration and b) The time required to come to a stop.
Solution: (a= - 2.0 m/s)a) From v 2 = v0
2 + 2ax we have v0 = 10 m/s. b) From v = v0 + at we have t = 5 s.
Example: An electron in cathode ray tube of a television set enters a region where it accelerates uniformly from a speed of to a speed of in a distance of 2 cm (a) for what length of time is the electron in this region where accelerate?
s/m103 4
s/m105 6
sec1095.7105103
1022)xx(2t 9
64
2
0
0
)b (what is the acceleration of the electron in this region?
214
9
460 s/m1025.6
1095.7
103105
ta
Freely Falling ObjectsFreely Falling ObjectsNear the earthNear the earth’’s surface, the acceleration dues surface, the acceleration due
to gravity to gravity gg is roughly constant is roughly constant::
gg = = aaEarthEarth’’s surfaces surface = 9.81 m/s = 9.81 m/s22 toward toward
the center of the earththe center of the earth
A A freely falling objectfreely falling object is an object that moves under is an object that moves under the influence of gravity only. Neglecting air resistance.the influence of gravity only. Neglecting air resistance.
An object is in free fall as soon as it is released, whether An object is in free fall as soon as it is released, whether it is dropped from rest, thrown downward, or thrown it is dropped from rest, thrown downward, or thrown upwardupward
Question: What about the mass of an object?Question: What about the mass of an object?Answer: The acceleration of gravity is the same for all Answer: The acceleration of gravity is the same for all
objects near the surface of the Earth, regardless of objects near the surface of the Earth, regardless of mass.mass.
Free Fall -- an Object Free Fall -- an Object DroppedDropped
Initial velocity is zeroInitial velocity is zero Frame: let up be Frame: let up be
positivepositive Use the kinematic Use the kinematic
equationsequations Generally use y instead Generally use y instead
of x since verticalof x since vertical
vo= 0
a = g
2
2
8.9
2
1
sma
aty
y
x
Free Fall -- an Object Free Fall -- an Object Thrown DownwardThrown Downward
a = ga = g With upward being With upward being
positive, acceleration positive, acceleration will be negative, g = -will be negative, g = -9.8 m/s9.8 m/s²²
Initial velocity Initial velocity 0 0 With upward being With upward being
positive, initial velocity positive, initial velocity will be negativewill be negative
Free Fall -- object thrown upwardFree Fall -- object thrown upward
Initial velocity is Initial velocity is upward, so positiveupward, so positive
The instantaneous The instantaneous velocity at the velocity at the maximum height is maximum height is zerozero
a = g everywhere a = g everywhere in the motionin the motion g is always g is always
downward, negativedownward, negative
v = 0
Thrown upwardThrown upward
The motion may be symmetricalThe motion may be symmetrical then tthen tupup = t = tdowndown
then vthen vff = -v = -voo
The motion may not be symmetricalThe motion may not be symmetrical Break the motion into various partsBreak the motion into various parts
generally up and downgenerally up and down
Non-symmetrical Non-symmetrical Free FallFree Fall
Need to divide the Need to divide the motion into motion into segmentssegments
Possibilities Possibilities includeinclude Upward and Upward and
downward portionsdownward portions The symmetrical The symmetrical
portion back to the portion back to the release point and then release point and then the non-symmetrical the non-symmetrical portionportion
Example: A stone is dropped from rest from the top of a building, as shown in Figure. After 3s of free fall, what is the displacement y
of the stone?
Solution: 20 0
1
2y y t gt
20 0 (9.8)(3) 44.1y m
Example: A student throws a set of keys vertically upward to another student in a window 4m above as shown in Figure. The keys are caught 1.5s later by the student.(a) With what initial velocity were the keys thrown?(b) What was the velocity of the keys just before they were caught?
Solution:
a) Let yo= 0 and y = 4m at t =1.5s then we find
y = yo + vo t - 1/2 g t2
4 = 0 + 1.5 vo - 4.9 (1.5)2
vo = 10 m/s
b) The velocity at any time t > 0 is given by
v = vo - gt
v = 10 - 9.8 (1.5) = -4.68 m/s
Graphical example: A ball is thrown Graphical example: A ball is thrown upward from the ground levelupward from the ground level..
x = ball’s height above the ground
velocity is positive when the ball is moving upward
Why is acceleration negative?
a
aa
b
b
b
Discussion
[1 ]The position versus time for a certain particle moving along the x axis is shown in Figure. Find the average velocity in the time intervals
)a (0 to 2 s, (b) 2 to 4 s, (c) 4 s to 5 s, (d) 4 s to 7 s, (e) 0 to 8 s.
s/m2
5
t
x
s/m5t
x
0t
x
m10010x
m5105x
055x m1055x
s/m3.33
10
a) 0 to 2 s
b) 2 to 4 s
c) 4 to 5 s
d) 4 to 7 s
[2 ]A position-time graph for a particle moving along the x axis is shown in Figure (a) Find the average velocity in the time interval t = 1.50 s to t = 4 s.
(b) At what value of t is the velocity zero ?
Solution:
)a( s/m4.25.14
82
)b( At time t = 4 sec
[3 ]A particle moves along the x axis. Its position is given by the equation x= 2 t + 3t 2 with x in meters and t in seconds. Calculate the velocity and
acceleration at t=3s .
Solution:
s/m20362v
t62dt
dxv
3t
First, the instantaneous velocity at t=3s
2
3ts/m6a
6dt
dva
The instantaneous acceleration at t=3s
[4 ]Consider the motion of the object whose velocity-time graph is given in the diagram
1 -What is the acceleration of the object between times t=0 to 2s?2 -What is the acceleration of the object between times t=10 to 12s?
3 -What is the net displacement of the object between times 0 to 16s?
Solution:
equals the area under the v-t graph =100m2 how? the net displacement
1-
2-
3-
[5 ]A particle moves along the x axis according to the equation x = 2+ 3t - t 2 where , x is in meters and t is in seconds. At t= 3 s, find (a) the position of the
particle, (b) its velocity, and (c) its acceleration .
Solution:
)a (the position of the particle m291332x s3t
)b (its velocity,
s/m9v
t23dt
dxv
3t
)c (its acceleration
s/m2a
2dt
dva
3t
[6 ]A truck covers 40m in 8.5 s while smoothly slowing down to a final speed of 2.80 m/s. (a) Find its original speed. (b) Find its acceleration .
Solution:
)a(
t)2
(xx 00
5.8)2
8.2(040 0
s/m61.60
Let the origin be at the initial position x0=0
)b(
20
0
s/m45.05.8
61.68.2
ta
at
]7[ A boy throws a ball upward from the top of a building with an initial velocity of 20.0 m/s. The building is 50 meters high. Determine (a) the time needed for the ball to reach its maximum height (b) the maximum height (c) the time needed for the ball to reach the level of the thrower (d) the velocity of the stone at this instant (e) the velocity and position of the ball after 5.00 s (f) the velocity
of the ball when it reaches the bottom of the building
)a (At maximum height v = 0, so using
Given Information:g = 9.8 m/s2
vo = 20 m/syo= 50 m
s04.2t
)t)(8.9(200
gtvv 0
Solution:
)b (the maximum height
m4.70y
)04.2(8.92
104.22050y
gt2
1tyy
2
200
)c (the time needed for the ball to reach the level of the thrower
Note :the time to reach the top and back down to the thrower's level is the same so 2(2.04) = 4.08 seconds
)d (the velocity of the ball at this instant
s/m20v
)04.2)(8.9(0v
gtvv 0
Which is the same speed of the
original throw upward, except now the negative implies that the ball is moving with a speed of 20 m/s
downward .
)e (the velocity and position of the ball after 5.00 s
s/m29v
)5)(8.9(20v
gtvv 0
m5.27y
)5(8.92
152050y
gt2
1tyy
2
200
)f (The velocity of the ball just at the moment it reaches the ground ,
s/m1.37v
)500)(8.9(2)20(v
)yy(g222
020
2
But, the context of the problem would have us choose, -37.1 m/s because the ball is moving downward just as it hits the ground .
]8[ Suppose a spacecraft is traveling with a speed of 3250 m/s, and it slows down by firing its retro rockets, so that a= -10 m/s2. What is the velocity of the
spacecraft after it has traveled 215km ?
s/m105.2v
)010215)(10(2)3250(v
)yy(a2
3
322
020
2
Solution:
[9] A car can brake to a stop in a distance of 121 ft. from a speed of 60.0 mi/h. To brake to a stop from a speed of 80.0 mi/h requires a stopping distance of 211 ft. What is the average braking acceleration for (a) 60 mi/h to rest, (b) 80 mi/h to rest, (c) 80 mi/h to 60 mi/h? Express the answers m/s2.
1 mile = 1609 m = 1.609 km 1 ft = 0.304 8 m = 30.48 cm
1 m = 39.37 in. = 3.281 ft 1 in. = 0.025 4 m = 2.54 cm
121 121 .3048 36.8808ft m m
211 211 .3048 64.31ft m m
160960 60 26.82
60 60
mil m m
h s s
160980 80 35.76
60 60
mil m m
h s s
2 20
2
2
2
0 (26.82) 2 36.88
9.75
( ) v v a x
s
a
a
ma
2 20
2
2
2
0 (35.76) 2 64.31
9.94
( ) v v a x
s
b
a
ma
060 / 0( ) & 8 /v mil h v hc mil H.W
Additional Questions
[1 ]The position versus time for a certain particle moving along the x axis is shown in Figure. Find the average velocity in the time intervals
)a (0 to 2 s, (b) 2 to 4 s, (c) 4 s to 5 s, (d) 4 s to 7 s, (e) 0 to 8 s.
[2 ]A particle moves along the x axis. Its position is given by the equation x= 2 t + 3t 2 with x in meters and t in seconds. Calculate the velocity and
acceleration at t=3s .
[3 ]A position-time graph for a particle moving along the x axis is shown in Figure (a) Find the average velocity in the time interval t = 1.50 s to t = 4 s.
(b) At what value of t is the velocity zero ?
[4 ]A particle moves along the x axis according to the equation x = 2+ 3t - t 2 where , x is in meters and t is in seconds. At t= 3 s, find (a) the position of the
particle, (b) its velocity, and (c) its acceleration .
[5 ]Consider the motion of the object whose velocity-time graph is given in the diagram
1 -What is the acceleration of the object between times t=0 to 2s?2 -What is the acceleration of the object between times t=10 to 12s?
3 -What is the net displacement of the object between times 0 to 16s?
[6 ]A truck covers 40m in 8.5 s while smoothly slowing down to a final speed of 2.80 m/s. (a) Find its original speed. (b) Find its acceleration .
]7[ A boy throws a ball upward from the top of a building with an initial velocity of 20.0 m/s. The building is 50 meters high. Determine (a) the time needed for the ball to reach its maximum height (b) the maximum height (c) the time needed for the ball to reach the level of the thrower (d) the velocity of the stone at this instant (e) the velocity and position of the ball after 5.00 s (f) the velocity
of the ball when it reaches the bottom of the building
]8[ Suppose a spacecraft is traveling with a speed of 3250 m/s, and it slows down by firing its retro rockets, so that a= -10 m/s2. What is the velocity of the
spacecraft after it has traveled 215km ?
[9] A car can brake to a stop in a distance of 121 ft. from a speed of 60.0 mi/h. To brake to a stop from a speed of 80.0 mi/h requires a stopping distance of 211 ft. What is the average braking acceleration for (a) 60 mi/h to rest, (b) 80 mi/h to rest, (c) 80 mi/h to 60 mi/h? Express the answers m/s2.
[10 ]A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The building is 50.0 m high, and the stone just misses the edge of the roof on its way down, as shown in Figure. Using tA = 0 as the time the stone leaves the thrower’s hand at position A, determine
)A (the time at which the stone reaches its maximum height,
)B (the maximum height ,)C (the time at which the stone returns to the height
from which it was thrown,)D (the velocity of the stone at this instant, and
)E (the velocity and position of the stone at t = 5.00 s.
]11 [A car moved 20 km East and 60 km West in 2 hours. What is its average velocity? [ 20m/s ]
]12 [A car moved 20 km East and 70 km West. What is the distance? [ 90m/s ]
]13 [If a car accelerates from 5 m/s to 15 m/s in 2 seconds, what is the car's average acceleration? Answer [ 5m/s2 ]
]14 [How long does it take to accelerate an object from rest to 10 m/s if the acceleration was 2 m/s? Answer [ 5s ]
]15 [If a car accelerated from 5 m/s to 25 m/s in 10 seconds, how far will it travel? Answer [ 150m ]
]16 [What is the displacement of a car whose initial velocity is 5 m/s and then accelerated 2 m/s2 for 10 seconds? Answer [ 150m ]
[17 ]What is the final velocity of a car that accelerated 10 m/s2 from rest and traveled 180m? Answer ] 60m/s [
[18 ]How long will it take for an apple falling from a 29.4m-tall tree to hit the ground? Answer ] 2.4s [
[19 ]A particle moves along the x axis according to the equation x = 2.00 + 3.00t -1.00t 2, where x is in meters and t is in seconds. At t = 3.00 s, find (a) the position of the particle, (b) its velocity, and (c) its acceleration.
[20 ]An object moves along the x axis according to the equation x(t) = (3.00t2 -2.00t + 3.00) m. Determine (a) the average speed between t = 2.00 s and t = 3.00 s, (b) the instantaneous speed at t = 2.00 s and at t =3.00 s, (c) the average acceleration between t = 2.00 s and t = 3.00 s, and (d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s.
[21 ]A particle moves along the x axis. Its position is given by the equation x =2 + 3t - 4t 2 with x in meters and t in seconds. Determine (a) its position when it changes direction and (b) its velocity when it returns to the position it had at t = 0.
[22 ]A ball is thrown directly downward, with an initial speed of 8.00 m/s, from a height of 30.0 m. After what time interval does the ball strike the ground?
[23 ]A ball is dropped from rest from a height h above the ground. Another ball is thrown vertically upwards from the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground.
Two Dimensional Kinematics
Two Dimensional Two Dimensional KinematicsKinematicsMotion in the x direction is independent from motion
in the y direction. We use the same equations from Lecture 5, but for each dimension separately.
There are not really any new equations in this Lecture.
The position vector for a particle moving in the xy plane can be written
r xi yj
The velocity of the particle is given by
Motion in two dimension with constant acceleration
(1)
(2)x y
dr dx dyv i j
dt dt dt
v v i v j
Since the acceleration is constant then its component also constant , therefore we can apply the equations of motion to both the x and y components of the velocity vector
ta x0xx
ta y0yy
t)jaia()ji(
j)ta(i)ta(
yx0y0x
y0yx0x
ta0
ji yx
Velocity x- component
Velocity y- component
(3)
But from (2)
(5)
(4)
This result states that the velocity of a particle at some time t equals the vector sum of its initial velocity v0 and the additional velocity at ,
Similarly,
2x2
10x0 tatvxx
2y2
10y0 tatvyy
Substituting these expressions into (1)
221
0r0
2yx0y0x00
tatvrr
t)jaia(2
1t)jviv()jyix(r
(8)
(7)
(6)
Motion in 2-dimensions with constant Motion in 2-dimensions with constant accelerationacceleration..
0x20
2x
2x2
10x0
x0x21
0
x0xx
xxa2vv
tatvxx
tvvxx
tavv
“Horizontal” (x) “Vertical” (y)
0y20
2y
2y2
10y0
y0y21
0
y0yy
yya2vv
tatvyy
tvvyy
tavv
x = x-coordinate at time t
x0 = [x-coordinate at time t=t0]
vx = velocity component in x-direction at time t .
vx0 = velocity component in x-direction at time t=t0
y = y-coordinate at time t
y0 = [x-coordinate at time t=t0]
vy = velocity component in y-direction at time t .
vy0 = velocity component in y-direction at time t=t0
Example: A particle starts from the origin at t = 0 with an initial velocity having an x component of 20 m/s and a y component of -15 m/s. The particle moves in the xy plane with an x component of acceleration only given by , ax = 4.0 m/s2.
s/m)t420(ta x0xx
s/m15ta y0yy
s/m]j15i)t420[(ji yx
(A) Determine the components of the velocity vector at any time and the total velocity vector at any time.
Solution:
21v
vtan
x
y1
s/m43vvv 2y
2x
B) Calculate the velocity and speed of the particle at t = 5.0 s.
s/m)j15i40(j15i)5420(v
To determine the angle that v makes with +ve x axis
And the speed is the magnitude of v
(C) Determine the x and y coordinates of the particle at any time t and the position vector at this time
m)t2t20(x
t42
1t200x
tatvxx
2
2
2x2
10x0
m)t15(y
0t150y
tatvyy 2y2
10y0
m]j15i)t2t20[(yjxir 2
then., the position vector at any time is
Projectile MotionProjectile Motion Assume that acceleration of gravity is constant, Assume that acceleration of gravity is constant,
downward and has a magnitude of g = 9.81 m/sdownward and has a magnitude of g = 9.81 m/s22
Air resistance is ignoredAir resistance is ignored The EarthThe Earth’’s rotation is ignoreds rotation is ignored The variation of gravity as a function of The variation of gravity as a function of
distance from the center of the earth is distance from the center of the earth is ignoredignored
The The ““Acceleration of gravityAcceleration of gravity”” means the means the acceleration of an object that is in free fall, acceleration of an object that is in free fall, with no other important forces .with no other important forces .
Horizontal velocity is constant: Horizontal velocity is constant: aaxx = 0 = 0 Vertical motion governed by constant Vertical motion governed by constant
acceleration of gravityacceleration of gravity
To show that the trajectory of a projectile is a parabola, let us choose our reference frame such that the y direction is vertical and positive is upward. Furthermore, let us assume that at t = 0, the projectile leaves the origin with speed v0, as shown in Figure. The vector makes an angle with the horizontal.
cosvvv
vcos 00x
0
0x
sinvvv
vsin 00y
0
0y
Now, the projectile motion can be analyzed as a two component motion : Horizontal component with uniform horizontal velocity and vertical component with uniform acceleration due to gravity of Earth.
The horizontal velocity component (remain constant in time and equal to initial x component of velocity ,since no horizontal
component of acceleration )
ttanconscosvvv 00xx
The vertical velocity component
gtsinvgtvv 00yy
Using (3)
Using (4)
(9)
(10)
The horizontal position component
t)cosv(vx 00x
The vertical position component
20
20y gt
2
1t)sinv(gt
2
1tvy
From (11)
cosv
xt
0
Substitute into (12) 222
0
x)cosv2
g(tanxy
This is of the form 2bxaxy Which is the equation of a parabola
Using (6)
Using (7)
(11)
(12)
Horizontal Range and Maximum Height of a Projectile
The distance R is called the horizontal range of the projectile,and the distance h is its maximum height.
From eq. (10) vy=0 at peakg
sinvt 0
The time t at which the projectile reaches the peak:
Note: time of flight is 02 sin2
vt
g
Substitute into (12)
g2
sinvh
)g
sinv(g
2
1
g
sinv)sinv(yh
220
2000max
The maximum height.
The range R is the horizontal position of the projectile at a time that is twice the time at which it reaches its peak, that is, at time 2t.
Using eq. (11) then
)t2)(cosv(xR 0
g
2sinv
g
cossinv2R
)g
sinv2)(cosv(R
20
20
00
Substitute with value of t
RangeRange
The range R of a projectile is the horizontal distance it travels before landing.
2sin
g
vR
20
The maximum value of R is
g
vR
20
max
This result follows from the fact that the maximum value of 2sin
which occurs when 902 Therefore, R is a maximum when 45
QuizQuizThree projectiles (a, b and c) are launched with the same initial speed but with different launch angles, as shown. List the projectiles in order of increasing horizontal component of initial velocity
1. (a, b,c)2. (c, b, a)3. (b, a, c) or (b, c, a)4. (a, c, b) or (c, a, b)
(smallest to largest)
Problem: A golfer hits a ball with an initial speed of 40.3 m/s at an angle of 32.0° from the horizontal. (a) How long does the golf ball take to reach its maximum height? (b) How far does the ball go and how long is it in the air? (c) What is the maximum height of the golf ball? (d) What is its speed when it hits the ground?
ProblemProblemAn Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers. The plane is traveling horizontally at 40.0 m/s at a height of 100 m above the ground. Where does the package strike the ground relative to the point at which it was released?
d
Circular Motion & Relative Velocity
Circular MotionCircular MotionConsider an object moving at constant speed in a circle. The direction of motion is changing, so the velocity is changing (even though speed is constant).
Therefore, the object is accelerating.
The direction of the acceleration is toward the center of the circle and so we call it centripetal acceleration.
The magnitude of the acceleration is
r
vac
2
Centripetal AccelerationCentripetal Acceleration
radiansin measured if
2 travelledDistance
ˆ)sin(ˆ)cos(
ˆ)sin(ˆ)cos(
2
1
r
yvxvv
yvxvv
The best estimate of the The best estimate of the acceleration at P is found by acceleration at P is found by calculating the average acceleration calculating the average acceleration for the symmetric interval 1for the symmetric interval 12.2.
2
2
Elapsed time t d/v 2
Components of Acceler
cos cos0
2sin sin sin
ation
2
x
y
y
θr/
v va
θr/vv v v
aθr/v r
va
r
v
0if
r
vac
2
Then,
Example: What is the centripetal acceleration of the Earth as it moves in its orbit around the Sun?
Solution:
rac
2
t
r 2But
smt
rac /1093.5
4 32
2
Then
yeart
mearthr
1
10496.1)( 11
Tangential acceleration
The tangential acceleration component causes the change in the speed of the particle. This component is parallel to the instantaneous velocity, and is given by
Tangential and Radial acceleration
dt
dat
Note: If the speed is constant then the tangential acceleration is zero (uniform Circular Motion)
raa cr
2
The radial acceleration component arises from the change in direction of the velocity vector and is given by
Radial acceleration
Total acceleration
The total acceleration vector a can be written as the vector sum of the component vectors:
rt
rt
aaa
aaa
22
Since the component perpendicular to other
Example: A car exhibits a constant acceleration of 0.300 m/s2 parallel to the roadway. The car passes over a rise in the roadway such that the top of the rise is shaped like a circle of radius 500 m. At the moment the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of 6.00 m/s. What is the direction of the total acceleration vector for the car at this instant?
If the angle between
22
/072.0500
36sm
rar
222 /309.0 smaaa rt
5.13tan 1
t
r
a
a
Problem: A train slows down as it rounds a sharp horizontal turn, slowing from 90km/h to 50km/h in the 15s that it takes to round the bend. The radius of the curve is 150m. Compute the
acceleration at the train.
Example: A particle moves in a circular path 0.4m in radius with constant speed. If the particle makes five revolution in each second of its motion, find (a) the speed of the particle
and (b) its acceleration.
(a) Since r =0.4m, the particle travels a distance 0f 2 r = 2.51m in each revolution. Therefore, it travels a distance of 12.57m in each second (since it makes 5 rev. in the second).v = 12.57m/1sec = 12.6 m/s
2
397ar
(b)
Centripetal ForceCentripetal Force
A string cannot A string cannot push sideways or push sideways or lengthwise.lengthwise.
A string in tension A string in tension only pulls.only pulls.
The string pulls the The string pulls the ball ball inwardinward toward toward the center of the the center of the circle circle
What if we cut the sting?
The ball should move off with constant velocityThis means the ball will continue along the tangent to the circle.
Centripetal ForceCentripetal ForceIf there is a centripetal acceleration, then If there is a centripetal acceleration, then the net force must also be athe net force must also be a centripetal centripetal forceforce::
r
vmmaF cc
2
The Conical The Conical PendulumPendulum
As the ball As the ball revolves faster, revolves faster, the angle the angle increasesincreases
WhatWhat’’s the speed s the speed for a given angle?for a given angle?
Example:
2
2
sin (1)
cos (2)
tan
tan
( sin )
sin tan
mvT
rT mg
then
v
rg
v rg
but r L
Lg
Problem: I rotate a ball at an angle of 30o. What is the centripetal acceleration? If the string is 1 meter long, how fast is it rotating?
ProblemProblemDriving in your car with a constant speed of 12 m/s, you encounter a bump in the road that has a circular cross section, as indicated in the Figure. If the radius of curvature of the bump is 35 m, find the apparent weight of a 67-kg person in your car as you pass over the top of the bump.
Nmg
a=v2/r
Relative VelocityRelative Velocity TwoTwo observers moving relative to each other generally do not observers moving relative to each other generally do not
agree on the outcome of an experimentagree on the outcome of an experiment For example, observers A and B below see different paths for the For example, observers A and B below see different paths for the
ballball
Relative Velocity equationsRelative Velocity equations The positions as seen from the two reference frames The positions as seen from the two reference frames
are related through the velocity are related through the velocity
The derivative of the position equation will give the The derivative of the position equation will give the velocity equationvelocity equation
These are called the These are called the Galilean transformation Galilean transformation equationsequations
tvrr 0
0vvr
Central concept for problem solving: Central concept for problem solving: ““xx”” and and ““yy”” components of motion treated components of motion treated
independentlyindependently.. Again: man on the cart tosses a ball straight up in the Again: man on the cart tosses a ball straight up in the
air.air. You can view the trajectory from two reference You can view the trajectory from two reference
frames:frames:
Reference frame
on the ground.
Reference frame
on the moving train.
y)t) motion governed by 1) a = -g y
2) vy = v0y – g t3) y = y0 + v0y – g t2/2
x motion: x = vxt
Net motion: R = x)t) i + y)t) j (vector)
Acceleration in Different Frames of Acceleration in Different Frames of ReferenceReference
The derivative of the velocity equation will give The derivative of the velocity equation will give the acceleration equationthe acceleration equation
– vv’’ = = vv –– vvo o
– aa’’ = = aa
The acceleration of the particle measured by an The acceleration of the particle measured by an observer in one frame of reference is the same observer in one frame of reference is the same as that measured by any other observer moving as that measured by any other observer moving at a at a constant velocityconstant velocity relative to the first frame. relative to the first frame.
QuestionsQuestions[1]You are on a train traveling 40 mph North. If you walk 5 [1]You are on a train traveling 40 mph North. If you walk 5
mph toward the front of the train, what is your speed mph toward the front of the train, what is your speed relative to the ground?relative to the ground?
A) 45 mphA) 45 mph B) 40 mphB) 40 mph C) 35 mphC) 35 mph
[2]You are on a train traveling 40 mph North. If you walk 5 [2]You are on a train traveling 40 mph North. If you walk 5 mph toward the rear of the train, what is your speed mph toward the rear of the train, what is your speed relative to the ground?relative to the ground?
A) 45 mphA) 45 mph B) 40 mphB) 40 mph C) 35 mphC) 35 mph
[3]You are on a train traveling 40 mph North. If you walk 5 [3]You are on a train traveling 40 mph North. If you walk 5 mph sideways across the car, what is your speed relative mph sideways across the car, what is your speed relative to the ground?to the ground?
A) < 40 mphA) < 40 mph B) 40 mphB) 40 mph C) >40 mphC) >40 mph
Discussion
[1]: Along jumper leave the ground at an angle of 20to the horizontal and at a speed of 11m/s
(1) How far does he jump?
s/m34.1020cos11cosvv 00x
s/m76.320sin11sinvv 00y
s84.3t
t8.976.30
gtvv 0yy
Solution:
The initial x and y component are
The time t at which the projectile reaches the peak:
m94.7)768.0)(20)(cos11(R
)t2)(cosv(xR 0
(2) What is the maximum height reached?
m72.0)384.0(8.92
1384.076.3y
gt2
1tvy
2max
20y
The range is
[2]: A stone is thrown from the top of a building upward at an angle of 30.0 ° to the horizontal with an initial speed of 20.0 m/s, as shown in Figure. If the height of the building is 45.0 m,
(A) how long does it take the stone to reach the ground?
s/m1030sin20sinvv 00y
s/m3.1730cos20cosvv 00x
s22.4t
045t10t9.4
t8.92
1t1045
gt2
1tvy
2
2
20y
By using the equation
(B) What is the speed of the stone just before it strikes the ground?
s/m3.17vv 0xx
s/m4.3122.48.910gtvv 0yy
s/m9.35vvv 2y
2x
[3:] A ball from a tower of height 30 m is projected down at an angle of 30° from the horizontal with a speed of 10 m/s. How long does it take the ball
to reach the ground. (Consider g = 10 m/s2 ):
smvvy /530sin10sin00
st
tt
tt
gttvy y
3
06
102
1530
2
1
2
2
20
By using the equation
Neglecting negative value of time, t = -2 s
Solution:
]4[:
On a hot summer day a young girl swings on a rope above the local swimming hole. When she lets go of the rope her initial velocity is 2.25 m/s at an angle of 35.0° above the horizontal. If she is in flight for 1.60 s, how high above the water was she when she let go of the rope?
v0 = 2.25 m/s
v0x = v0 cos(35) = 1.8 m/s
v0y = v0 sin(35) = 1.3 m/s
Solution:
y = y0 +v0y t – (1/2) g t2
y-y0 = - 10.5 m
y =0
y0 = +10.5 m
By using
[5:]A projectile is launched horizontally with an initial speed v0 from a height y0. Neglect aerodynamic friction.
v0 = 3.5 m/sy0 = 4.0 m
)1 (What is the time t for the projectile to reach the ground ?
v0
x
y
y0
20 0
0
0
1
24.0
0
y
y
y y v t gt
y m
v
90.0
816.0)81.9/()0.4(2
2
10.40
0
2
2
t
t
gt
y
Solution:
(a)2 (What is the horizontal component of the projectile’s velocity when it hits the ground?
smv
a
smv
tavv
x
x
x
xxx
/5.3
0
/5.30
0
smv
smssmgtvv
smv
vvv
yy
x
yx
/53.9
/86.8)90.0)(/81.9(
/5.32
0
22
)3 (What is the speed of the projectile when it hits the ground?
[1]:Along jumper leave the ground at an angle of 20to the horizontal and at a speed of 11m/s
[3:] A ball from a tower of height 30 m is projected down at an angle of 30° from the horizontal with a speed of 10 m/s. How long does it take
the ball to reach the ground. (Consider g = 10 m/s2 ):
[2]: A stone is thrown from the top of a building upward at an angle of 30.0 ° to the horizontal with an initial speed of 20.0 m/s, as shown in Figure. If the height of the building is 45.0 m,
]4[:On a hot summer day a young girl swings on a rope above the local swimming hole. When she lets go of the rope her initial velocity is 2.25 m/s at an angle of 35.0° above the horizontal. If she is in flight for 1.60 s, how high above the water was she when she let go of the rope?
[5 ]A projectile is launched horizontally with an initial speed v0 from a height y0. Neglect aerodynamic friction.
v0 = 3.5 m/sy0 = 4.0 m
[[66]] A projectile is fired in such a way that its A projectile is fired in such a way that its horizontal range is equal to three times its horizontal range is equal to three times its maximum height. What is the angle of maximum height. What is the angle of projectionprojection??
]7[ The vector position of a particle varies in time according to the expression r =(3.00ˆi -6.00t2 ˆj ) m. (a) Find expressions for the velocity and acceleration as functions of time. (b) Determine the particle’s position and velocity at t = 1.00 s.
[8] [8] A projectile is fired in such a way that its horizontal A projectile is fired in such a way that its horizontal range is equal to three times its maximum height. What range is equal to three times its maximum height. What is the angle of projection?is the angle of projection?
]9[ A fireworks rocket explodes at height h, the peak of its vertical trajectory. It throws out burning fragments in all directions, but all at the same speed v. Pellets of solidified metal fall to the ground without air resistance. Find the smallest angle that the final velocity of an impacting fragment makes with the horizontal.
]10[ An automobile whose speed is increasing at a rate of 0.600 m/s2 travels along a circular road of radius 20.0 m. When the instantaneous speed of the automobile is 4.00 m/s, find (a) the tangential accelerationcomponent, (b) the centripetal acceleration component, and (c) the magnitude and direction of the total acceleration.
[11] A pendulum with a cord of length [11] A pendulum with a cord of length r r =1.00 m swings in a =1.00 m swings in a vertical plane . When the pendulum is in the two horizontal vertical plane . When the pendulum is in the two horizontal positions positions θθ=90.0° and =90.0° and θθ= 270°, its speed is 5.00 m/s. (a) Find = 270°, its speed is 5.00 m/s. (a) Find the magnitude of the radial acceleration and tangential the magnitude of the radial acceleration and tangential acceleration for these positions. (b) Draw vector diagrams to acceleration for these positions. (b) Draw vector diagrams to determine the direction of the total acceleration for these two determine the direction of the total acceleration for these two positions. (c) Calculate the magnitude and direction of the total positions. (c) Calculate the magnitude and direction of the total acceleration.acceleration.
Newton’s Laws of Motion
NewtonNewton’’s Laws of Motions Laws of Motion In this chapter we will study Newton's In this chapter we will study Newton's
laws of motion.laws of motion. These are some of the These are some of the most fundamental and important most fundamental and important principles in physicsprinciples in physics..
Intuitively, we know that force is a Intuitively, we know that force is a ““pushpush”” or or ““pullpull””..
Idea:Idea: Force is the cause of motion in classical Force is the cause of motion in classical mechanics. mechanics.
Forces come in different classes (types):Forces come in different classes (types):1.1. Contact Forces : Contact Forces : involve physical contact involve physical contact
between objects between objects Example: friction, viscosity etcExample: friction, viscosity etc……2.2. Field Forces :Field Forces :don't involve physical contact don't involve physical contact
between objects between objects Examples: Gravity, ElectromagnetismExamples: Gravity, Electromagnetism
What is the force
Fundamental force in nature
Force F is a vector quantity: You push or pull in a specific direction
[Q] If force has direction, what is it’s measure?
In 1686, Newton presented hisIn 1686, Newton presented his Three Laws of Three Laws of MotionMotion::
NewtonNewton’’s First Laws First LawAn object at rest remains at rest, and an An object at rest remains at rest, and an object in motion continues in motion object in motion continues in motion with constant velocity, unless it with constant velocity, unless it experiences a net forceexperiences a net force..Velocity = constant (i.e. acceleration = 0) if there is no force (or if
all forces add to zero).Remember:• Velocity = constant, does not mean velocity = 0.• Velocity = constant means constant magnitude AND direction
Inertia Inertia FramesFrames
The tendency of an object to resist a change The tendency of an object to resist a change in its velocity is calledin its velocity is called inertiainertia..
The measure of inertia isThe measure of inertia is massmass..– SI units measure mass as multiples of the SI units measure mass as multiples of the
standard kilogram (kg=1000g) stored at the standard kilogram (kg=1000g) stored at the International Bureau of Weights and Measures International Bureau of Weights and Measures in Sin Sèèvres, France.vres, France.
NewtonNewton’’s First Law: If s First Law: If F=0F=0, then , then a=0a=0..What if What if FF 0? 0?
Newton's first law sometimes called the law of inertia
NewtonNewton’’s Second Laws Second Law The acceleration of an object is directly The acceleration of an object is directly
proportional to the resultant force acting on it proportional to the resultant force acting on it and inversely proportional to its mass. The and inversely proportional to its mass. The direction of the acceleration is the direction of direction of the acceleration is the direction of the resultant forcethe resultant force..
Force is a vectorForce is a vector– The net force is the vector sum of all forces acting on the The net force is the vector sum of all forces acting on the
object m.object m. Mass is a scalar:Mass is a scalar:
– The value of the mass of an object does not change with The value of the mass of an object does not change with the direction of the acceleration.the direction of the acceleration.
– The equation The equation FF=m=maa is also a definition of mass. is also a definition of mass. Mass is invariant:Mass is invariant:
– If two objects are put together (or separated) , the mass If two objects are put together (or separated) , the mass of the combined object is simply the arithmetic sum of of the combined object is simply the arithmetic sum of the two masses m = mthe two masses m = m11+m+m22..
The SI unit of force is )N(Newtonss
m.kg2
One Newton is the force required to accelerate one kg one meter per second. Note that the first law is a special case of the second.
0 0 0F a Is constant or zero
The Vector Nature of The Vector Nature of ForcesForces
In the formula In the formula FF = m = maa, , FF is the total is the total (net) force acting on the object. We (net) force acting on the object. We must consider the vector sum of all must consider the vector sum of all forces acting on an object. We can also forces acting on an object. We can also consider each dimension separatelyconsider each dimension separately::
zz
yy
xx
maF
maF
maF
An object of mass 5 kg undergoes an acceleration An object of mass 5 kg undergoes an acceleration of of aa = (8 m/s = (8 m/s22) ) ŷŷ = = 8 m/s8 m/s2 2 in + y direction in + y direction
What is the force on that objectWhat is the force on that object??
FF = m = maa ) = ) = 55 kgkg)()(8 m/s8 m/s22 ( (ŷ = 40 kgŷ = 40 kgm/sm/s22 ŷ ŷ
ŷ = vector on unit length (no dimensions) in +y ŷ = vector on unit length (no dimensions) in +y
directiondirection..The force is in the same direction as the accelerationThe force is in the same direction as the acceleration..
Example
Example: Two forces, F1=45.0N and F2=25.0N act on a 5.00kg block sitting on a table as shown. What is the horizontal acceleration (magnitude and direction) of the block?
F1x= F1cos(65.0) = 19.0 N
xx maF
F2x= F2 = 25.0 N
19.0 - 25.0 = (5.00)ax
ax = -1.2 m/s2
Solution:
Example: What is the average force exerted by a shot putter on a 7.0 kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s.
NmaFs
ma
a
xxavv
210307
30
8.221690
)(2
2
020
2
Solution:
NewtonNewton’’s Third Laws Third LawIf object 1 exerts a force If object 1 exerts a force FF on object 2, on object 2,
then object 2 exerts a force then object 2 exerts a force ––FF on on object 1.object 1.– Forces come in pairs.Forces come in pairs.– The force pairs act on The force pairs act on differentdifferent objects. objects.– The forces have the same magnitude but The forces have the same magnitude but
opposite direction.opposite direction.
Example: I push on the wall with a force of 20 N. The wall pushes back on me with a force of 20 N in the opposite direction.
2on11on2 FF
WeightWeightThe weight of any object on the Earth is the The weight of any object on the Earth is the gravitational force exerted on it by the Earthgravitational force exerted on it by the Earth::
WW = = mmgg
NoteNote : :
Weight is a forceWeight is a force (and therefore a vector) (and therefore a vector)..
Weight is not equivalent to massWeight is not equivalent to mass..
The weight of an object is different on the earth and on the moon since the strength of the gravitational field is different ( g e g m ).
Newton’s Laws of Motion (Applications)
If the net for force exerted on an object is zero, the acceleration of the object is zero and its velocity remains constant. That is, if the net force acting on the object is zero, the object either remains at rest or continues to move with constant velocity. When the velocity of an object is constant (including when the object is at rest), the object is said to be in equilibrium.
Definition of equilibrium
The first condition of equilibrium
0Fx
0Fy
Example: A traffic light weighing 122 N hangs from a cable tied to two other cables fastened to a support, as in Figure The upper cables make angles of 37.0 ° and 53.0 ° with the horizontal. Find the tension in the three cables.
Solution:
Note that the traffic light in equilibrium. Then
0Fx
0Fy
N122T0F 3y since
37cosT53cosT0F 12x
0T8.0T6.0 12
12237sinT53sinT0F 12y
N4.73T1 N4.97T2
From the condition of equilibrium
0122T6.0T8.0 12
By solve the eqs.
Example: Suppose the two angles are equal. What would be the relationship between T1 and T2?
Example: A person weighs a fish of mass m on a spring scale attached to the ceiling of an elevator, as illustrated in Figure. Show that if the elevator accelerates either upward or downward, the spring scale gives a reading that is different from ward, the weight of the fish.
Solution:
We chose upward as the positive y direction
(1) If the elevator is either at rest or moving at constant velocity, the fish does not accelerate, and so,
0mgTFy mgT Or
(Remember that the scalar mg is the weight of the fish.)
(2) If the elevator moves with an acceleration a relative to an observer standing outside the elevator in an inertial frame , Newton’s second law applied to the fish gives the net force on the fish:
yy mamgTF (*)
For example: if the weight of the fish is 40.0 N and a upward, so that ward, ay = 2.00 m/ s2, the scale reading from (*) is
N2.48208.440T
mamgT y
(2) If a is downward so that ay = - 2.00 m/s2, then (*) gives us,
N84.31)2(08.440T
mamgT y
Thus, we conclude from (*) that the scale reading T is greater than the weight mg if a is upward, so that ay isPositive, and that the reading is less than mg if a is downward, so that, ay is negative.
Normal Force
Example: A car of mass m is on an icy driveway inclined at an angle as in Figure.
Find the acceleration of the car, assuming that the, driveway is frictionless.
Solution:
Now we apply Newton Newton’s second law in component form, noting that ay = 0:
0cosmgnF
masinmgF
y
xx (1)
(2)
From (1) singa x
From (2) cosmgn
Example: solve the example if the mass 1000 kg and the angle 30.
Example: When two objects of unequal mass are hung vertically over a frictionless pulley of negligible mass, as in Figure, the arrangement is called an Atwood machine. The device is sometimes used in the laboratory to measure the free-fall acceleration. Determine the magnitude of the acceleration of the two objects and the tension in the lightweight cord.
When Newton’s second law is applied to object m1, weobtain
y11y amgmTF
Solution:
Similarly , for object m2 we , find
y22y amgmTF
)mm
mm(a
21
12y
When (2) is added to (1), T cancels and we have
g)mm
mm2(T
21
21
Example: find the acceleration and the tension of Atwood's machine in which m1=2kg, m2=4kg.
Then we have
example: A ball of mass m1 and a block of mass m2 are attached by a lightweight cord that passes over a frictionless pulley of negligible mass, as in Figure. The block lies on frictionless incline of angle . Find the magnitude of the acceleration of the two objects and the tension in the cord.
Equations of motion for m1
amgmTF
F
y
x
11
0
(1)
(2)
Solution:
Equations of motion for m2
0cos
sin
2
22
gmNF
amTgmF
y
x
(4)
(3)
Now, solve (2) , (3) simultaneously . The acceleration
21
12 sin
mm
gmgma
When this is substituted into (2) we find the normal force
21
21 )sin1(
mm
gmmT
Example: if m1=10kg, m2=5kg and =45 find a and T.
Example: A 3.0 kg mass hangs at one end of a rope that is attached to a support on a railroad car. When the car accelerates
to the right, the rope makes an angle of 4.0° with the vertical. Find the acceleration of the car.
In the vertical direction:
Solution:
In the horizontal direction:
Problem: A train is given an acceleration of 5 m/s2. What is the tension between the two cars with a mass of 2000 kg.
Forces of Friction
In most cases we need to include friction. For many surfaces the force of sliding friction is proportional to the normal force. That is, as the normal force increases, the force of friction increases linearly. When this is true, we
write
with the proportionality constant called , the coefficient of kinetic friction. This is the friction when the object is moving.
nfr FF
k
nf kk
If I push a table and it doesn't move, there must also be a force pushing back. That force comes from friction, as well. It is the friction when the object is not moving, or static, called the coefficient of static friction. It is always greater than or equal to the coefficient of kinetic friction. It is often harder to start something sliding than to keep it sliding, due to the difference in these coefficients.
nf ss
The Static force of friction ( fs ) is the force of friction between two objects when there is no motion .
The Kinetic force of friction ( fk ) is the force of friction between two objects when there is motion.
Example: An 18.0 kg box is released on a 37.0° incline and accelerates down the incline at 0.27 m/s2. Find the coefficient of friction and the frictional force. (Draw diagram)
Forces in y-direction are:
Nn
mgn
maF yy
14137cos8.918
0cos
Solution:
Nf
mafmg
maF
x
xx
10127.01837sin8.918
)(sin
Forces in x-direction are:
Now, we can relate these to the coefficient of kinetic friction ,
716.0 kk nf
Additional Questions
]1:[ Suppose we have a situation as we did in a lab experiment, in which a hanging mass and cart are connected by a string over a pulley.
The hanging mass is 0.20 kg, but we don't know the mass of the cart. After the cart is released and the objects are moving, the tension in the string is found to be 1.80 N.
)A (Find the acceleration of the hanging mass. First, draw a free-body diagram
)B (Find the mass of the cart, assuming that there is no friction.)C (find the mass of the cart assuming that there is a 0.50 N frictional force
acting on the cart.
)A (Find the acceleration of the hanging mass. First, draw a free-body diagram
)B (Find the mass of the cart, assuming that there is no friction .
)C (find the mass of the cart assuming that there is a 0.50 N frictional force acting on the cart.
]2:[ Find the tensions in each string in the figure at right. They are attached to a wall and ceiling, as shown, and are holding up a mass of 10 kg
]3:[ We show below two blocks connected by a light string and pulled by a string in front. The “front” block has a mass of 5.0 kg and the other block’s mass is 2.0 kg. The large block slides on frictionless surface. The back, smaller block has a rough bottom surface which creates a frictional force of 6.0 N. The tension T1 in the front string is 20.0 N. Determine the tension T2 in the middle string
]2:[
Knowing T1, we can find T2:
]3: [
]4:[ A 20-kg box is sitting on the floor and someone is pulling on it with a rope. The tension in the rope is 60 N at an angle 60◦ above the horizontal. However, the box does not move. It just sits there, held by static friction.(a)Draw a free-body diagram of the box.
)b( Find the normal force that the floor exerts on the box.
]5:[ An inclined plane makes an angle 20◦ with the horizontal. A 0.20-kg bar of soap slides down the incline, and experiences a frictional force of 0.20 N. (A) Draw a free-body diagram of the bar of soap. (B) Find the net force on the bar of soap. (C) Calculate the rate of acceleration of the bar of soap
[4]:
]6:[ Two forces F1 and F2 act on a 5.00-kg object. If F1 = 20.0 N and F2 = 15.0 N, find the accelerations in (a) and (b) of Figure
]7:[ A 3.00-kg object is moving in a plane, with its x and y coordinates given by x = 5t2 + 1 and y =3t 3 + 2, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.00 s
[8]: Two blocks of mass 3.50 kg and 8.00 kg are connected by amassless string that passes over a frictionless pulley. The inclines are frictionless. Find (a) the magnitude of the acceleration of each block and (b) the tension in the string
]9:[ Forces are being applied to a box sitting on a surface with friction. Will the box move horizontally (along the surface)? F1=50N, F2=50N, Mass of the block 10kg, and ms=0.4.
Hint: must be greater or equal to xF sf
]10:[ A 20 kg sled is being pulled across a horizontal surface at a constant velocity. The pulling force has a magnitude of 80.0 N and is directed at an angle or 30.0° above the horizontal. Determine the coefficient of kinetic friction.
[11 ]A 3.00-kg object undergoes an acceleration given by a = (2.00ˆi + 5.00ˆj) m/s2. Find the resultant force acting on it and the magnitude of the resultant force.
[12 ]A 1.00-kg object is observed to have an acceleration of 10.0 m/s2 in a direction 30.0° north of east . The force F2 acting on the object has a magnitude of 5.00 N and is directed north. Determine the magnitude and direction of the force F1 acting on the object.
[13 ]A 25.0-kg block is initially at rest on a horizontal surface. A horizontal force of 75.0 N is required to set the block in motion. After it is in motion, a horizontal force of 60.0 N is required to keep the block moving with constant speed. Find the coefficients of static and kinetic friction from this information.
[14 ]A 5.00-kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00-kg object, as in Figure. Draw free-body diagrams of both objects. Find the acceleration of the two objects and the tension in the string.
General Physics 1, Lec 12, By/ T.A. Eleyan 229
.
Energy and Energy Transfer
230
When an object undergoes displacement under force then work is said to be done by the force and the amount of work (W) is defined as
the product of the component of force along the direction of
displacement times the magnitude of the displacement.
where is the angle between the direction of the force and the direction of the motion .
Fd
cos. FddFW
Work Done by a Constant Force
231
The SI unit of work is (N×m) which is given the name (Joule).1 Newton×meter = 1 Joule
If I push on a wall and the wall does not move (no displacement), the work is (0J) because the displacement is (0m).
Work is a scalar Work has only magnitude, no direction.
Note that if F is in the same direction as d, then = 0, and W= Fd
232
No work is done on the bucket to move horizontally (why?)
Work is done in lifting the box (why?)
233
Negative Work and Total WorkNegative Work and Total Work
Work can be positive, negative or zero depending on the angle between the force and the displacement.
If there is more than one force, each force can do work. The total work is calculated from the total (or net) force:
Wtotal = Ftotald cos
234
Example : Suppose I pull a package with a force of 98 N at an angle of 55° above the horizontal ground for a distance of 60m. What
is the total work done by me on the package?
Note that (F cos is the component of the force along the direction of motion. (Along the direction of the package's displacement.)
Solution:
JWork
mNWork
FdWork
3490
55cos6098
cos
235
Problem: A force F = (6ˆi - 2ˆj) N acts on a particle that undergoes a displacement r = (3ˆi +ˆj) m. Find the work done by the force on the particle
Example : A 0.23 kg block slides down an incline of 25° at a constant velocity. The block slides 1.5 m. What is the work done by the normal force, by gravity, and by friction? What is the total work done on the
block?
General Physics 1, Lec 12, By/ T.A. Eleyan 236
The Normal Force
The friction Force
NmgF
mgF
maF
rf
rf
95.025sin8.923.0sin
0sin
NmgF
mgF
maF
n
n
04.225cos8.923.0cos
0cos
General Physics 1, Lec 12, By/ T.A. Eleyan 237
Now what is the work done by each individual force.
Work done by the Normal Force:
Work done by the Frictional Force:
Work done by the Gravitational Force
Now, to finally determine the "net-Work"
JdFW nn 090cos5.104.2cos
JdFW frfr 40.1180cos5.195.0cos
JmgddFW gg 40.165cos5.18.923.0coscos
JWWWW gfrnnet 0
General Physics 1, Lec 12, By/ T.A. Eleyan 238
Force - Displacement GraphForce - Displacement Graph
The work done by a force can be found from the area between the force curve and the x-axis (remember, area below the x-axis is negative):
Constant force
General Physics 1, Lec 12, By/ T.A. Eleyan 239
Work done by a varying force
xFWf
i
x
xx
0lim
f
f
i
i
xx
x xxxx
F x F dx
f
i
x
x
x
W F dx
240
Example: A force acting on a particle varies with x, as shown in Figure .Calculate the work done by the force as the particle moves from x
= 0 to x = 6.0 m .
Solution The work done by the force is equal to the area under the curve from x= 0 to x= 6.0 m. This area is equal to the area of the rectangular section from 0 to 4 plus the area of the triangular section from 4 to 6. The area of the rectangle is (5.0 N)(4.0 m) = 20 J, and the area of the triangle is ½(2m)(5N)=5J. Therefore, the total work =25J
General Physics 1, Lec 12, By/ T.A. Eleyan 241
Problem: An object is acted on by the force shown in the Figure. What is the work from 0 to 1.00m
General Physics 1, Lec 12, By/ T.A. Eleyan 242
Work Done by a Spring
sF kxWhere k is a positive constant called (the spring constant)
2max
1
2sW kx
General Physics 1, Lec 12, By/ T.A. Eleyan 243
Kinetic EnergyKinetic Energy
An object in motion has kinetic energyAn object in motion has kinetic energy::
mm = mass = mass
vv = speed (magnitude of velocity) = speed (magnitude of velocity)
The unit of kinetic energy is Joules (J)The unit of kinetic energy is Joules (J)..
Kinetic energy is a scalar (magnitude only)Kinetic energy is a scalar (magnitude only)
Kinetic energy is non-negative (zero or positiveKinetic energy is non-negative (zero or positive))
221 mvK
General Physics 1, Lec 12, By/ T.A. Eleyan 244
Work-Energy TheoremWork-Energy Theorem
The net (total) work done on an object by the The net (total) work done on an object by the total force acting on it is equal to the change total force acting on it is equal to the change in the kinetic energy of the objectin the kinetic energy of the object::
WWtotaltotal = = KE = KEKE = KEfinalfinal - KE - KEinitialinitial
2 21 1
2 2total f iW mv mv
245
Example : How much work does it take to stop a 1000 kg car traveling at 28 m/s?
Solution:
Problem: A 0.600-kg particle has a speed of 2.00 m/s at point (A) and kinetic energy of 7.50 J at point (B). What is
)a (its kinetic energy at (A) ?)b (its speed at (B) ?
)c (the total work done on the particle as it moves from A to B?
JW
W
mvmvKEW if
5
22
22
109.3
)28)(1000(2
1)0)(1000(
2
12
1
2
1
General Physics 1, Lec 12, By/ T.A. Eleyan 246
Example : A 58-kg skier is coasting down a slope inclined at 25° above the horizontal. A kinetic frictional force of 70 N opposes her motion. Near the top of the slope the skier's speed is 3.6 m/s. What is her speed at a point which is 57 m downhill?
the net-work = Wf+Wg :
Solution:
JW
dfW
f
rf
399015770
180cos
JW
dmgW
g
g
136905742.08.958
0cos.25sin
JWnet 9700
General Physics 1, Lec 12, By/ T.A. Eleyan 247
Now use the Work-Energy Theorem:
smv
vm
Wv
mvWmv
mvmvKEW
f
if
if
if
/6.18)6.3(58
97002
2
2
1
2
12
1
2
1
2
2
22
22
General Physics 1, Lec 12, By/ T.A. Eleyan 248
Example: A 65-kg bicyclist rides his 10.0-kg bicycle with a speed of 12 m/s.
(a) How much work must be done by the brakes to bring the bike and rider to a stop?
(b) How far does the bicycle travel if it takes 4.0 s to come to rest?
(c) What is the magnitude of the braking force?
a) Friction = only horizontal forceWnet = Kf -Ki <0Wnet = 0 – (1/2) m vi
2
Wnet = (1/2) (10+65) (12)2
Wnet = 5400 kg m 2 /s 2 = 5400 J
Solution:
General Physics 1, Lec 12, By/ T.A. Eleyan 249
c) Braking forceWnet = Fnet d = FFriction d
FFriction = Wnet /d = ()/(24) = 225 N
b) Find acceleration firstv = v0+at0 = v0 + at,a= v0/ta = 12/4 = 3 m/s2
x= x0+v0 t+(1/2)at2
x= 0 + (12)(4) + (1/2)(-3)(4)2
x= 48 -24 = 24 m
250
Power is defined as the rate at which work is done.
Power
The SI unit of power is "watts" (W).
Power can also be written as;
Whenever you want to determine power, you must first determine the force and the velocity or the work being done and the time.
Time
WorkPower
WattSecond
Joule1
1
1
Fvt
dF
Time
WorkPower )(
Additional Questions
[1] The force acting on a particle varies as in Figure . Find the work done by the force on the particle as it moves (a) from x = 0 to x = 8.00 m, (b) from x = 8.00 m to x = 12.0 m, and (c) from x = 0 to x = 12.0 m.
]2 [A 4.00-kg particle is subject to a total force that varies with position as shown in Figure. The particle start from rest at x = 0. What is its speed at (a) x = 5.00
m, (b) x = 10.0 m, (c) x = 15.0 m?
[3 ]A 0.300-kg ball has a speed of 15.0 m/s. (a) What is its kinetic energy? (b) What If? If its speed were doubled, what would be its kinetic
energy?
[4 ]A 40.0-kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130 N. If the coefficient of friction between box and floor is 0.300, find (a) the work done by the applied force, (b) the increase in internal energy in the box-floor system due to friction, (c) the work done by the normal force, (d) the work done by the gravitational force, (e) the change in kinetic energy of the box, and (f) the final speed of the box.
[5 ]The electric motor of a model train accelerates the train from rest to 0.620 m/s in 21.0 ms. The total mass of the train is 875 g. Find the average power delivered to the train during the acceleration.
[6 ]A 700-N Marine in basic training climbs a 10.0-m vertical rope at a constant speed in 8.00 s. What is his power output?
[7] A 1500-kg car accelerates uniformly from rest to a speed of 10 m/s in 3s. Find (a) the work done on the car in this time, (b) the average power delivered the engine in the first 3s, and (c) the instantaneous power delivered by the engine at t = 2s.
[8] A mechanic pushes a 2500kg car from rest to a speed v doing 5000J of work in the process. During this time, the car moves 25m. Neglecting friction between the car and the road, (a) What is the final speed, v, of the car? (b) What is the horizontal force exerted on the car?
[9] A 200kg cart is pulled along a level surface by an engine. The coefficient of friction between the carte and surface is 0.4. (a) How much power must the engine deliver to move the carte at constant speed of 5m/s? (b) How much work is done by the engine in 3min?
[10] A 65kg woman climbs a flight of 20 stairs, each 23 cm high. How much work was done against the force of gravity in the process?
[11 ]A skier of mass 70.0 kg is pulled up a slope by a motor driven cable. (a) How much work is required to pull him a distance of 60.0 m up a 30.0° slope (assumed frictionless) at a constant speed of 2.00
m/s? (b) A motor of what power is required to perform this task?
[13] A horizontal force of 150 N is used to push a 40-kg box on a rough horizontal surface through a distance of 6m. If the box moves at constant speed, find (a) the work done by the 150-N force, (b) the work done by friction.
[14] When a 4-kg mass is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.5cm. If the 4-kg mass is removed, (a) how far will the spring stretch if a 1.5-kg mass is hung on it, and (b) how much work must an external agent do to stretch the same spring 4.0 cm from its unstretched position?
[12] If an applied force varies with position according to F.= 3x3 - 5, where x is in m, how much work is done by this force on an object that moves from x = 4 m to x = 7 m?
[15 ]A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruising speed of 1.75 m/s. (a) What is the average power of the elevator motor during this period? (b) How does this power compare with the motor power when the elevator
moves at its cruising speed?
[16 ]A 4.00-kg particle moves along the x axis. Its position varies with time according to x = t + 2.0t 3, where x is in meters and t is in seconds. Find (a) the kinetic energy at any time t, (b) the acceleration of the particle and the force acting on it at time t, (c) the power being delivered to the particle at time t, and (d) the work done on the particle in the interval t = 0 to t=2.00 s.
[17] If a man lifts a 20-kg bucket from a well and does 6 kJ of work, how deep is the well? Assume the speed of the bucket remains constant as it is lifted.
Potential Energy
There are two types of forces:
• conservative (gravity, spring force)•All microscopic forces are conservative:
Gravity, Electro-Magnetism, Weak Nuclear Force, Strong Nuclear Force
• nonconservative (friction, tension)Macroscopic forces are non-conservative,
Conservative ForcesConservative Forces
A force is A force is conservativeconservative if the work it does on an if the work it does on an object moving between two points is object moving between two points is independent independent of the path takenof the path taken..
work done depends only on ri and rf
( 1) ( 2)PQ PQW along W along
( 1) ( 2)
( 1) ( 2) 0PQ QP
PQ QP
W along W along
W along W along
If an object moves in a closed path (ri = rf) then total work done by the force is zero.
Nonconservative ForcesNonconservative Forces
( 1) ( 2)PQ PQW along W along
non-conservative forces dissipate energy
work done by the force depends on the path
( 1) ( 2)
( 1) ( 2) 0PQ QP
PQ QP
W along W along
W along W along
Work Done by Conservative Work Done by Conservative ForcesForces
Potential Energy:Potential Energy: Energy associated with the Energy associated with the positionposition of an object of an object..
For example:For example: When you lift a ball a distance When you lift a ball a distance yy, , gravity does negative work on the ball. This gravity does negative work on the ball. This work can be recovered as kinetic energy if work can be recovered as kinetic energy if we let the ball fall. The energy that was we let the ball fall. The energy that was ““storedstored”” in the ball is in the ball is potential energypotential energy..
WWcc = - = -U =-U =-UUfinalfinal –– U Uinitialinitial]]
WWcc = work done by a conservative force = work done by a conservative force
U = change in potential energyU = change in potential energy
Gravitational Potential EnergyGravitational Potential EnergyGravitational potential energy U = mg(y-y0), where, y =
height
U=0 at y=y0 (e.g. surface of earth).
Work done by gravity:
Wg = -mg y = -mg (y- y0)
Spring Potential EnergySpring Potential Energy
Uf – Ui = [Work done by spring on mass]
Mass m starts at x=0 (Ui =0) and moves until spring is stretched to position x.
WorkSpring = - ½ kx2
U(x) – 0 = 1/2 kx2)
USpring(x) = ½ kx2
x = displacement from equilibrium position
x
F=kx
Area in triangle= kx times increment in x= Work done by spring
Conservation of EnergyConservation of Energy Energy is neither created nor destroyedEnergy is neither created nor destroyed The energy of an isolated system of The energy of an isolated system of
objects remains constant.objects remains constant.
Mechanical energyMechanical energy EE is the sum of the potential is the sum of the potential and kinetic energies of an objectand kinetic energies of an object..
E = U + KE = U + K
The total mechanical energy in any isolated The total mechanical energy in any isolated system of objects remains constantsystem of objects remains constant ifif the the objects interact only through conservative objects interact only through conservative forcesforces::
E = constantE = constant
EEf f == EEi i UUf f + K+ Kff = U = Uii+ K+ Kii
U + U + K = K = E = 0E = 0
Mechanical Energy (Conservative Forces)
Example: A ball of mass m is dropped from a height h above the ground, as shown in Figure
(A) Neglecting air resistance, determine the speed of the ball when it is at a height y above the ground.
2 2
2
1 1
2 21
02
2 ( )
f i
f f i i
f f i i
f
f
E E
U K U K
mgy mv mgy mv
mgy mv mgh
v mg h y
(B) Neglecting air resistance, determine the speed of the ball when it is at the ground.
Example: A 0.5 kg block is used to compresses a spring with a spring constant of 80.0 N/m a distance of 2.0 cm. After the spring is released, what is the final speed of the block?
2 2 2 2
2 2
2 2
1 1 1 1
2 2 2 21 1
0 02 2
80 (0.02)0.25 /
5
f f i i
f f i i
f i
if
U K U K
kx mv kx mv
mv kx
kxv m s
m
Solution
Example: A particle of mass m = 5.00 kg is released from point A and slides on the frictionless track shown in Figure.Determine (a) the particle’s speed at points B and C and(b) the net work done by the gravitational force in moving the particle from A to C.
2 2
2
1 1
2 21
5 9.8 3.20 5 5 9.8 5 02
245 156.85.94 /
2.5
B B A A
B B A A
B
B
U K U K
mgy mv mgy mv
v
v m s
(a) the particle’s speed at points B
Find the particle’s speed at points C then find the work from the relation W U
Work Done by Nonconservative Work Done by Nonconservative ForcesForcesNonconservative forces change the amount of mechanical
energy in a system.
0
0f i
nc
E
E E
E W
Wnc = work done by nonconservative force
Problem: Children and sled with mass of 50 kg slide down a hill with a height of 0.46 m. If the sled starts from rest and has a speed of 2.6 m/s at the bottom, how much thermal energy is lost due to friction (i.e. what is the work that friction does)? If the hill has an angle of 20° above the horizontal what was the frictional force.
2 2
( ) ( )
1 1
2 2
f f i i nc
f f i i
U K U K W
mgh mv mgh mv Wnc
Since vi = 0, and hf = 0,
JW
mghmvW
nc
fnc
5646.08.9506.2502
12
11
2
The force done by friction is determined from;
Jd
WF
dFW
fr
fr
42180cos24.1
56
cos
cos
Example: A child of mass m rides on an irregularly curved slide of height h = 2.00 m, as shown in Figure. The child starts from rest at the top.
(A) Determine his speed at the bottom, assuming no friction is present.
210 0
2
2 6.26 /
f f i i
f
f
U K U K
mv mgh
v gh m s
(B) If a force of kinetic friction acts on the child, how much mechanical energy does the system lose? Assume that vf =3.00 m/s and m = 20.0 kg.
2
2
( )
10 ( 0)
21
21
20 9 20 9.8 2 3022
f f i i
f
f
E U K U K
E mv mgh
E mv mgh
E J
Example: A block having a mass of 0.80kg is given an initial velocity vA = 1.2m/s to the right and collides with a spring of negligible mass and force constant k = 50N/m, as shown in Figure.
(A) Assuming the surface to be frictionless, calculate the maximum compression of the spring after the collision.
2 2max
2
max
1 1
2 2
0.15
C A
A
A
E E
kx mv
mvx m
k
(B) Suppose a constant force of kinetic friction acts between the block and the surface, with = 0.50. If the speed of the block at the moment it collides with the spring is vA = 1.2 m/s, what is the maximum compression xC in the spring?
k
2 21 1............(1)
2 2
......................(2)
f i
C A
k C
k C
E E E
E kx mv
But
E f x
E mgx
2 2
2 2
2
1 10
2 21 1
50 (0.8)(1.2) (0.50)(0.8)(9.8) 02 2
25 3.92 0.576 0
C A k C
C C
C C
then
kx mv mgx
x x
x x
Additional Questions
[1] A 1300 kg car drives up a 17.0 m hill. During the drive, two nonconservative forces do work on the car:
(i) the force of friction, and
(ii) the force generated by the car’s engine.
The work done by friction is –3.31 105 J;
the work done by the engine is +6.34 105 J.
Find the change in the car’s kinetic energy from the bottom of the hill to the top of the hill.
[2] A single conservative force Fx = (2x + 4) N acts on a 5kg particle, where x is in m. As the particle moves along the x axis from x = 1 m to x = 5 m, calculate (a) the work done by this force, (b) the change in the potential energy of the particle, and (c) its kinetic energy at x = 5 m if its speed at x =1m is 3 m/s.
[3] Use conservation of energy to determine the final speed of a mass of 5.0kg attached to a light cord over a massless, frictionless pulley and attached to another mass of 3.5 kg when the 5.0 kg mass has fallen (starting from rest) a distance of 2.5 m as shown in Figure
[4] A 5kg block is set into motion up an inclined plane as in Figure with an initial speed of 8 m/s. The block comes to rest after travelling 3 m along the plane, as shown in the diagram. The plane is inclined at an angle of 30' to the horizontal. (a) Determine the change in kinetic energy. (b) Determine the change in potential energy. (c) Determine the frictional force on the block. (d) What is the coefficient of kinetic friction?
[5] A block with a mass of 3 kg starts at a height h = 60 cm on a plane with an inclination angle of 30', as shown in Figure. Upon reaching the bottom of the ramp, the block slides along a horizontal surface. If the coefficient of friction on both surfaces is ,Uk = 0.20, how far will the block slide on the horizontal surface before coming to rest?
[6] Initially sliding with a speed of 1.7 m/s, a 1.7 kg block collides with a spring and compresses it 0.35 m before coming to rest. What is the force constant of the spring?
[7 ]A horizontal force of 100 N is applied to move a 45-kg cart across a 9.0-m level surface. What work is done by the 100-N force?
]8[ A Hooke’s law spring is compressed 12.0 cm from equilibrium and the potential energy stored is 72.0 J. What is the spring constant in this case?
]9[ An object of mass m = 2.0 kg is released from rest at the top of a frictionless incline of height 3 m and length 5 m. Taking g = 10 m/s, use energy considerations to find the velocity of the object at the bottom of the incline.
]10[ A block of mass 1.0 kg is placed at the top of an incline of length 125 m and height 62.5 m. The plane has a rough surface. When the block arrives at the bottom of the plane it has a velocity of 25 m/ss. What is the magnitude of the constant frictional force acting on the block? Take g = 10m/ss
[11] A 0.2-kg pendulum bob is swinging back and forth. If the speed of the bob at its lowest point is 0.65 m/s, how high does the bob go above its minimum height?
[12] Two objects are connected by a light string passing over a light frictionless pulley as shown in Figure. The object of mass 5.00 kg is released from rest. Using the principle of conservation of energy, (a) determine the speed of the 3.00-kg object just as the 5.00-kg object hits the ground. (b) Find the maximum height to which the 3.00-kg object rises.
[13] A 5.00-kg block is set into motion up an inclined plane with an initial speed of 8.00 m/s. The block comes to rest after traveling 3.00m along the plane, which is inclined at an angle of 30.0° to the horizontal. For this motion determine
(a) the change in the block’s kinetic energy, (b) (b) the change in the potential energy of the
block–Earth system,(c) the friction force exerted on the block
(assumed to be constant). (d) What is the coefficient of kinetic friction?
[14 ]A block of mass 0.250 kg is placed on top of a light vertical spring of force constant 5 000 N/m and pushed downward so that the spring is compressed by 0.100 m. After the block is released from rest, it travels upward and thenleaves the spring. To what maximum height above the point of release
does it rise?
[15 ]A single constant force acts on a 4.00-kg particle. (a) Calculate the work done by this force if the particle moves from the origin to the point having the vector position . Does this result depend on the path? Explain. (b) What is the speed of the particle at r if its speed at
the origin is 4.00 m/s? (c) What is the change in the potential energy?
[16 ]A potential-energy function for a two-dimensional force is of the form U = 3x3y + 7x. Find the force that acts at the point (x, y)
[17 ]The coefficient of friction between the 3.00-kg block and the surface in Figure is 0.400. The system starts from rest. What is the speed of the 5.00kg ball when it has fallen 1.50 m
[18] A 5-kg mass is attached to a light string of length 2m to form a pendulum as shown in Figure. The mass is given an initial speed of 4m/s at its lowest position. When the string makes an angle of 37o with the vertical, find (a) the change in the potential energy of the mass, (b) the speed of the mass, and (c) the tension in the string. (d) What is the maximum height reached by the mass above its lowest position?
[19 ]A 20.0-kg block is connected to a 30.0-kg block by a string that passes over a light frictionless pulley. The 30.0-kg block is connected to a spring that has negligible mass and a force constant of 250 N/m, as shown in Figure.The spring is unstretched when the system is as shown in the figure, and the incline is frictionless. The 20.0-kg block is pulled 20.0 cm down the incline (so that the 30.0-kgblock is 40.0 cm above the floor) and released from rest. Find the speed of each block when the 30.0-kg block is 20.0 cm above the floor (that is, when the spring is unstretched).
[20] find the particle’s speed at points B, where the particle released from point A and slides on the frictionless track
[21] Find the distance x.
[22] An object of mass m starts from rest and slides a distance d down a frictionless incline of angleθ . While sliding, it contacts an unstressed spring of negligible mass as shown in Figure. The object slides an additional distance x as it is brought momentarily to rest by compression of the spring (of force constant k). Find the initial separation d between object and spring.
[23] A 1.9-kg block slides down a frictionless ramp, as shown in the Figure. The top of the ramp is 1.5 m above the ground; the bottom of the ramp is h = 0.25 m above the ground. The block leaves the ramp moving horizontally, and lands a horizontal distance d away. Find the distance d.
[24] A skateboard track has the form of a circular arc with a 4.00 m radius, extending to an angle of 90.0° relative to the vertical on either side of the lowest point, as shown in the Figure. A 57.0 kg skateboarder starts from rest at the top of the circular arc. What is the normal force exerted on the skateboarder at the bottom of the circular arc? What is wrong with this picture?
At bottom, a = v2/rWhich direction?
292
Discussion
293
[[11 ] ]A force F = (4xA force F = (4x ˆ̂i + 3yi + 3yˆ̂j) N acts on an object as the j) N acts on an object as the object moves in the x direction from the origin to x object moves in the x direction from the origin to x = 5.00 m. Find the work done on the object by the = 5.00 m. Find the work done on the object by the forceforce..
Jx
dxxidxjyixdrFWf
i
r
r
50)2
(4)4(ˆ).ˆ3ˆ4(.5
0
25
0
5
0
[2 ]A 3.00-kg object has a velocity (6.00ˆi - 2.00ˆj) m/s. (a) What is its kinetic energy at this time? (b) Find the total work done on the object if its velocity changes to (8.00ˆi + 4.00ˆj) m/s .
JvvmKEWb
JmvKEa
vv
if
ii
fi
60)4080(2
3)(
2
1][
60)40)(3(2
1
2
1][
801664,40436
22
2
294
[3] A 1300 kg car drives up a 17.0 m hill. During the drive, two nonconservative forces do work on the car:
(i) the force of friction, and
(ii) the force generated by the car’s engine.
The work done by friction is –3.31 105 J;
the work done by the engine is +6.34 105 J.
Find the change in the car’s kinetic energy from the bottom of the hill to the top of the hill.
5 5
5 5 5
-3.31 10 6.34 10 1300 9.8 17
3.03 10 2.16 10 6.5 10
ncW k u
k
k J
295
[4] A 1500-kg car accelerates uniformly from rest to a speed of 10 m/s in 3s. Find (a) the work done on the car in this time, (b) the average power delivered the engine in the first 3s, and
2 21( )
21
1500 (100 0) 750002
f i
f i
W k k k
W m
W J
750002500
3
WP wat
t
a)
b)
296
[5] When a 4kg mass is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.5cm. If the 4kg mass is removed, (a) how far will the spring stretch if a 1.5kg mass is hung on it, and (b) how much work must an external agent do to stretch the same spring 4.0 cm from its unstretched position?
1568
0.9
mgF kx k N m
xmg
F kx x cmk
a)
b)
طريق عن يحلوتناسب نسبة
General Physics 1, Lec14, By/ T.A. Eleyan 297
[6] A single conservative force Fx = (2x + 4) N acts on a 5kg particle, where x is in m. As the particle moves along the x axis from x = 1 m to x = 5 m, calculate (a) the work done by this force, (b) the change in the potential energy of the particle, and (c) its kinetic energy at x = 5 m if its speed at x =1m is 3 m/s.
552
11
(2 4) 4 40W x dx x x J
40
W u
u J
40
W K
K J
c)
b)
a)
General Physics 1, Lec14, By/ T.A. Eleyan 298
[7] Use conservation of energy to determine the final speed of a mass of 5.0kg attached to a light cord over a massless, frictionless pulley and attached to another mass of 3.5 kg when the 5.0 kg mass has fallen (starting from rest) a distance of 2.5 m as shown in Figure
2 21 2 1 2
0
1 1
2 2
f f i i
E
k u k u
m m m gh m gh
Substitute and find v
General Physics 1, Lec14, By/ T.A. Eleyan 299
[8] A 5kg block is set into motion up an inclined plane as in Figure with an initial speed of 8 m/s. The block comes to rest after travelling 3 m along the plane, as shown in the diagram. The plane is inclined at an angle of 30' to the horizontal. (a) Determine the change in kinetic energy. (b) Determine the change in potential energy. (c) Determine the frictional force on the block. (d) What is the coefficient of kinetic friction?
General Physics 1, Lec14, By/ T.A. Eleyan 300
2
k
10
2
( ) ( )
f = mg cos30
f i
i
f i
f i
nc f i
f f i i
k k k
k m
u u u
u mgh mgh
W E E
fs k u k u
a)
b)
c)
d)
301
[9] A block with a mass of 3 kg starts at a height h = 60 cm on a plane with an inclination angle of 30', as shown in Figure. Upon reaching the bottom of the ramp, the block slides along a horizontal surface. If the coefficient of friction on both surfaces is ,Uk = 0.20, how far will the block slide on the horizontal surface before coming to rest?
302
21
22
22
1( 0) (0 )2
1(0 0) ( 0)
21
= 2
nc f i
f
i
i
W E E
fs mv mgh
fs mv
fs mv
And find v
And find s2
General Physics 1, Lec14, By/ T.A. Eleyan 303
[10] Initially sliding with a speed of 1.7 m/s, a 1.7 kg block collides with a spring and compresses it 0.35 m before coming to rest. What is the force constant of the spring?
2 21 10 0
2 240.1 /
f i
spring
E E
kx m
K N m
General Physics 1, Lec14, By/ T.A. Eleyan 304
[[1111]]The ambulance (mass 3000kg) shown in the Figure(2) The ambulance (mass 3000kg) shown in the Figure(2) slides (wheels locked) down a frictionless incline that is slides (wheels locked) down a frictionless incline that is 10 m long. It starts from rest at point A, and continues 10 m long. It starts from rest at point A, and continues along a rough surface until it comes to a complete stop along a rough surface until it comes to a complete stop at point C. If the coefficient of kinetic friction between at point C. If the coefficient of kinetic friction between the ambulance and the horizontal rough surface is 0.1. the ambulance and the horizontal rough surface is 0.1. (a) Calculate the speed of the ambulance at point B.(b) (a) Calculate the speed of the ambulance at point B.(b) Compute the distance d the ambulance slides on the Compute the distance d the ambulance slides on the horizontal rough surface before stoppinghorizontal rough surface before stopping..
305
Linear Momentum and Collisions
306
Linear momentumLinear momentum is defined as is defined as::pp = = mmvv
Momentum is given by mass times velocity. Momentum is given by mass times velocity. Momentum is a vector.Momentum is a vector. The units of momentum are (no special unit):The units of momentum are (no special unit): [p] = kg[p] = kg··m/sm/s
Linear MomentumLinear Momentum
Since p is a vector, we can also consider the components of momentum:
px = mvx, py = mvy, pz = mvz
Note: momentum is “large” if m and/or v is large. (define large, meaning hard for you to stop).
307
Recall thatRecall that
dt
d
dt
mddt
d m
pv F
vaaF
and
Another way of writing Newton’s Second Law is
F = dp/dt= rate of change of momentum
This form is valid even if the mass is changing.
This form is valid even in Relativity and Quantum Mechanics.
308
f
i
t
t
dtFI
Impulse of a Force
Exercise: Show that impulse and momentum have the same units.
)t,(t intervalan over F force a of I Impulse fi
309
dtFPdFdt
Pd
dtFpdf
i
f
i
p
p
t
t
IPPP if
The impulse of a force during a certain The impulse of a force during a certain interval of time is the change in momentum interval of time is the change in momentum
that the force producesthat the force produces
Impulse & Momentum
310
F-t The GraphF-t The Graph
Impulse is a vector Impulse is a vector quantityquantity
The magnitude of the The magnitude of the impulse is equal to the impulse is equal to the area under the force-time area under the force-time curvecurve
Dimensions of impulse are Dimensions of impulse are M L / TM L / T
Impulse is not a property Impulse is not a property of the particle, but a of the particle, but a measure of the change in measure of the change in momentum of the particlemomentum of the particle
311
Example: In a particular crash test, a car of mass 1500 kg collides with a wall, as shown in Figure. The initial and final velocities of the car are , respectively. If the collision lasts for 0.15 s, find the impulse caused by the collision and the average force exerted on the car.
15 / , 2.6 /i fm s m s
1) The impulse
2) The average force
24
12
/.1064.2)156.2(1500
)(
smkg
vvmPI
Nt
PF 4
4
106.1715.0
1064.2
312
Example: A 100-g ball is dropped from 2.00 m above the ground. It rebounds to a height of 1.50 m. What was the average force exerted by the floor if the ball was in contact with the floor for 1×10-2 s.
212
21
/.17.1)26.642.5(1.0)(,
?,/24.5,/26.6
smkgvvmPNow
Howsmvandsmv
The force acting on the golf ball can be determined from
Nt
PF 2
21017.1
101
17.1
313
Problem: A 0.3-kg hockey puck moves on frictionless ice at 8 m/s toward the wall. It bounces back away from the wall at 5 m/s. The puck is in contact with the wall for 0.2 s.
(a) What is the change in momentum of the hockey puck during the bounce?
(b) What is the impulse on the hockey puck during the bounce?
(c) What is the average force of the wall on the hockey puck during the bounce?
314
Example: A 325-gm ball with a speed v of 6.22 m/s strikes a wall at an angle of 33.00 and then rebounds with the same speed and angle. It is in contact with the wall for 10.4 ms. (a) What impulse was experienced by the ball ? What was the average force exerted by the ball on the wall ?
x
y
315
)ˆsinˆ(cos jimvPi
)ˆsinˆcos( jimvPf
imvP ˆcos2
(a)
Nit
IFball
ˆ0.326
smkgiPI /ˆ40.3
(b)
NiFF ballwallˆ0.326
ال على كلو yو xيوثرموجب
على y+وx–يوثر
316
Principle of Conservation of Linear Principle of Conservation of Linear MomentumMomentumWhen the net external force acting on a system is zero, When the net external force acting on a system is zero,
the total linear momentum of the system remains the total linear momentum of the system remains constantconstant..
21 PPP
Proof (Two-Body Systems)
2121 FF
dt
dP
dt
Pd
dt
Pd
21ext2212ext11 FFFFFF
,, ;
2112 FF
However,
317
exttotext2ext1 FFFdt
Pd,,,
In the absence of a net external In the absence of a net external forceforce,,
.,, ConstPor0dt
Pd
318
Example: A firecracker with a mass of 100g, initially at rest, explodes into 3 parts. One part with a mass of 25g moves along the x-axis at 75m/s. One part with mass of 34g moves along the y-axis at 52m/s. What is the velocity of the third part?
The third part has a mass of 100-34-25 = 41g.
Along x-direction:
smm
vmv
vmvm
PP
xx
xx
xfxi
/4641
)75)(25(
00
3
113
3311
319
CollisionsCollisionsIn general, a In general, a ““collisioncollision”” is an interaction in which is an interaction in which two objects strike one anothertwo objects strike one another the net external impulse is zero or negligibly the net external impulse is zero or negligibly
small (momentum is conserved)small (momentum is conserved)
Examples: Examples: Two billiard balls colliding on a billiards tableTwo billiard balls colliding on a billiards table An alpha particle colliding with a heavy atomAn alpha particle colliding with a heavy atom Two galaxies collidingTwo galaxies colliding
320
Types of CollisionsTypes of Collisions
[1] [1] In an In an elasticelastic collision collision, momentum and kinetic , momentum and kinetic energy are conservedenergy are conserved– Perfectly elastic collisions occur on a microscopic levelPerfectly elastic collisions occur on a microscopic level– In macroscopic collisions, only approximately elastic In macroscopic collisions, only approximately elastic
collisions actually occurcollisions actually occur
[2] [2] In an In an inelastic inelastic collisioncollision, kinetic energy is not , kinetic energy is not conserved although momentum is still conservedconserved although momentum is still conserved– If the objects stick together after the collision, it is a If the objects stick together after the collision, it is a
perfectly inelastic perfectly inelastic collisioncollision
In an inelastic collision, some kinetic energy is lost, but the objects do not stick together
Elastic and perfectly inelastic collisions are limiting cases, most actual collisions fall in between these two types
321
Elastic CollisionsElastic Collisions
Both momentum Both momentum and kinetic energy and kinetic energy are conservedare conserved
1 1 2 2
1 1 2 2
2 21 1 2 2
2 21 1 2 2
1 1
2 21 1
2 2
i i
f f
i i
f f
m m
m m
m m
m m
v v
v v
v v
v v
322
Perfectly Inelastic CollisionsPerfectly Inelastic Collisions
Since the objects Since the objects stick together, they stick together, they share the same share the same velocity after the velocity after the collisioncollision
fii vmmvmvm )( 212211
323
Example: A ball with a mass of 1.2 kg moving to the right at 2.0 m/s collides with a ball of mass 1.8 kg moving at 1.5 m/s to the left. If the collision is an elastic collision, what are the velocities of the balls after the collision?
324
1 1, 2 2, 1 1, 2 2,
1, 2,
1, 2,
2 2 2 21 1, 2 2, 1 1, 2 2,
2 21, 2,
1.2 2 1.8 1.5 1.2 1.8
1.2 1.8 0.3................................(1)
1 1 1 1
2 2 2 21 1 1 1
1.2 4 1.8 2.25 1.2 1.82 2 2 2
0.6
i i f f
f f
f f
i i f f
f f
m m m m
m v m v m v m v
v v
v v v v
v v
v v
2 21, 2,0.9 4.425..............................(2)f fv v
Solve (1)&(2) to find velocities of the balls after the collision?
We can apply conservation of momentum and conservation of kinetic energy:
325
Example: A 3kg sphere makes a perfectly inelastic collision with a second sphere initially at rest. The composite system moves with a speed equal to 1/3rd the original speed of the 3 kg sphere. What is the mass of the second sphere?
326
we can apply conservation of momentum:
kgm
m
m
vmv
vmmv
vmmvmvm
ii
ii
fii
6
39
)3
1)(3(3
)3
1)(3(3
)3
1)(3()0(3
)(
2
2
2
121
1221
212211
327
Linear Momentum and Collisions
328
Example: A 1 500-kg car traveling east with a speed of 25.0 m/s collides at an intersection with a 2500-kg van traveling north at a speed of 20.0 m/s, as shown in Figure. Find the direction and magnitude of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together)
m/s kg 1075.3 4carcarix vmp
cos)( fvancarfx vmmp
fxix pp
)1.....(coskg) 1000.4(m/s kg 1075.3 34 fv
From the x axis component
329
m/s kg 1000.5 4vanvaniy vmp
sin)( fvancarfy vmmp
fyiy pp
)2.......(sinkg) 1000.4(m/s kg 1000.5 34 fv
33.1m/s kg 1075.3
m/s kg 1000.5tan
4
4
1.53
m/s 6.151.53sin)kg 1000.4(
m/s kg 1000.53
4
fv
From the y axis component
From (1) & (2)
330
Example: In the ballistic pendulum experiment, a bullet of mass .06 kg is fired horizontally into a wooden block of mass .2 kg. The wooden block is suspended from the ceiling by a long string as shown in the diagram. The collision is perfectly inelastic and after impact the bullet and the block swing together until the block is 12 m above it's initial position. Finda) the velocity of the bullet and block just after impact and
b) the velocity of the bullet just before impact
Energy is conserved between points B and C
21 2 1 2
1( ) ( )
2
2 1.53 /
c A
B
B
U K
m m gh m m v
v gh m s
B
331
Momentum is conserved between A and B
1 1 2( )
2 6.6 /
A B
A B
A
p p
m v m m v
v gh m s
332
ExampleExample::Two metal spheres, suspended by vertical cords, initially Two metal spheres, suspended by vertical cords, initially just touch, as shown in Fig. Sphere 1, with mass just touch, as shown in Fig. Sphere 1, with mass mm11=30 g, is pulled to the left to height h=30 g, is pulled to the left to height h11=8.0cm, and =8.0cm, and then released from rest. After swinging down, it then released from rest. After swinging down, it undergoes an elastic collision with sphere 2, whose undergoes an elastic collision with sphere 2, whose mass mmass m22=75 g. What is the velocity =75 g. What is the velocity vv11ff of sphere 1 just of sphere 1 just
after the collisionafter the collision ? ?
333
222
211
211
221111
2
1
2
1
2
1ffi
ffi
vmvmvm
vmvmvm
where 11 2ghv i
222
21
211
22111
)(
)(
ffi
ffi
vmvvm
vmvvm
ffi vvv 211
if
fifi
vmm
mmv
vvmvvm
121
211
112111 )()(
General Physics I, Lec 16 By/ T.A. Eleyan 334
Example: The figure below shows an elastic collision of two pucks on a frictionless air table. Puck A has mass mA = 0.500 kg, and puck B has mass mB = 0.300 kg. Puck A has an initial velocity of 4.00 m⁄s in the positive x−direction and a final velocity of 2.00 m ⁄s in an unknown direction. Puck B is initially at rest. We want to find the final speed VB2 of puck B and the angles α and β as
illustrated in the figure.
2 2 21 2 2
22
2
1 1 1
2 2 2
0.5 16 0.5 4 0.3
4.47 /
A A A A B B
B
B
m v m v m v
v
v m s
Because the collision is elastic, the initial and final kinetic energies are equal
General Physics I, Lec 16 By/ T.A. Eleyan 335
1 2 2cos cos
0.5 4 0.5 2cos 0.3 4.47cos
2 cos 1.34cos ...............(1)
xi xf
A A A A B B
p p
m v m v m v
2 20 sin sin
0 0.5 2sin 0.3 4.47sin
0 sin 1.34sin ...............(2)
yi yf
A A B B
p p
m v m v
Conservation of the x− component of total momentum gives
Conservation of the y− component of total momentum gives
There are two simultaneous equations for α and β
Solve to find the angles
General Physics I, Lec 16 By/ T.A. Eleyan 336
Center of MassCenter of Mass
The center of mass (CM) of an object or a group of objects (system) is the “average” location of the mass in the system. The system behaves as if all of its mass were concentrated at the center of mass.
General Physics I, Lec 16 By/ T.A. Eleyan 337
Center of mass Center of mass –– point objects point objects –– 1 1 DD
n n
i i i i1 1 2 2 i 1 i 1
cm n1 2
ii 1
m x m xm x m x ...
xm m ... Mm
338
Center of mass Center of mass –– point objects point objects –– 2 2 DD
n
i ii 1
cm
m xx
M
n
i ii 1
cm
m yy
M
n
i ii 1
cm
m zz
M
n
i ii 1
cm
m rr
M
339
Center of mass Center of mass –– solid objects solid objects –– 1 D 1 D
n
i ii 1
cm
m xx
M
cm
xdmx
M
340
Center of mass Center of mass –– solid objects solid objects –– 2 2 DD
cm
xdmx
M
cm
ydmy
M
cm
zdmz
M
cm
rdmr
M
General Physics I, Lec 16 By/ T.A. Eleyan 341
A method for finding the center of mass of any object.
Hang object from two or more points.
Draw extension of suspension line.
Center of mass is at intercept of these lines.
General Physics I, Lec 16 By/ T.A. Eleyan 342
ProblemProblem::
Three particles of masses Three particles of masses mmAA = 1.2 kg, = 1.2 kg, mmBB = 2.5 kg, and = 2.5 kg, and mmCC = 3.4 kg form an = 3.4 kg form an equilateral triangle of edge length equilateral triangle of edge length aa = 140 cm. Where is the center of = 140 cm. Where is the center of
mass of this three particle systemmass of this three particle system ? ?
Three masses located in the xy plane have the following coordinates:2 kg at (3,-2)3 kg at (-2,4)1 kg at (2,2)
Find the location of the center of mass
Problem:
343
Motion of the Center of MassMotion of the Center of Mass
A system of objects behaves as if all its mass were located at the center of mass.
Mm
mmmm
cmvvv
V
......
21
2211
Velocity of CM:
Acceleration of CM:
Mm
mmmm
cmaaa
A
...
...
21
2211
MAcm = Fnet,extNewton's Laws for a System of Particles
344
ProblemProblem::
The three particles in Fig. The three particles in Fig. aa are initially at rest. Each are initially at rest. Each experiences an experiences an externalexternal force due to bodies outside force due to bodies outside the three-particle system. the three-particle system. The directions are The directions are indicated, and the indicated, and the magnitudes are magnitudes are FFAA=6 N =6 N , , FFBB=12 N =12 N , and , and FFCC=14 N. =14 N. What is the magnitude of What is the magnitude of the acceleration of the the acceleration of the center of mass of the center of mass of the system, and in what system, and in what
direction does it movedirection does it move ? ?
Additional Questions
]1 [A 3.00-kg particle has a velocity of (3.00i ˆ + 4.00j ˆ ) m/s. (a) Find its x and y components of momentum. (b) Find the magnitude and direction of its momentum.
[2 ]An estimated force–time curve for a baseball struck by a bat is shown in Figure. From this curve, determine
)a (the impulse delivered to the ball, (b) the average force exerted on the ball, and (c) the peak force exerted on the
ball..
[3 ]A 3.00-kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle (Fig. ). If the ball is in contact with the wall for 0.200 s, what is the average force exerted
by the wall on the ball?
[4 ]A 10.0-g bullet is fired into a stationary block of wood (m = 5.00 kg). The relative motion of the bullet stops inside the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.600 m/s. What was the original speed of the bullet?
[5 ]Two blocks are free to slide along the frictionless wooden track ABC shown in Figure. The block of mass m1 = 5.00 kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass m2 =10.0 kg, initially at rest. The two blocks never touch. Calculate
the maximum height to which m1 rises after the elastic collision..
]6[ A 12.0-g wad of sticky clay is hurled horizontally at a 100-g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between the block and the surface is 0.650, what was the speed of the clay immediately before impact?
[7) ]a (Three carts of masses 4.00 kg, 10.0 kg, and 3.00 kg move on a frictionless horizontal track with speeds of 5.00 m/s, 3.00 m/s, and 4.00 m/s, as shown in Figure . Velcro couplers make the carts stick together after colliding. Find the final velocity of the train of three carts. (b) What If? Does your answer require that all the carts collide and stick together at the same time? What if they collide in a different order?
[8 ]Two automobiles of equal mass approach an intersection. One vehicle is traveling with velocity 13.0 m/s toward the east, and the other is traveling north with speed v2i. Neither driver sees the other. The vehicles collide in the intersection and stick together, leaving parallel skid marks at an angle of 55.0° north of east. The speed limit for both roads is 35 mi/h, and the driver of the northward-moving vehicle claims he was within the speed limit when the collision occurred. Is he telling the truth?
[9 ]A billiard ball moving at 5.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves, at 4.33 m/s, at an angle of 30.0° with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck
ball’s velocity after the collision .
[10 ]Two particles with masses m and 3m are moving toward each other along the x axis with the same initial speeds vi . Particle m is traveling to the left, while particle 3m is traveling to the right. They undergo an elastic glancing collisionsuch that particle m is moving downward after the collision at right angles from its initial direction. (a) Find the final speeds of the two particles. (b) What is the angle & at which the particle 3m is scattered?
[11 ]Four objects are situated along the y axis as follows: a 2.00 kg object is at 3.00 m, a 3.00-kg object is at 2.50 m, a 2.50-kg object is at the origin, and a
4.00-kg object is at - 0.500 m. Where is the center of mass of these objects?.
[12 ]A 2.00-kg particle has a velocity (2.00ˆi - 3.00ˆ j) m/s, and a 3.00-kg particle has a velocity (1.00ˆi +6.00ˆ j) m/s. Find (a) the velocity of the center of mass and (b) the total momentum of the system.
[13 ]A 0.2 kg baseball is traveling at 40 m/s. After being hit by a bat, the ball's velocity is 50 m/s in the opposite direction. Find a) the impulse and b) the average force exerted by the bat if the ball and bat are in contact for 0.002 s
[14 ]A 1000 kg car traveling 22 m/s hits a concrete bridge support and comes to a stop in .5 s. a) Find the average force acting on the car and b) if the bridge support had been cushioned so that the stopping time was increased to 3 s, what would have been the average force?
[15 ]A 1.0 kg object traveling at 1.0 m/s collides head on with a 2.0 kg object initially at rest. Find the velocity of each object after impact if the collision is perfectly elastic
[[1616 ] ]Car A has a mass of 1000 kg and is traveling to the right at 5 Car A has a mass of 1000 kg and is traveling to the right at 5 m/s. Car B has a mass of 5000 kg and is traveling to the m/s. Car B has a mass of 5000 kg and is traveling to the right at 3 m/s. Car A read-ends Car B and they stick right at 3 m/s. Car A read-ends Car B and they stick togethertogether..
What is the velocity of the center of mass after the collisionWhat is the velocity of the center of mass after the collision??
))Before the collision, the velocity of the CM is 3.33 m/sBefore the collision, the velocity of the CM is 3.33 m/s((
[17 ]Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them (Fig.). A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s. (a) What is the speed of the block of mass M? (b) Find the original elastic potential energy in the spring if M = 0.350 kg.
Discussion
[1 ]An estimated force–time curve for a baseball struck by a bat is shown in Figure. From this curve, determine
)a (the impulse delivered to the ball, (b) the average force exerted on the ball, and (c) the peak force exerted on the
ball..
NFc
Ndt
dPFb
sN
curvetFunderarea
Fdta
3max
33
3
1018]
109105.1
5.13]
.5.1318000105.12
1
Impulse]
[2 ]Two blocks are free to slide along the frictionless wooden track ABC shown in Figure. The block of mass m1 = 5.00 kg is released from A. and makes a head on elastic collision with a block of mass m2 =10.0 kg, initially at rest at point B. Calculate the maximum height to which m1 rises after the
elastic collision..
mhand
vmghmNow
smvvmm
mmv
collisonelasticanfromvfindtoand
smvmvmgh
EEusevfindtoNow
lyrespectivecollisionafterandbeforemspeedofvv
f
fif
f
ii
BAi
fi
556.0,
02
10,
/3.3
,
/9.92
10
,
,
max
11max1
1121
211
1
11
1
111
[4]As shown in Figure , a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length L and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle?
glm
Mv
glMv
mmv
collisiontheinconservedissystemtheofMomentum
glglv
Mvlma
KUKU
EE
b
b
iiff
if
4
)2(2
24
2
100)2( 2
[5 ]Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them (Fig.). A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s. (a) What is the speed of the block of mass M? (b) Find the original elastic potential energy in the spring if M = 0.350 kg.
JU
UvMMv
KUKUorEEb
smvMMvMMv
PP
Pa
i
i
iiffif
if
4.8)2(35.032
1)6(35.0
2
1
00)3(2
1
2
10
]
/66023
0]
22
22
21
The formula for CM of a continuous object is
Lx
xCM xdmM
x0
1
Since the density of the rod () is constant;
[6] Show that the center of mass of a rod of mass M and length L lies in midway between its ends, assuming the rod has a uniform mass per unit length.
dxdm
LM /
The mass of a small segment
Therefore
CMx
Find the CM when the density of the rod non-uniform but varies linearly as a function of x, x
CMx
Lx
xxdx
M 0
1 Lx
x
xM
0
2
2
11
2
2
11L
M
ML
M 2
11
2
L
Lx
xxdx
M 0
1
Lx
xdxx
M 0
21 Lx
x
xM
0
3
3
11
3
3
11L
M
ML
M 3
21
3
2LCMx
Rotational MotionRotational Motion
For circular motion, the distance (arc length) For circular motion, the distance (arc length) ss, , the radius the radius rr, and the angle , and the angle are related by are related by::
r
s
DEGRAD 180
Note that is measured in radians:
Angular Position, θ
> 0 for counterclockwise rotation from reference line
The The angular displacementangular displacement is defined as the is defined as the angle the object rotates through during angle the object rotates through during some time intervalsome time interval
Every point on the disc undergoes the same Every point on the disc undergoes the same angular displacement in any given time angular displacement in any given time intervalinterval if
Notice that as the disk Notice that as the disk rotates, rotates, changes. We changes. We define the angular define the angular displacement, displacement, , as, as::
= = ff - - ii
which leads to the average which leads to the average angular speed angular speed avav
if
ifav ttt
Angular Velocity, ω
Instantaneous Angular VelocityInstantaneous Angular Velocity
As usual, we can define the instantaneous As usual, we can define the instantaneous angular velocity asangular velocity as::
tt
lim0
Note that the SI units of are: rad/s = s-1
> 0 for counterclockwise rotation < 0 for clockwise rotation
If v = speed of a an object traveling around a circle of radius r
= v / r
The period of rotation is the time it takes to complete one revolution.
T
22
T
Problem:
What is the period of the Earth’s rotation about its own axis?
What is the angular velocity of the Earth’s rotation about its own axis?
We can also define the average angular We can also define the average angular acceleration acceleration avav::
And instantaneous angular accelerationAnd instantaneous angular acceleration
if
ifav ttt
tt
lim0
Angular Acceleration, a
The SI units of are: rad/s2 = s-2
We will skip any detailed discussion of angular acceleration, except to note that angular acceleration is the time rate of change of angular velocity
Notes about angular kinematicsNotes about angular kinematics:: When a rigid object rotates about a fixed axis, When a rigid object rotates about a fixed axis,
every portion of the object has the same every portion of the object has the same angular speed and the same angular angular speed and the same angular accelerationacceleration
i.e. i.e. , and , and are are notnot dependent upon dependent upon rr, , distance form hub or axis of rotationdistance form hub or axis of rotation
ExamplesExamples::
1. Bicycle wheel turns 240 revolutions/min. What is its angular velocity in radians/second?
secradians1.25secradians8rev1
rads2
sec60
min1
min
rev240
2. If wheel slows down uniformly to rest in 5 seconds, what is the angular acceleration?
2secrad5sec5
secrad250
t
if
3. How many revolution does it turn in those 5 sec?
srevolution102
rev1rad5.62)(
rad5.62sec5secrad52
1sec5secrad25
2
1
2
20
rev
tt
Recall that for linear motion we had:
Perhaps something similar for angular quantities?
20 2
1attvx
Analogies Between Linear and Analogies Between Linear and Rotational MotionRotational Motion
Rotational Motion About Rotational Motion About a Fixed Axis with a Fixed Axis with Constant AccelerationConstant Acceleration
Linear Motion with Linear Motion with Constant AccelerationConstant Acceleration
ti
2
2
1tti
222i xavv i 222
2
2
1attvx i
atvv i
Relationship Between Angular and Relationship Between Angular and Linear QuantitiesLinear Quantities
DisplacementsDisplacements
SpeedsSpeeds
AccelerationsAccelerations
r
s
vr
t
s
rt1
1
ra
v r
Total linear acceleration is
Since the tangential speed v is
Tangential and the radial acceleration.
ta
The magnitude of tangential acceleration at is
The radial or centripetal acceleration ar is
ra
a
v
t
r
t
rt
r
r
v2
r
r 2
2r
22rt aa 222 rr 42 r
Example: (a) What is the linear speed of a child seated 1.2m from the center of a steadily rotating merry-go-around that makes one complete revolution in 4.0s? (b) What is her total linear acceleration?
v
1 21.6 /
4.0 4.0
rev radrad s
s s
Since the angular speed is constant, there is no angular acceleration.
Tangential acceleration is ta
Radial acceleration is ra
a
r 1.2 1.6 / 1.9 /m rad s m s
r 2 21.2 0 / 0 /m rad s m s 2r 2 21.2 1.6 / 3.1 /m rad s m s
2 2t ra a 2 20 3.1 3.1 /m s
Calculation of Moments of InertiaCalculation of Moments of InertiaMoments of inertia for large objects can be computed, if we assume that the object consists of small volume elements with mass, mi.
It is sometimes easier to compute moments of inertia in terms of volume of the elements rather than their mass
Using the volume density, , replace dm in the above equation with dV.
The moment of inertia for the large rigid object is:
iii
mmrI
i
2
0lim dmr 2
dV
dm
The moments of inertia becomes
dVrI 2
dVdm
The moment of inertia of a uniform hoop of mass M and radius R about an axis perpendicular to the plane of the hoop and passing through its center.
x
y
RO
dm
The moment of inertia is
dmrI 2
The moment of inertia for this object is the same as that of a point of mass M at the distance R.
dmR2 2MR
Example Example for Rigid Body Moment of Inertiafor Rigid Body Moment of InertiaCalculate the moment of inertia of a uniform rigid rod of length L and mass M about an axis perpendicular to the rod and passing through its center of mass.
The line density of the rod is L
M
so the mass is dm
I
dmrI 2
dx dxL
M
dmr 2 dxL
MxL
L2/
2/
2 2/
2/
3
3
1L
L
xL
M
33
223
LL
L
M
43
3L
L
M
12
2ML
dxL
MxL
0
2 L
xL
M
0
3
3
1
03
3 LL
M 3
3L
L
M
3
2ML
TorqueTorque: The ability of a force to rotate a : The ability of a force to rotate a body about some axisbody about some axis . .
Only the component of the force that is perpendicular to the radius causes a torque.
= r (Fsin)
Equivalently, only the perpendicular distance between the line of force and the axis of rotation, known as the moment arm r, can be used to calculate the torque.
= rF = (rsin)F
The net torque about a point O is the sum of all torques about O:
221121 dFdF
ProblemProblem::Calculate the net torque on the 0.6-m rod about the nail at the left. Three forces are acting on the rod as shown in the diagram.
30°
0.3 m
4 N
5 N
6 N
Torque & Angular AccelerationTorque & Angular Acceleration
Let’s consider a point object with mass m rotating on a circle.
The tangential force Ft is tt maF
The torque due to tangential force Ft is
rFt I
Torque acting on a particle is proportional to the angular acceleration.
Analogs to Newton’s 2nd law of motion in rotation.
mr
rmat 2mr I
How about a rigid object?
The external tangential force dFt is
tdF
The torque due to tangential force Ft is
The total torque is
d
tdma dmr
rdFt dmr 2
dmr 2 I
ExampleExample: :
A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane. The rod is released from rest in the horizontal position. What are the initial angular acceleration of the rod and the initial linear acceleration of its right end?
The only force generating torque is the gravitational force Mg
Using the relationship between tangential and angular acceleration
L
dmrI0
2
Since the moment of inertia of the rod when it rotates about one end
We obtain
ta The tip of the rod falls faster than an object undergoing a free fall.
Fd2
LF
2
LMg I
L
dxx0
2L
x
L
M
0
3
3
3
2ML
I
MgL
2
32 2ML
MgL
L
g
2
3
L2
3g
Rotational Kinetic EnergyRotational Kinetic EnergyKinetic energy of a rigid object that undergoing a circular motion:
Since a rigid body is a collection of mass, the total kinetic energy of the rigid object is
The moment of Inertia, I, is defined as
Kinetic energy of a mass mi, moving at a tangential speed, vi, is
iK
RK
i
iirmI 2
IKR 2
1The above expression is simplified to
2
2
1iivm 2
2
1iirm
i
iK i
iirm 22
1
iiirm 2
2
1
ExampleExample::In a system consists of four small spheres as shown in the figure, assuming the radii are negligible and the rods connecting the particles are massless, compute the moment of inertia and the rotational kinetic energy when the system rotates about the y-axis at .
x
y
M Ml l
m
m
b
bO
I
Since the rotation is about y axis, the moment of inertia about y axis, Iy, is
RK
Thus, the rotational kinetic energy is
Find the moment of inertia and rotational kinetic energy when the system rotates on the x-y plane about the z-axis that goes through the origin O.
Why are some 0s?
This is because the rotation is done about y axis, and the radii of the spheres are negligible.
2i
iirm 2Ml 22Ml
2
2
1 I 2222
1 Ml 22Ml
2Ml 20m 20m
Discussion
[1 ]During a certain period of time, the angular position of a swinging door is described by θ= 5.00 + 10.0t + 2.00t2, where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0 (b) at t = 3.00 s.
2
0
0
/44dt
don acceleratiangular
/10410dt
d speedangular
rad 5position angular
0at t a]
srad
sradt
t
t
[5 ]A disk 8.00 cm in radius rotates at a constant rate of 1 200 rev/min about its central axis. Determine (a) its angular speed, (b) the tangential speed at a point 3.00 cm from its center, (c) the radial acceleration of a point on the rim, and (d) the total distance a point on the rim moves in 2.00 s.
srt
f
1.202108126rs d]
???a
m/s 1260108(126)rωa c]
m/s 3.77108126ωr vb]
rad/s 126)60
1200(22
T
2πω a]
2-
r
222c
2-
Consider a car driving at 20 m/s on a 30° banked circular curve of radius 40.0 m. Assume the car’s mass is 1000 kg.
1. What is the magnitude of frictional force experienced by car’s tires?
2. What is the minimum coefficient of friction in order for the car to safely negotiate the turn?
1. Draw a free body diagram, introduce coordinate frame and consider vertical and horizontal projections
Nmgr
vmf
mgfr
vmFx
376030sin30cos
30sin30cos
2
2
Nmgr
vmN
mgNr
vmFy
42
2
103.130cos30sin
30cos30sin
2. Use definition of friction force:
28.0101.3
3760
is minimalthus,
4
s
N
N
N
f
Nf
ss
s
A uniform solid cylinder has a radius R, mass M, and length L. Calculate its moment of inertia about its central axis
Example: Uniform Solid Cylinder
The volume density of the shell is V
M
so the mass is dm
I
dV dVV
M
dmr 2 )2(20
2
LrdrL
MrR
2
0
2
2
1
2MR
rM
R
drrMR
0
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