PROGRAM OF “PHYSICS”webdirectory.hcmiu.edu.vn/Portals/25/Docs/.../PHYS1-CH1-KINEMATICS_New.pdf · Test 2 The graph shows position as a function of time for two trains running

PROGRAM OF “PHYSICS”

Lecturer: Dr. DO Xuan Hoi

Room A1. 503

E-mail : [email protected]

mailto:[email protected]

PHYSICS I (General Mechanics)

02 credits (30 periods)

Chapter 1 Bases of Kinematics

Motion in One Dimension

Motion in Two Dimensions

Chapter 2 The Laws of Motion

Chapter 3 Work and Mechanical Energy

Chapter 4 Linear Momentum and Collisions

Chapter 5 Rotation of a Rigid Object About a Fixed Axis

Chapter 6 Static Equilibrium

Chapter 7 Universal Gravitation

References :

Halliday D., Resnick R. and Walker, J. (2005), Fundamentals of Physics, Extended seventh edition. John Willey and Sons, Inc.

Alonso M. and Finn E.J. (1992). Physics, Addison-Wesley Publishing Company

Hecht, E. (2000). Physics. Calculus, Second Edition. Brooks/Cole.

Faughn/Serway (2006), Serway’s College Physics, Brooks/Cole.

Roger Muncaster (1994), A-Level Physics, Stanley Thornes.

http://ocw.mit.edu/OcwWeb/Physics/index.htm

http://www.opensourcephysics.org/index.html

http://hyperphysics.phy-

astr.gsu.edu/hbase/HFrame.html

http://www.practicalphysics.org/go/Default.ht

ml

http://www.msm.cam.ac.uk/

http://www.iop.org/index.html . . .

http://www.opensourcephysics.org/index.html

http://hyperphysics.phy-astr.gsu.edu/hbase/HFrame.html



http://www.practicalphysics.org/go/Default.html

http://www.practicalphysics.org/go/Default.html

http://www.msm.cam.ac.uk/

http://www.iop.org/index.html

PHYSICS I Chapter 1 Bases of Kinematics

A. Motion in One Dimension 1. Position, Velocity, and Acceleration

2. One-Dimensional Motion with Constant Acceleration

3. Freely Falling Objects

B. Motion in Two Dimensions 4. The Position, Velocity, and Acceleration Vectors

5. Two-Dimensional Motion with Constant Acceleration.

Projectile Motion

6. Circular Motion. Tangential and Radial Acceleration

7. Relative Velocity and Relative Acceleration

Dynamics

►The branch of physics involving the motion of an object and the relationship between that motion and other physics concepts

►Kinematics is a part of dynamics In kinematics, you are interested in the

description of motion

Not concerned with the cause of the motion

1. 1 Position and Displacement

O

M

x

1. 1 Position and Displacement

► Position is defined in terms of a frame of reference

A

B y’

x’ O’

xi’ xf’

Frame A: xi > 0 ; xf > 0

Frame B: x’I < 0 ; x’f > 0

►One dimensional, so generally the x- or y-axis

Position and Displacement

► Position is defined in terms of a frame of reference One dimensional, so

generally the x- or y-axis

►Displacement measures the change in position Represented as x (if

horizontal) or y (if vertical)

Vector quantity (i.e. needs directional information) ►+ or - is generally sufficient to

indicate direction for one-dimensional motion

Units

SI Meters (m)

CGS Centimeters (cm)

US Cust Feet (ft)

Displacement Displacement measures the change in position

represented as x or y

m

mm

xxx if

70

1080

1

m

mm

xxx if

60

8020

2

Distance or Displacement?

►Distance may be, but is not necessarily, the magnitude of the displacement

Distance (blue line)

Displacement (yellow line)

Position-time graphs

Note: position-time graph is not necessarily a straight line, even though the motion is along x-direction

Test 1

An object (say, car) goes from one point in space to another. After it arrives to its destination, its displacement is 1. either greater than or equal to 2. always greater than 3. always equal to 4. either smaller or equal to 5. either smaller or larger

than the distance it traveled.

a. Average Velocity

► It takes time for an object to undergo a displacement

► The average velocity is rate at which the displacement occurs

► Direction will be the same as the direction of the displacement (t is always positive)

A. Motion in One Dimension

1. 2 Velocity

f iaverage

x xxv

t t

More About Average Velocity

►Units of velocity:

►Note: other units may be given in a problem,

but generally will need to be converted to these

Units

SI Meters per second (m/s)

CGS Centimeters per second (cm/s)

US Customary Feet per second (ft/s)

Example: Suppose that in both cases truck

covers the distance in 10 seconds:

11

70

10

7

average

x mv

t s

m s

22

60

10

6

average

x mv

t s

m s

Speed

►Speed is a scalar quantity (no information about sign/direction is need)

same units as velocity

Average speed = total distance / total time

►Speed is the magnitude of the velocity

Graphical Interpretation of Average Velocity

►Velocity can be determined from a position-time graph

►Average velocity equals the slope of the line joining the initial and final positions

40

3.0

13

average

x mv

t s

m s

b. Instantaneous Velocity

► Instantaneous velocity is defined as the limit of the average velocity as the time interval becomes infinitesimally short, or as the time interval approaches zero

►The instantaneous velocity indicates what is happening at every point of time

0 0lim lim i

instt t

x xxv v

t t

0 0lim lim i

instt t

x xxv v

t t

0 0lim lim i

inst t t

x xxv v

t t

'tdx

v xdt

Instantaneous Velocity

;dx vdt

0

0 ;t

t

x x vdt

The instantaneous velocity equals the first

derivative of the position with respect to time

'tdx

v xdt

0 0

;x t

x t

dx vdt 0

0

;t

x

xt

x vdt

0

0

t

t

x x vdt

Uniform Velocity

►Uniform velocity is constant velocity :

v = const

►The instantaneous velocities are always the same

All the instantaneous velocities will also equal the average velocity

Uniform Velocity

►Uniform velocity is constant velocity :

v = const

0

0;

t

t

x x vdt 0

0;

t

t

x x v dt

0 0

( )x x v t t

0

0t 0

x x vt

Graphical Interpretation of Instantaneous Velocity

► Instantaneous velocity is the slope of the tangent to the curve at the time of interest

►The instantaneous speed is the magnitude of the instantaneous velocity

Average vs Instantaneous Velocity

Average velocity Instantaneous velocity

Test 2 The graph shows position as a function of time for two trains running on parallel tracks. Which of the following is true:

A B

time

position

tB

1. at time tB both trains have the same velocity 2. both trains speed up all the time 3. both trains have the same velocity at some time before tB 4. train A is longer than train B 5. all of the above statements are true

Note: the slope of curve B is parallel to line A at some point t< tB

a. Average Acceleration

►Changing velocity (non-uniform) means an acceleration is present

►Average acceleration is the rate of change of the velocity

►Average acceleration is a vector quantity (i.e. described by both magnitude and direction)

► A. Motion in One Dimension

► 1. 3 Acceleration

f iaverage

v vva

t t

Average Acceleration

►When the sign of the velocity and the acceleration are the same (either positive or negative), then the speed is increasing

►When the sign of the velocity and the acceleration are opposite, the speed is decreasing

Units

SI Meters per second squared (m/s2)

CGS Centimeters per second squared (cm/s2)

US Customary Feet per second squared (ft/s2)

b. Instantaneous Acceleration

►Instantaneous acceleration is the limit of the average acceleration as the time interval goes to zero

The instantaneous acceleration equals the first

derivative of the velocity and the second

derivative of the position with respect to time

0 0lim lim f i

inst t t

v vva a

t t

' '' ;t ta v x 2

2

dv d xa

dt dt

2. Uniform Acceleration

►When the instantaneous accelerations are always the same, the acceleration will be uniform

The instantaneous accelerations will all be equal to the average acceleration

' '' ;t ta v x 2

2

dv d xa

dt dt

Uniform Acceleration

a = const

WHAT ARE THE FORMULAE FOR V AND X ?

' '' ;t ta v x 2

2

dv d xa

dt dt


a = const

' '' ;t ta v x 2

2

dv d xa

dt dt

0

0;

t

t

v v a dt

0 0

( )v v a t t

0

0;

t

t

v v adt

0 0

0 0 0 0( )

t t

t t

x x vdt x v a t t dt


a = const

0

0 0 0( )

t

t

x x v a t t dt

0 0

0 0 0( )

t t

t t

x v dt a t t dt

2

0 0 0 0( ) ( )

2

ax v t t t t

0

0t 2

0 02

ax x v t t


a = const

WHAT IS THE FORMULA BETWEEN V AND X INDEPENDENT OF t ?


a = const

a = const

;dx

vdv adtdt

;dv adt ;vdv adx

0 0

;v x

v x

vdv adx 0

220

2 2

x

x

vvadx

0

220

2 2

x

x

vva dx

2 2

0 02 ( )v v a x x

Graphical Interpretation of Acceleration

► Average acceleration is the slope of the line connecting the initial and final velocities on a velocity-time graph

► Instantaneous acceleration is the slope of the tangent to the curve of the velocity-time graph

Example 1: Motion Diagrams

►Uniform velocity (shown by red arrows maintaining the same size)

►Acceleration equals zero

Example 2:

► Velocity and acceleration are in the same direction

► Acceleration is uniform (blue arrows maintain the same length)

► Velocity is increasing (red arrows are getting longer)

Example 3:

► Acceleration and velocity are in opposite directions

► Acceleration is uniform (blue arrows maintain the same length)

► Velocity is decreasing (red arrows are getting shorter)

One-dimensional Motion With Constant Acceleration

► If acceleration is uniform

thus:

Shows velocity as a function of acceleration and time

ov v at

'a v

One-dimensional Motion With Constant Acceleration

►Used in situations with uniform acceleration

ov v at

'x v

21

2o ox x v t at

2 2 2ov v a x

Summary of kinematic equations

EXAMPLE 1 The velocity of a particle moving along the x

axis varies in time according to the expression

, where t is in seconds.

(a) Find the average acceleration in the time

interval t = 0 to t = 2.0 s.

(a) 2(40 5 0 ) / 40 /Av m s m s

2(40 5 2 ) / 20 /Bv m s m s

220 / 40 /10 /

2,0 0B A

average

B A

v v m s m sa m s

t t

2(40 5 ) /v t m s

(b)

EXAMPLE 1

2(40 5 ) / v t m s

The velocity of a particle moving along the x

axis varies in time according to the expression

, where t is in seconds.

(b) Determine the acceleration at t = 2.0 s.

2(40 5 ) /v t m s

2' ( 10 ) /a v t m s

2 2( 10 2.0 ) / 20 /s m s m s

3. Free Fall

►All objects moving under the influence of only gravity are said to be in free fall

►All objects falling near the earth’s surface fall with a constant acceleration

►This acceleration is called the acceleration due to gravity, and indicated by g

Acceleration due to Gravity

► Symbolized by g

► g = 9.8 m/s² (can use g = 10 m/s² for estimates)

► g is always directed downward

toward the center of the earth

Free Fall -- an Object Dropped

► Initial velocity is zero

► Frame: let up be positive

►Use the kinematic equations

Generally use y instead

of x since vertical

vo = 0

y

x

a = g

2

2

10

2

9.8

y y at

a m s

Free Fall -- an Object Thrown Downward

► a = g With upward being positive,

acceleration will be negative, g = -9.8 m/s²

► Initial velocity vo 0 With upward being positive,

initial velocity will be negative

vo 0

y

x

a = g

v > vo

2

0

2

10

2

9.8

y y y v at

a m s

Free Fall -- object thrown upward

► Initial velocity is upward, so positive

►The instantaneous velocity at the maximum height is zero

► a = g everywhere in the motion g is always downward,

negative

v = 0

vo 0

a = g

Thrown upward

►The motion may be symmetrical

then tup = tdown

then vf = -vo

►The motion may not be symmetrical

Break the motion into various parts

►generally up and down

Non-symmetrical Free Fall

►Need to divide the motion into segments

► Possibilities include

Upward and downward portions

The symmetrical portion back to the release point and then the non-symmetrical portion

Combination Motions

Test 3 A person standing at the edge of a cliff throws one ball straight up and another ball straight down at the same initial speed. Neglecting air resistance, the ball to hit ground below the cliff with greater speed is the one initially thrown 1. upward 2. downward 3. neither – they both hit at the same speed

Note: upon the descent, the velocity of an object thrown straight up with an initial velocity v is exactly –v when it passes the point at which it was first released.

EXAMPLE 2 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 as the time the stone leaves the

thrower’s hand at position , determine

(a) the time at which the stone reaches

its maximum height

(a)

2

0+ = 20.0 m/s + (- 9.80 m/s ) = 0v v at t

2

20.0 /2.04

9.80 /

m st s

m s







(b) the maximum height

(b)

2 2

1y = +

2

1(20.0 m/s) (2.04 s) + (-9.80 m/s ) (2.04 s)

2

2B 0v t at

=

20.4By m







(b) the maximum height

Other method : 2 2

0 02 ( )v v a y y

2 2

00 (20.0) 2( 9.80)( )y y

0

20.4y y m







(c) the time at which the stone returns

to the height from which it was thrown

(c) 2

0 0

10

2Cy y v t at

0t

0

2

2 2 20.0 /4.08

9.80 /

v m st s

a m s







(d) the velocity of the stone at the stone

returns to the height from which it was thrown

(d)

C 0

2

= +

= 20.0 m/s + (-9.80 m/s ) (4.08 s)

v v at

20.0 /m s







(d) the velocity of the stone at the stone

returns to the height from which it was thrown

Other method : 2 2

0 02 ( )v v a y y

2 2

0 0 02 ( ) 0v v a y y

0

20.0 /v v m s







(e) the velocity and position of the stone at

t = 5.00 s

(e)

0

2

= +

= 20.0 m/s + (-9.80 m/s ) (5.00 s)

v v at

29.0 /m s







(e) the velocity and position of the stone at

t = 5.00 s

(e)

22.5y m

2 2

1y = +

2

1(20.0 m/s) (5.00 s) + (-9.80 m/s ) (5.00 s)

2

22.5

2B 0v t at

=

m

PROBLEM 1

A person walks first at a constant speed of 5.00 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.00 m/s. What are (a) her average speed over the entire trip and (b) her average velocity over the entire trip?

SOLUTION

From A to B : From B to A :

(a)

Average speed over the entire trip :

1

1

ABv

t

2

2

ABv

t

1 2

2 2AB ABv

t t t

1 2 1 2

2 2

/ / 1 / 1 /

AB

AB v AB v v v

2

1 / 5.00 / 1 / 3.00 /m s m s1.88 /m s

PROBLEM 1

A person walks first at a constant speed of 5.00 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.00 m/s. What are (a) her average speed over the entire trip and (b) her average velocity over the entire trip?

SOLUTION (b)

Average velocity over the entire trip :

xv

t

00

t

PROBLEM 2

A car starts from rest and accelerates at 0.500 m/s2

while moving down an inclined plane 9.00 m long. When it reaches the bottom, the car rolls up another plane, where, after moving 15.0 m, it comes to rest. (a) What is the speed of the car at the bottom of the first plane ?

SOLUTION

(a)

2 2

0 02 ( )v v a x x

2 2 20 2 0.5 9.00/v m s m

3.00 /v m s

PROBLEM 2

A car starts from rest and accelerates at 0.500 m/s2 while moving down an inclined plane 9.00 m long. When it reaches the bottom, the car rolls up another plane, where, after moving 15.0 m, it comes to rest. (b) How long does it take to roll down the first plane?

SOLUTION (b)

0 = + v v at

0

v vt

a

2

3.00 / 0

0.05 /

m s

m s 60.0 s

PROBLEM 2

A car starts from rest and accelerates at 0.500 m/s2 while moving down an inclined plane 9.00 m long. When it reaches the bottom, the car rolls up another plane, where, after moving 15.0 m, it comes to rest. (c) What is the acceleration along the second plane?

SOLUTION (c)

Velocity is uniformly decreasing

2 2

0' 2 '( ' ' )v v a x x

2 2

0

''

2( ' ' )

v va

x x

2 20 (3.00 / )

2 15.0

m s

m

20.300 /m s

PROBLEM 2

A car starts from rest and accelerates at 0.500 m/s2 while moving down an inclined plane 9.00 m long. When it reaches the bottom, the car rolls up another plane, where, after moving 15.0 m, it comes to rest. (d) What is the car’s speed 8.00 m along the second plane?

SOLUTION (d)

2 2

1 1 02 '( ' )v v a x x

2 2 2

1(3.00 / ) 2 ( 0.300 / ) 8.00v m s m s m

1

2.05 /v m s


B. Motion in Two Dimensions

4. The Position, Velocity, and Acceleration Vectors

5. Two-Dimensional Motion with Constant Acceleration. Projectile Motion





4. The Position, Velocity, and Acceleration Vectors

4.1 Displacement

►The position of an object is described by its position vector:

x x

y

M

O i

j

r

r xi yj

4.1 Displacement

►The position of an object is described by its position vector

►The displacement of the object is defined as the change in its position

f ir r r

r

4.2 Velocity

The instantaneous velocity is the limit of the average velocity as Δt approaches zero The direction of the instantaneous velocity is along a

line that is tangent to the path of the particle and in the direction of motion

v

0limt

r d rv

t dt

d

xi yjdt

dx dy

i jdt dt

X Yv v i v j

4.3 Acceleration

The instantaneous acceleration is the limit of

the average acceleration as Δt approaches

zero

2 2

2 2

d x d yi j

dt dt

0limt

v dva

t dt

d dx dyi j

dt dt dt

X Ydv dvi j

dt dt

X Ya a i a j




5.1 Two-Dimensional Motion with Constant Acceleration

a const

Vector acceleration is a const :

;X Ya a i a j const ;X Ya const a const

;X YX Y

dv dva a

dt dt

0 0

;X X X Y Y Yv a t v v a t v

0 0

( ) ( )X X Y Yv a t v i a t v j

0 0

( ) ( )X Y X Yv a i a j t v i v j

0

v at v

We can demonstrate :

( To compare with motion in one dimension :

)

2

0 02

ar r v t t

21

2o ox x v t at




5.2 Projectile Motion

5.1 Two-Dimensional Motion with Constant Acceleration

a g const


a g const

Rules of Projectile Motion

► Introduce coordinate frame: y is up

► The x- and y-components of motion can be treated independently

► Velocities (incl. initial velocity) can be broken down into its x- and y-components

► The x-direction is uniform motion

ax = 0

► The y-direction is free fall

|ay|= g

►x-direction

ax = 0

x = vxot

►This is the only operative equation in the x-direction since there is uniform velocity in that direction

cos constantxo o o xv v v

►y-direction

take the positive direction as upward

then: free fall problem

►only then: ay = -g (in general, |ay|= g)

uniformly accelerated motion, so the motion equations all hold

ooy o sinvv sinyo o ov v

sinyo o ov v

On Ox and Oy :

On Ox and Oy :

0r

Rules of Projectile Motion

0 0

v at v gt v

0 0

0X X Xv t v v

0Y Yv gt v

2 2

0 0 0 02 2

a gr r v t t r v t t

0 0Xx x v t

2

0 0

1

2Yy y v t gt

►The velocity of the projectile at any point of its motion is the vector sum of its x and y components at that point

Velocity of the Projectile

2 2 1tany

x y

x

vv v v and

v


a g constOn Ox and Oy :

0 0 0X Xx x v t v t

2 2

0 0 0

1 1

2 2Y Yy y v t gt v t gt

2 2002 2

0 0 0 0

1(tan )

2 ( ) 2( cos )Y

X X

v g gx x x x

v v v

2

0 2

0 0

(tan )2( cos )

gy x x

v

trajectory: parabola

The horizontal range R:

R

2

0 2

0 0

(tan ) 02( cos )

gy x x

v

0x

2

00

sin2v

x Rg

R max: 0 = 45o


a g const

Maximum height H:

H

2

0 2

0 0

'' (tan )

2( cos )

gy x x

v

2

00

sin22 2

vRx

g

0 2

0 0

'tan 0 ;

( cos )

gx

v

2

0 2

0 0

(tan )2( cos )

gy x x

v

22 2

0 00 0 02

0 0

(tan ) sin2 sin22 22( cos )

v vgH

g gv

2

0 0( sin )

2

vH

g

Free fall:

2 2 2 2

0 0 0 0( sin ) 0 ( sin ) 2v v v gH

►An object may be fired horizontally

►The initial velocity is all in the x-direction

vo = vx and vy = 0

►All the general rules of projectile motion apply

Examples of Projectile Motion:

Non-Symmetrical Projectile Motion

► Follow the general rules for projectile motion

►Break the y-direction into parts up and down

symmetrical back to initial height and then the rest of the height

Test 4

Consider the situation depicted here. A gun is accurately aimed at a dangerous criminal hanging from the gutter of a building. The target is well within the gun’s range, but the instant the gun is fired and the bullet moves with a speed vo, the criminal lets go and drops to the ground. What happens? The bullet

1. hits the criminal regardless of the value of vo.

2. hits the criminal only if vo is large enough.

3. misses the criminal.

Test 4

Consider the situation depicted here. A gun is accurately aimed at a dangerous criminal hanging from the gutter of a building. The target is well within the gun’s range, but the instant the gun is fired and the bullet moves with a speed vo, the criminal lets go and drops to the ground. What happens? The bullet

1. hits the criminal regardless of the value of vo.

2. hits the criminal only if vo is large enough.

3. misses the criminal.

Note: The downward acceleration of the bullet and the criminal are identical,

so the bullet will hit the target – they both “fall” the same distance!

EXAMPLE 3 A long-jumper leaves the ground at an angle of 20.0°

above the horizontal and at a speed of 11.0 m/s.

(a) How far does he jump in the horizontal direction ?

(Assume his motion is equivalent to that of a particle.)

(a)

At the top:

cosxo o ov v

0 0

=(x 0x v t v cos )t

siny o ov v gt

sin 0y o ov v gt





At the top:

sin ' 0y o ov v gt

sin' o ov

tg

0

2

(11.0 / ) sin20 00.384

9.80 /

m ss

m s

EXAMPLE 3

At the ground:

2 ' 2 0.384 0.768t t s s

A long-jumper leaves the ground at an angle of 20.0°




0 0 0

0

( cos )

11.0 / cos 22 0 0.768 7.94

xx v t v t

m s s m



(b) What is the maximum height reached ?

(b) At the top:

21( sin )

2MAX o oy v t gt

0

2 2

(11.0 / ) sin20.0 0.384

19.80 / (0.384 ) 0.722

2

m s s

m s s m

PROBLEM 3

A stone is thrown from the top of a building upward at an angle of 30.0° to the horizontal and with an initial speed of 20.0 m/s, as shown in the figure. If the height of the building is 45.0 m, (a) how long is it before the stone hits the ground?

SOLUTION (a)

0

0 0sin 20 / sin30 10.0 /Yv v m s m s

0

0 0cos 20 / cos30 17.3 /Xv v m s m s

2

0 0

1

2Yy y v t gt

2 2145.0 0 (10.0 / ) (9.80 / ) ;

2m m s t m s t 4.22t s

PROBLEM 3

A stone is thrown from the top of a building upward at an angle of 30.0° to the horizontal and with an initial speed of 20.0 m/s, as shown in the figure. If the height of the building is 45.0 m, (b) What is the speed of the stone just before it strikes the ground?

SOLUTION (b)

0

17.3 /Xv m s 0

10.0 /Yv m s

0

17.3 /X Xv v m s

2

09.80 / 4.22 10.0 / 31.4 /Y Yv gt v m s s m s m s

2 2 2 2(17.3 / ) ( 31.4 / ) 35.9 /x yv v v m s m s m s

PROBLEM 4

An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers, as shown in the figure. (a) If the plane is traveling horizontally at 40.0 m/s and is 100 m above the ground, where does the package strike the ground relative to the point at which it was released?

SOLUTION (a) x

y

SOLUTION

(40.0 / )x vt m s t

2 2 21( 4.90 / )

2y gt m s t

2 2100 ( 4.90 / ) ;m m s t 4.52t s

(40.0 / ) 4.52 181x m s s m

PROBLEM 4

An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers, as shown in the figure. (b) What are the horizontal and vertical components of the velocity of the package just before it hits the ground?

SOLUTION (b) x

y

SOLUTION

40.0 /Xv m s

2

09.80 / 4.52 0

44.3 /

Y Yv gt v m s s

m s

2 2 2 2(40.0 / ) ( 44.3 / )

59.9 /

x yv v v m s m s

m s

PROBLEM 4

An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers, as shown in the figure. (c) Where is the plane when the package hits the ground?

SOLUTION (c) x

y

SOLUTION

The plane is directly over the package.

(40.0 / ) ;x vt m s t 4.52t s

(40.0 / ) 4.52 59.9 /x vt m s s m s


A. Motion in One Dimension

B. Motion in Two Dimensions 4. The Position, Velocity, and Acceleration Vectors

5. Two-Dimensional Motion with Constant Acceleration.

Projectile Motion


6. Circular Motion.

Tangential and Radial Acceleration

6.1 Review of uniform circular motion

v v

R

Magnitude :

aR

22

R

va R const

R

Acceleration vector : perpendicular to the path and

always points toward the center of the circle

centripetal (center-seeking) acceleration

Magnitude of velocity does not change

6. Circular Motion.

Tangential and Radial Acceleration

6.2 Motion along a curved path

Velocity vector : changes both in magnitude and

in direction at every point

Acceleration vector two component vectors:

a radial component vector aR

and a tangential component vector aT

R Ta a a

v v

v

v

v

The tangential acceleration causes the change in the

speed of the particle: parallel to the instantaneous velocity,

magnitude :

T

d va

dt

The radial acceleration arises from the change in direction of the velocity vector : perpendicular to the path, magnitude : 2

R

va

R

(R : radius of curvature of the path at the point)

T

d va

dt

The magnitude of the acceleration vector :

2

R

va

R

2 2

T Ra a a

Uniform circular motion (v is constant ) : aT = 0 the acceleration is always completely radial

Unit vectors

r is a unit vector lying along the radius vector and directed radially outward from the center of the circle is a unit vector tangent to the circle in the direction of increasing

Both and move along with the particle r

2

ˆ ˆR T

d v va a a r

dt R

r

EXAMPLE 4 A ball tied to the end of a string 0.50 m in length swings in

A vertical circle under the influence of gravity, as shown in

figure. When the string makes an angle 20° with the

vertical, the ball has a speed of 1.5 m/s.

(a) Find the magnitude of the radial component of

acceleration at this instant.

(a)

2

R

va

R

2

2(1.8 / )4.5 /

0.50

m sm s

m


a vertical circle under the influence of gravity, as shown in



(b) What is the magnitude of the tangential acceleration

when = 20°?

(b)

ma mg T

sin 0Tma mg

sinTa g

2 0 2(9.8 / ) sin20 3.4 /m s m s


a vertical circle under the influence of gravity, as shown in



(c) Find the magnitude and direction of the total

acceleration at = 20°.

(c)

2 2 24.5 3.4 /m s

2 2

T Ra a a

25.6 /m s

1tan T

R

a

a 037 1 3.4

tan4.5

PROBLEM 5

The figure shows the total acceleration and velocity of a particle moving clockwise in a circle of radius 2.50 at a given instant of time. At this instant, find (a) the radial acceleration

SOLUTION (a)

(b) the speed of the particle

2

;R

va

R(b) Rv Ra

(c) its tangential acceleration

(c)

cosRa a

2 0 215.0 / cos 30 2.3 /m s m s

sinTa a

2 015.0 / sin30m s 27.50 /m s

TaRa

2.5 2.3 2.4 /m s

PHYSICS I ► Chapter 1 Bases of Kinematics

► A. Motion in One Dimension ► 1. Position, Velocity, and Acceleration

► 2. One-Dimensional Motion with Constant Acceleration

► 3. Freely Falling Objects

► B. Motion in Two Dimensions ► 4. The Position, Velocity, and Acceleration Vectors

► 5. Two-Dimensional Motion with Constant Acceleration.

Projectile Motion

► 6. Circular Motion. Tangential and Radial Acceleration

► 7. Relative Velocity and Relative Acceleration


7.1 Relative velocity

Consider a particle at point A

A

O x

y

r

0v

An observer in reference frame S

An observer in reference frame S’ moving to the right relative to S with a constant velocity v0 .

'r

O’ x’

y’

' ' ;OA OO O A

0'OO v t

' ' ;r r OO

0'r r v t

Differentiate with respect to time : 0

'd r d rv

dt dt

0'v v v

A

O x

y

r

0v

'r

O’ x’

y’

0'r r v t

0'v v vGalilean transformation equations

The woman standing on the beltway sees the walking man pass by at a slower speed than the woman standing on the stationary floor does

PRACITCE : 0'v v v

' 'AO AO O Ov v v

AOv

A O’

O

'AOv

'O Ov

Transformation for acceleration

The acceleration of the particle measured by an observer in the frame of reference S is the same as that measured by any other observer moving with constant velocity relative to the frame S.

0' d vd v d v

dt dt dt 0'v v v d v

dt

'a a

Test 5

A passenger in a car traveling at 60 km/h pours a cup of coffee for the driver. The path of the coffee as it moves from a Thermos bottle into a cup as seen by the passenger is

1. nearly vertical into the cup, just as if the passenger were standing on the ground pouring it

2. a parabolic path

3. an oblique line

Test 6

A passenger in a car traveling at 60 km/h pours a cup of coffee for the driver. The path of the coffee as it moves from a Thermos bottle into a cup as seen by someone standing beside the road and looking in the window of the car as it drives past is

1. nearly vertical into the cup, just as if the passenger were standing on the ground pouring it

2. a parabolic path

3. an oblique line

Observer A on a moving vehicle throws a ball upward and sees it rise and fall in a straight-line path.

Stationary observer B sees a parabolic path for the same ball.

EXAMPLE 5 A boat heading due north crosses a wide river with a speed

of 10.0 km/h relative to the water. The water in the river

has a uniform speed of 5.00 km/h due east relative to the

Earth.

(a) Determine the velocity of the boat relative to an

observer standing on either bank.

(a) 2 2

bE br rEv v v

2 2(10.0 / ) (5.00 / )bEv km h km h

vbr

vrE

vbE

;bE br rEv v v

11.2 /km h

Direction of bEv

tan rE

br

v

v

5.00 /;

10.0 /

km h

km h 026.6




Earth.

(b) If the width of the river is 3.0 km, find the time it takes the boat to cross it.

(b)

cos

ABAC

vbr

vrE

vbE

cos br

bE

v

v

10.0 /;

11.2 /

km h

km h

A

B C

3.00

(10.0 / ) /(11.2 / )

km

km h km h

3.36 km

bE

ACt

v

3.36

11.2 /

km

km h 0.30 18.0 minh




Earth.

(c) If this boat is to travel due north, as shown in the figure, what should its heading be? Find the time it takes the boat to cross the river. The width of the river is 3.0 km

(c)

sin rE

br

v

v

5.00 /;

10.0 /

km h

km h 030.0

The boat must head upstream in order to

travel directly northward across the river

The boat must steer a course 30.0° west of north




Earth.

(c) If this boat is to travel due north, as shown in the figure, what should its heading be? Find the time it takes the boat to cross the river. The width of the river is 3.0 km

(c) cosbE brv v

0(10.0 / ) cos 30.0km h 8.66 /km h

bE

ABt

v

3.00

8.66 /

km

km h 21.0 min

A

B

PROBLEM 6

The compass of an airplane indicates that it is headed due north, and its airspeed indicator shows that it is moving through the air at 240 km/h. There is a wind of 100 km/h from west to east. (a) What is the velocity of the airplane relative to the earth?

SOLUTION

(a)

PROBLEM 6

The compass of an airplane indicates that it is headed due north, and its airspeed indicator shows that it is moving through the air at 240 km/h. There is a wind of 100 km/h from west to east. (b) In what direction should the pilot head to travel due north? What will be her velocity relative to the earth?

SOLUTION

(b)

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PROGRAM OF “PHYSICS”webdirectory.hcmiu.edu.vn/Portals/25/Docs/.../PHYS1-CH1-KINEMATICS_New.pdf · Test 2 The graph shows position as a function of time for two trains running