Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)1 General Physics I Part 2 (lecture 12-19)...
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Gneral Physics I, Lecture Not Gneral Physics I, Lecture Not e, Part 1 (Lecture 1-11) e, Part 1 (Lecture 1-11) 1 General Physics I Part 2 (lecture 12-19) Instructor Tamer A. Eleyan 2009/2010
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)1 General Physics I Part 2 (lecture 12-19) Instructor Tamer A. Eleyan 2009/2010
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)1 General
Physics I Part 2 (lecture 12-19) Instructor Tamer A. Eleyan
2009/2010
Slide 2
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)2 General
Physics 1, Lec 12, By/ T.A. Eleyan 2 Lecture 12 Energy and Energy
Transfer
Slide 3
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)3 General
Physics 1, Lec 12, By/ T.A. Eleyan 3 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. Work Done
by a Constant Force
Slide 4
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)4 General
Physics 1, Lec 12, By/ T.A. Eleyan 4 The SI unit of work is (Nm)
which is given the name (Joule). 1 Newtonmeter = 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
Slide 5
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)5 General
Physics 1, Lec 12, By/ T.A. Eleyan 5 No work is done on the bucket
to move horizontally (why?) Work is done in lifting the box
(why?)
Slide 6
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)6 General
Physics 1, Lec 12, By/ T.A. Eleyan 6 Negative 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: W total = F total d cos
Slide 7
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)7 General
Physics 1, Lec 12, By/ T.A. Eleyan 7 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 :
Slide 8
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)8 General
Physics 1, Lec 12, By/ T.A. Eleyan 8 Problem: A force F = (6i - 2j)
N acts on a particle that undergoes a displacement r = (3i +j) 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?
Slide 9
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)9 General
Physics 1, Lec 12, By/ T.A. Eleyan 9 The Normal Force The friction
Force
Slide 10
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)10 General
Physics 1, Lec 12, By/ T.A. Eleyan 10 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"
Slide 11
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)11 General
Physics 1, Lec 12, By/ T.A. Eleyan 11 Force - 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
Slide 12
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)12 General
Physics 1, Lec 12, By/ T.A. Eleyan 12 Work done by a varying
force
Slide 13
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)13 General
Physics 1, Lec 12, By/ T.A. Eleyan 13 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 (2m)(5N)=5J. Therefore, the
total work =25J
Slide 14
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)14 General
Physics 1, Lec 12, By/ T.A. Eleyan 14 Problem: An object is acted
on by the force shown in the Figure. What is the work from 0 to
1.00m
Slide 15
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)15 General
Physics 1, Lec 12, By/ T.A. Eleyan 15 Work Done by a Spring Where k
is a positive constant called (the spring constant)
Slide 16
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)16 General
Physics 1, Lec 12, By/ T.A. Eleyan 16 Kinetic Energy An object in
motion has kinetic energy: m = mass v = speed (magnitude of
velocity) The unit of kinetic energy is Joules (J). Kinetic energy
is a scalar (magnitude only) Kinetic energy is non-negative (zero
or positive )
Slide 17
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)17 General
Physics 1, Lec 12, By/ T.A. Eleyan 17 Work-Energy Theorem The net
(total) work done on an object by the total force acting on it is
equal to the change in the kinetic energy of the object: W total =
KE = KE final - KE initial
Slide 18
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)18 General
Physics 1, Lec 12, By/ T.A. Eleyan 18 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?
Slide 19
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)19 General
Physics 1, Lec 12, By/ T.A. Eleyan 19 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 = W f +W g : Solution
:
Slide 20
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)20 General
Physics 1, Lec 12, By/ T.A. Eleyan 20 Now use the Work-Energy
Theorem:
Slide 21
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)21 General
Physics 1, Lec 12, By/ T.A. Eleyan 21 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 force W net = K f -K i
Slide 22
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)22 General
Physics 1, Lec 12, By/ T.A. Eleyan 22 c)Braking force W net = F net
d = F Friction d F Friction = W net /d = ( )/(24) = 225 N b)Find
acceleration first v = v 0 +at 0 = v 0 + at, a= v 0 /t a = 12/4 = 3
m/s 2 x= x 0 +v 0 t+(1/2)at 2 x= 0 + (12)(4) + (1/2)(-3)(4) 2 x= 48
-24 = 24 m
Slide 23
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)23 General
Physics 1, Lec 12, By/ T.A. Eleyan 23 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.
Slide 24
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)24
Additional Questions (Lecture 12)
Slide 25
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)25 [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?
Slide 26
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)26 [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?
Slide 27
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)27 [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?
Slide 28
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)28 [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?
Slide 29
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)29 [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.
Slide 30
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)30 Lecture
13 Potential Energy
Slide 31
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)31 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,
Slide 32
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)32
Conservative Forces A force is conservative if the work it does on
an object moving between two points is independent of the path
taken. work done depends only on r i and r f
Slide 33
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)33 If an
object moves in a closed path (r i = r f ) then total work done by
the force is zero.
Slide 34
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)34
Nonconservative Forces non-conservative forces dissipate energy
work done by the force depends on the path
Slide 35
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)35 Work
Done by Conservative Forces Potential Energy: Energy associated
with the position of an object. For example: When you lift a ball a
distance y, gravity does negative work on the ball. This work can
be recovered as kinetic energy if we let the ball fall. The energy
that was stored in the ball is potential energy. W c = - U =- U
final U initial ] W c = work done by a conservative force U =
change in potential energy U = change in potential energy
Slide 36
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)36
Gravitational Potential Energy Gravitational potential energy U =
mg(y-y 0 ), where, y = height U=0 at y=y 0 (e.g. surface of earth).
Work done by gravity: W g = -mg y = -mg (y- y 0 )
Slide 37
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)37 Spring
Potential Energy U f U i = [Work done by spring on mass] Mass m
starts at x=0 (U i =0) and moves until spring is stretched to
position x. Work Spring = - kx 2 U(x) 0 = 1/2 kx 2 ) U Spring (x) =
kx 2 x = displacement from equilibrium position x F= kx Area in
triangle = kx times increment in x = Work done by spring
Slide 38
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)38
Conservation of Energy Energy is neither created nor destroyed
Energy is neither created nor destroyed The energy of an isolated
system of objects remains constant. The energy of an isolated
system of objects remains constant.
Slide 39
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)39
Mechanical energy E is the sum of the potential and kinetic
energies of an object. E = U + K E = U + K The total mechanical
energy in any isolated system of objects remains constant if the
objects interact only through conservative forces: E = constant E =
constant E f = E i U f + K f = U i + K i U + K = E = 0 Mechanical
Energy (Conservative Forces)
Slide 40
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)40
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.
(B) Neglecting air resistance, determine the speed of the ball when
it is at the ground.
Slide 41
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)41
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? Solution
Slide 42
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)42
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 particles speed at points B and C and (b) the net work done by
the gravitational force in moving the particle from A to C.
Slide 43
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)43 (a) the
particles speed at points B Find the particles speed at points C
then find the work from the relation
Slide 44
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)44 Work
Done by Nonconservative Forces Nonconservative forces change the
amount of mechanical energy in a system. W nc = work done by
nonconservative force
Slide 45
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)45
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. Since v i = 0, and h f = 0,
Slide 46
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)46 The
force done by friction is determined from;
Slide 47
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)47
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.
Slide 48
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)48 (B) If
a force of kinetic friction acts on the child, how much mechanical
energy does the system lose? Assume that v f =3.00 m/s and m = 20.0
kg.
Slide 49
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)49
Example: A block having a mass of 0.80kg is given an initial
velocity v A = 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.
Slide 50
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)50 (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 v A = 1.2 m/s, what is the
maximum compression x C in the spring?
Slide 51
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)51
Slide 52
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)52
Additional Questions (Lecture 13)
Slide 53
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)53 [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 cars 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 cars kinetic energy from the
bottom of the hill to the top of the hill. [2] A single
conservative force F x = (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.
Slide 54
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)54 [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?
Slide 55
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)55 [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,U k =
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?
Slide 56
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)56 [8] A
Hookes 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
Slide 57
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)57 [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.
Slide 58
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)58 [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 blocks
kinetic energy, (b)(b) the change in the potential energy of the
blockEarth system, (c) the friction force exerted on the block
(assumed to be constant). (d) What is the coefficient of kinetic
friction?
Slide 59
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)59 [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 then leaves 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 = 3x 3 y + 7x. Find the force that acts at
the point (x, y)
Slide 60
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)60 [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?
Slide 61
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)61 [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-kg block 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).
Slide 62
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)62 [20]
find the particles speed at points B, where the particle released
from point A and slides on the frictionless track [21] Find the
distance x.
Slide 63
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)63 [22] An
object of mass m starts from rest and slides a distance d down a
frictionless incline of angle. 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.
Slide 64
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)64 [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 = v
2 /r Which direction?
Slide 65
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)65 General
Physics 1, Lec14, By/ T.A. Eleyan 65 Lecture 14 Discussion
Slide 66
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)66 General
Physics 1, Lec14, By/ T.A. Eleyan 66 [1] A force F = (4x i + 3y j)
N acts on an object as the object moves in the x direction from the
origin to x = 5.00 m. Find the work done on the object by the
force. [2] A 3.00-kg object has a velocity (6.00 i - 2.00 j) 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 (8.00 i + 4.00
j) m/s.
Slide 67
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)67 General
Physics 1, Lec14, By/ T.A. Eleyan 67 [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 cars engine. The work done by friction is 3.31 10 5 J; the
work done by the engine is +6.34 10 5 J. Find the change in the
cars kinetic energy from the bottom of the hill to the top of the
hill.
Slide 68
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)68 General
Physics 1, Lec14, By/ T.A. Eleyan 68 [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 a) b)
Slide 69
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)69 General
Physics 1, Lec14, By/ T.A. Eleyan 69 [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? a) b)
Slide 70
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)70 General
Physics 1, Lec14, By/ T.A. Eleyan 70 [6] A single conservative
force F x = (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. c) b) a)
Slide 71
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)71 General
Physics 1, Lec14, By/ T.A. Eleyan 71 [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 Substitute and find v
Slide 72
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)72 General
Physics 1, Lec14, By/ T.A. Eleyan 72 [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 ?
Slide 73
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)73 General
Physics 1, Lec14, By/ T.A. Eleyan 73 a) b) c) d)
Slide 74
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)74 General
Physics 1, Lec14, By/ T.A. Eleyan 74 [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?
Slide 75
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)75 General
Physics 1, Lec14, By/ T.A. Eleyan 75 And find v And find s2
Slide 76
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)76 General
Physics 1, Lec14, By/ T.A. Eleyan 76 [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?
Slide 77
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)77 General
Physics 1, Lec14, By/ T.A. Eleyan 77 [11]The ambulance (mass
3000kg) shown in the Figure(2) slides (wheels locked) down a
frictionless incline that is 10 m long. It starts from rest at
point A, and continues along a rough surface until it comes to a
complete stop at point C. If the coefficient of kinetic friction
between the ambulance and the horizontal rough surface is 0.1. (a)
Calculate the speed of the ambulance at point B.(b) Compute the
distance d the ambulance slides on the horizontal rough surface
before stopping.
Slide 78
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)78 General
Physics I, Lec 15 By/ T.A. Eleyaan 78 Lecture 15 Linear Momentum
and Collisions
Slide 79
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)79 General
Physics I, Lec 15 By/ T.A. Eleyaan 79 Linear momentum is defined
as: p = mv 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 m/s [p] = kg m/s Linear
Momentum Since p is a vector, we can also consider the components
of momentum: p x = mv x, p y = mv y, p z = mv z Note: momentum is
large if m and/or v is large. (define large, meaning hard for you
to stop).
Slide 80
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)80 General
Physics I, Lec 15 By/ T.A. Eleyaan 80 Recall that Another way of
writing Newtons 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.
Slide 81
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)81 General
Physics I, Lec 15 By/ T.A. Eleyaan 81 Impulse of a Force Exercise:
Show that impulse and momentum have the same units.
Slide 82
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)82 General
Physics I, Lec 15 By/ T.A. Eleyaan 82 The impulse of a force during
a certain interval of time is the change in momentum that the force
produces Impulse & Momentum
Slide 83
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)83 General
Physics I, Lec 15 By/ T.A. Eleyaan 83 F-t The Graph F-t The Graph
Impulse is a vector quantity Impulse is a vector quantity The
magnitude of the impulse is equal to the area under the force-time
curve The magnitude of the impulse is equal to the area under the
force-time curve Dimensions of impulse are M L / T Dimensions of
impulse are M L / T Impulse is not a property of the particle, but
a measure of the change in momentum of the particle Impulse is not
a property of the particle, but a measure of the change in momentum
of the particle
Slide 84
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)84 General
Physics I, Lec 15 By/ T.A. Eleyaan 84 The impulse can also be found
by using the time averaged force The impulse can also be found by
using the time averaged force I = t I = t This would give the same
impulse as the time varying force does This would give the same
impulse as the time varying force does F-t The Graph F-t The
Graph
Slide 85
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)85 General
Physics I, Lec 15 By/ T.A. Eleyaan 85 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.
1) The impulse 2) The average force
Slide 86
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)86 General
Physics I, Lec 15 By/ T.A. Eleyaan 86 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 110 -2 s. The force acting on the
golf ball can be determined from
Slide 87
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)87 General
Physics I, Lec 15 By/ T.A. Eleyaan 87 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?
Slide 88
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)88 General
Physics I, Lec 15 By/ T.A. Eleyaan 88 Example: A 325-gm ball with a
speed v of 6.22 m/s strikes a wall at an angle of 33.0 0 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
Slide 89
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)89 General
Physics I, Lec 15 By/ T.A. Eleyaan 89 (a) (b) x y x+y
Slide 90
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)90 General
Physics I, Lec 15 By/ T.A. Eleyaan 90 Principle of Conservation of
Linear Momentum When the net external force acting on a system is
zero, the total linear momentum of the system remains constant.
Proof (Two-Body Systems) However,
Slide 91
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)91 General
Physics I, Lec 15 By/ T.A. Eleyaan 91 In the absence of a net
external force,
Slide 92
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)92 General
Physics I, Lec 15 By/ T.A. Eleyaan 92 Look at two billiard balls
colliding: Before collision: After collision: From the law of the
conservation of momentum. It states: The Total Momentum of an
isolated system of bodies remains constant.
Slide 93
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)93 General
Physics I, Lec 15 By/ T.A. Eleyaan 93 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:
Slide 94
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)94 General
Physics I, Lec 15 By/ T.A. Eleyaan 94 Along y-direction: So the
magnitude of the velocity of the third piece is; And the direction
of v 3 is given by
Slide 95
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)95 General
Physics I, Lec 15 By/ T.A. Eleyaan 95 Collisions In general, a
collision is an interaction in which two objects strike one another
two objects strike one another the net external impulse is zero or
negligibly small (momentum is conserved) the net external impulse
is zero or negligibly small (momentum is conserved)Examples: Two
billiard balls colliding on a billiards table An alpha particle
colliding with a heavy atom Two galaxies colliding
Slide 96
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)96 General
Physics I, Lec 15 By/ T.A. Eleyaan 96 Types of Collisions [1] In an
elastic collision, momentum and kinetic energy are conserved
Perfectly elastic collisions occur on a microscopic level In
macroscopic collisions, only approximately elastic collisions
actually occur [2] In an inelastic collision, kinetic energy is not
conserved although momentum is still conserved If the objects stick
together after the collision, it is a perfectly inelastic collision
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
Slide 97
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)97 General
Physics I, Lec 15 By/ T.A. Eleyaan 97 Elastic Collisions Both
momentum and kinetic energy are conserved Both momentum and kinetic
energy are conserved
Slide 98
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)98 General
Physics I, Lec 15 By/ T.A. Eleyaan 98 Perfectly Inelastic
Collisions Since the objects stick together, they share the same
velocity after the collision Since the objects stick together, they
share the same velocity after the collision
Slide 99
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)99 General
Physics I, Lec 15 By/ T.A. Eleyaan 99 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?
Slide 100
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)100
General Physics I, Lec 15 By/ T.A. Eleyaan 100 Solve (1)&(2) to
find velocities of the balls after the collision? We can apply
conservation of momentum and conservation of kinetic energy:
Slide 101
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)101
General Physics I, Lec 15 By/ T.A. Eleyaan 101 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
1/3 rd the original speed of the 3 kg sphere. What is the mass of
the second sphere?
Slide 102
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)102
General Physics I, Lec 15 By/ T.A. Eleyaan 102 we can apply
conservation of momentum:
Slide 103
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)103
General Physics I, Lec 16 By/ T.A. Eleyan 103 Lecture 16 Linear
Momentum and Collisions
Slide 104
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)104
General Physics I, Lec 16 By/ T.A. Eleyan 104 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) From the x axis component
Slide 105
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)105
General Physics I, Lec 16 By/ T.A. Eleyan 105 From the y axis
component From (1) & (2)
Slide 106
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)106
General Physics I, Lec 16 By/ T.A. Eleyan 106 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. Find a) 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 B
Slide 107
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)107
General Physics I, Lec 16 By/ T.A. Eleyan 107 Momentum is conserved
between A and B
Slide 108
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)108
General Physics I, Lec 16 By/ T.A. Eleyan 108 Example: Two metal
spheres, suspended by vertical cords, initially just touch, as
shown in Fig. Sphere 1, with mass m 1 =30 g, is pulled to the left
to height h 1 =8.0cm, and then released from rest. After swinging
down, it undergoes an elastic collision with sphere 2, whose mass m
2 =75 g. What is the velocity v 1f of sphere 1 just after the
collision?
Slide 109
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)109
General Physics I, Lec 16 By/ T.A. Eleyan 109 where
Slide 110
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)110
General Physics I, Lec 16 By/ T.A. Eleyan 110 Example: The figure
below shows an elastic collision of two pucks on a frictionless air
table. Puck A has mass m A = 0.500 kg, and puck B has mass m B =
0.300 kg. Puck A has an initial velocity of 4.00 ms in the positive
xdirection 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 V B2 of puck B and the angles and as illustrated in the
figure. Because the collision is elastic, the initial and final
kinetic energies are equal
Slide 111
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)111
General Physics I, Lec 16 By/ T.A. Eleyan 111 Conservation of the x
component of total momentum gives Conservation of the y component
of total momentum gives There are two simultaneous equations for
and Solve to find the angles
Slide 112
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)112
General Physics I, Lec 16 By/ T.A. Eleyan 112 Center 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.
Slide 113
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)113
General Physics I, Lec 16 By/ T.A. Eleyan 113 Center of mass point
objects 1 D
Slide 114
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)114
General Physics I, Lec 16 By/ T.A. Eleyan 114 Center of mass point
objects 2 D
Slide 115
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)115
General Physics I, Lec 16 By/ T.A. Eleyan 115 Center of mass solid
objects 1 D
Slide 116
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)116
General Physics I, Lec 16 By/ T.A. Eleyan 116 Center of mass solid
objects 2 D
Slide 117
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)117
General Physics I, Lec 16 By/ T.A. Eleyan 117 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.
Slide 118
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)118
General Physics I, Lec 16 By/ T.A. Eleyan 118 Problem: Problem:
Three particles of masses m A = 1.2 kg, m B = 2.5 kg, and m C = 3.4
kg form an equilateral triangle of edge length a = 140 cm. Where is
the center of mass 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:
Slide 119
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)119
General Physics I, Lec 16 By/ T.A. Eleyan 119 Motion of the Center
of Mass A system of objects behaves as if all its mass were located
at the center of mass. Velocity of CM: Acceleration of CM: MA cm =
F net,ext Newton's Laws for a System of Particles
Slide 120
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)120
General Physics I, Lec 16 By/ T.A. Eleyan 120 Problem : The three
particles in Fig. a are initially at rest. Each experiences an
external force due to bodies outside the three-particle system. The
directions are indicated, and the magnitudes are F A =6 N, F B =12
N, and F C =14 N. What is the magnitude of the acceleration of the
center of mass of the system, and in what direction does it move?
The three particles in Fig. a are initially at rest. Each
experiences an external force due to bodies outside the
three-particle system. The directions are indicated, and the
magnitudes are F A =6 N, F B =12 N, and F C =14 N. What is the
magnitude of the acceleration of the center of mass of the system,
and in what direction does it move?
Slide 121
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)121
Additional Questions (Lecture 15,16)
Slide 122
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)122 [1] A
3.00-kg particle has a velocity of (3.00i + 4.00j ) m/s. (a) Find
its x and y components of momentum. (b) Find the magnitude and
direction of its momentum. [2] An estimated forcetime 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..
Slide 123
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)123 [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?
Slide 124
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)124 [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..
Slide 125
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)125 [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?
Slide 126
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)126 [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 v 2i. 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 balls velocity after the collision.
Slide 127
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)127 [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 collision such 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 (2.00i - 3.00 j) m/s, and a 3.00-kg
particle has a velocity (1.00i +6.00 j) m/s. Find (a) the velocity
of the center of mass and (b) the total momentum of the
system.
Slide 128
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)128 [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 [16] 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 right at 3 m/s. Car A read-ends Car B and
they stick together. What is the velocity of the center of mass
after the collision? (Before the collision, the velocity of the CM
is 3.33 m/s)
Slide 129
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)129 [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.
Slide 130
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)130
Lecture 17 Discussion
Slide 131
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)131 [1] An
estimated forcetime 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..
Slide 132
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)132 [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..
Slide 133
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)133
Slide 134
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)134 [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?
Slide 135
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)135
Slide 136
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)136 [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.
Slide 137
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)137
Slide 138
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)138 The
formula for CM of a continuous object is 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. The mass of a small
segment
Slide 139
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)139
Therefore Find the CM when the density of the rod non-uniform but
varies linearly as a function of x, x
Slide 140
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)140
Lecture 18 Rotational Motion
Slide 141
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)141 For
circular motion, the distance (arc length) s, the radius r, and the
angle are related by: Note that is measured in radians: Angular
Position, > 0 for counterclockwise rotation from reference
line
Slide 142
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)142 The
angular displacement is defined as the angle the object rotates
through during some time interval The angular displacement is
defined as the angle the object rotates through during some time
interval Every point on the disc undergoes the same angular
displacement in any given time interval Every point on the disc
undergoes the same angular displacement in any given time
interval
Slide 143
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)143 Notice
that as the disk rotates, changes. We define the angular
displacement, , as: = f - i which leads to the average angular
speed av Angular Velocity,
Slide 144
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)144
Instantaneous Angular Velocity As usual, we can define the
instantaneous angular velocity as: 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
Slide 145
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)145 The
period of rotation is the time it takes to complete one revolution.
Problem: What is the period of the Earths rotation about its own
axis? What is the angular velocity of the Earths rotation about its
own axis?
Slide 146
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)146 We can
also define the average angular acceleration av : And instantaneous
angular acceleration Angular Acceleration, a The SI units of are:
rad/s 2 = 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
Slide 147
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)147 Notes
about angular kinematics: When a rigid object rotates about a fixed
axis, every portion of the object has the same angular speed and
the same angular acceleration When a rigid object rotates about a
fixed axis, every portion of the object has the same angular speed
and the same angular acceleration i.e. , and are not dependent upon
r, distance form hub or axis of rotation
Slide 148
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)148
Examples: 1. Bicycle wheel turns 240 revolutions/min. What is its
angular velocity in radians/second? 2. If wheel slows down
uniformly to rest in 5 seconds, what is the angular
acceleration?
Slide 149
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)149 3. How
many revolution does it turn in those 5 sec? Recall that for linear
motion we had: Perhaps something similar for angular
quantities?
Slide 150
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)150
Analogies Between Linear and Rotational Motion Rotational Motion
About a Fixed Axis with Constant Acceleration Linear Motion with
Constant Acceleration
Slide 151
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)151
Relationship Between Angular and Linear Quantities Displacements
Displacements Speeds Speeds Accelerations Accelerations
Slide 152
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)152 Total
linear acceleration is Since the tangential speed v is Tangential
and the radial acceleration. The magnitude of tangential
acceleration a t is The radial or centripetal acceleration a r
is
Slide 153
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)153
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? Since the angular speed is constant, there is no
angular acceleration. Tangential acceleration is Radial
acceleration is
Slide 154
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)154
Calculation of Moments of Inertia Moments of inertia for large
objects can be computed, if we assume that the object consists of
small volume elements with mass, mi.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 dm in
the above equation with dV. The moment of inertia for the large
rigid object is: The moments of inertia becomes
Slide 155
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)155 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 R O dm The moment of inertia is The moment of inertia
for this object is the same as that of a point of mass M at the
distance R.
Slide 156
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)156
Example for Rigid Body Moment of Inertia Calculate 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 so the mass is
Slide 157
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)157 Torque
: The ability of a force to rotate a body 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. = r F = (rsin
)F
Slide 158
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)158 The
net torque about a point O is the sum of all torques about O:
Slide 159
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)159
Problem: 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
Slide 160
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)160 Torque
& Angular Acceleration Lets consider a point object with mass m
rotating on a circle. The tangential force Ft Ft is The torque due
to tangential force Ft Ft is Torque acting on a particle is
proportional to the angular acceleration. Analogs to Newtons 2 nd
law of motion in rotation.
Slide 161
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)161 How
about a rigid object? The external tangential force dFt dFt is The
torque due to tangential force Ft Ft is The total torque is
Slide 162
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)162
Example : 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?
Slide 163
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)163 The
only force generating torque is the gravitational force MgMg Using
the relationship between tangential and angular acceleration Since
the moment of inertia of the rod when it rotates about one end We
obtain The tip of the rod falls faster than an object undergoing a
free fall.
Slide 164
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)164
Rotational Kinetic Energy Kinetic 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 m i, moving at a
tangential speed, v i, is The above expression is simplified
to
Slide 165
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)165
Example: 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 MM l l m m b b O
Slide 166
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)166 Since
the rotation is about y axis, the moment of inertia about y axis, I
y, is 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.
Slide 167
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)167
Lecture 19 Discussion
Slide 168
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)168 [1]
During a certain period of time, the angular position of a swinging
door is described by = 5.00 + 10.0t + 2.00t 2, where 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.
Slide 169
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)169 [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.
Slide 170
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)170
Consider a car driving at 20 m/s on a level circular turn of radius
40.0 m. Assume the cars mass is 1000 kg. 1. What is the magnitude
of frictional force experienced by cars tires? 2. What is the
minimum coefficient of friction in order for the car to safely
negotiate the turn?
Slide 171
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)171 1.
Draw a free body diagram, introduce coordinate frame and consider
vertical and horizontal projections 2. Use definition of friction
force:
Slide 172
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)172
Consider a car driving at 20 m/s on a 30 banked circular curve of
radius 40.0 m. Assume the cars mass is 1000 kg. 1. What is the
magnitude of frictional force experienced by cars tires? 2. What is
the minimum coefficient of friction in order for the car to safely
negotiate the turn?
Slide 173
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)173 1.
Draw a free body diagram, introduce coordinate frame and consider
vertical and horizontal projections 2. Use definition of friction
force:
Slide 174
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)174 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 so the
mass is
Slide 175
Gneral Physics I, Lecture Note, Part 1 (Lecture 1-11)175