PROGRAM OF PROGRAM OF “PHYSICS”“PHYSICS”Lecturer: Dr. DO Xuan Hoi

Room 413E-mail : dxhoi@hcmiu.edu.vn

PHYSICS I PHYSICS I (General Mechanics) (General Mechanics)

02 credits (30 periods)02 credits (30 periods)

Chapter 1 Bases of KinematicsChapter 1 Bases of Kinematics

Motion in One Dimension Motion in One Dimension

Motion in Two DimensionsMotion in Two Dimensions

Chapter 2 The Laws of Motion Chapter 2 The Laws of Motion

Chapter 3 Work and Mechanical EnergyChapter 3 Work and Mechanical Energy

Chapter 4 Linear Momentum and CollisionsChapter 4 Linear Momentum and Collisions

Chapter 5 Rotation of a Rigid Object Chapter 5 Rotation of a Rigid Object About a Fixed Axis About a Fixed Axis

Chapter 6 Static EquilibriumChapter 6 Static Equilibrium

Chapter 7 Universal GravitationChapter 7 Universal Gravitation

References :References :

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

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

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

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

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

http://ocw.mit.edu/OcwWeb/Physics/index.htmhttp://www.opensourcephysics.org/index.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/HFrame.htmlhttp://www.practicalphysics.org/go/Default.htmlhttp://www.msm.cam.ac.uk/http://www.iop.org/index.html...

PHYSICS IPHYSICS IChapter 5Chapter 5

Rotation of a Rigid Object About a Fixed Rotation of a Rigid Object About a Fixed AxisAxis

Rotational KinematicsTorque and Angular Acceleration Moments of InertiaRotational Kinetic EnergyRolling Motion of a Rigid Object Angular Momentum of a Rotating Rigid Object Conservation of Angular Momentum

1 1 Rotational Kinematics

A rigid object is one that is nondeformable—that is, it is an object in which the separations between all pairs ofparticles remain constant We treat the rotation of a rigid object about a fixed axis Every point on the Every point on the object undergoes object undergoes circular motioncircular motion about about the point Othe point O

Every point of the Every point of the object undergoes the object undergoes the same anglesame anglein any given time intervalin any given time interval

How to determine the position of a rotating object??

Angle Angle : coordinate for : coordinate for rotationrotation

O x

x

P0P

Fixed plane

M0

MO

Rotating plane

OM0

M

Angular coordinate :

(OM0 , OM) =

+

The The angular displacementangular displacement is defined as the is defined as the angle the object rotates through during angle the object rotates through during some time intervalsome time interval

The The average angular velocity (speed)average angular velocity (speed), , ω ω , of a , of a rotating rigid object is the rotating rigid object is the ratio of the angular ratio of the angular displacement to the time intervaldisplacement to the time interval

M0

M

i

+

O

if

ttt if

if

The The instantaneousinstantaneous angular velocity (speed) angular velocity (speed) is is defined as the limit of the average speed as the defined as the limit of the average speed as the time interval approaches zerotime interval approaches zero

radians/secradians/sec (rad/s)(rad/s)

M0

M

i

+

O

0limt

dt dt

Angular speed will be Angular speed will be

positivepositive if θ is increasing ( if θ is increasing (counterclockwisecounterclockwise))

negativenegative if θ is decreasing ( if θ is decreasing (clockwiseclockwise))

The average angular acceleration of a rotating object isdefined as the ratio of the change in the angular speed to the time interval t :

(rad/s(rad/s22))

The instantaneous angular acceleration is defined as thelimit of the ratio /t as t approaches zero :

ttt if

if

2

20limt

d dt dt dt

Notes about angular kinematics:Notes about angular kinematics:

When rotating about a fixed axis, every particle on a rigid object rotates through the same angle and has the same angular speed and the same angular acceleration.

TestTestA ladybug sits at the outer edge of a merry-go-round, and

a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second.The gentleman bug’s angular speed is

1. half the ladybug’s.2. the same as the ladybug’s.3. twice the ladybug’s.4. impossible to determine

Note: both insects have an angular speed of 1 rev/s

UNIFORM ROTATIONAL MOTION

const

0

0t

' ;const

ROTATIONAL MOTION WITHCONSTANT ANGULAR ACCELERATION

const 0t

20 02

t t

. 0 Increasing speed

. 0

2 20 02 ( )

Decreasing speed

Analogies Between Linear and Analogies Between Linear and Rotational MotionRotational Motion

Rotational Motion About Rotational Motion About a Fixed Axis with a Fixed Axis with Constant AccelerationConstant Acceleration

Linear Motion with Linear Motion with Constant AccelerationConstant Acceleration

0 t

20

12

t t

2 20 2 2 2

0 2v v a x

20

12

x v t at

0v v at

EXAMPLE 1EXAMPLE 1

1. Bicycle wheel turns 240 revolutions/min. What is its angular velocity in radians/second?

rev 1min 2 rads240 8 rad s 25.1rad s

min 60s 1rev

2. If wheel slows down uniformly to rest in 5 seconds, what is the angular acceleration?

20 25rad s5rad s

5sfi

t

EXAMPLE 1EXAMPLE 1

3. How many revolution does it turn in those 5 sec?

20

12

t t

2125rad sec 5sec 5rad sec 5sec 62.5rad

2

1rev( ) 62.5rad 10 revolutions

2rev

Relationship between linear and angular quantities Linear and angular position

0 O

s

M

v

Linear speed vector :

Tangential to the trajectory

v Direction of motionMagnitude : v R

s R Linear (tangential speed)

'

'sR

vR

Linear acceleration

M

v

Radial component :

R Ta a a

Tangential component

Ta R

22

Rv

a RR

Change of direction of

v

'v

O

Change of magnitude of

v

M

v

O

Na

Ta

a

2tan T

R

aa

2 2R Ta a a

PROBLEM 1

A discus thrower moves the discus in a circle of radius 80.0 cm. At a certain instant, the thrower is spinning at an angular speed of 10.0 rad/s and the angular speed is increasing at 50.0 rad/s2 At this instant, find the tangential and centripetal components of the acceleration of the discus and the magnitude of the acceleration.

SOLUTION

2 2 Torque and Angular Acceleration

a. a. Torque

Torque characterizes the tendency of a force to rotate an object about some axis

d ( moment arm or lever arm ):

the perpendicular distancefrom the pivot point to the

lineof action of F

Force F

d

Axe of rotation

N

Fd

mN.m

Torque as a Vector

r : the distance between the pivot point and the point

of application of F

Fd F r F r

If two or more forces are acting on a rigid object ?

1 2 1 1 2 2F d F d

Convention : the sign of the torque is positive if the turning tendency of the force is counterclockwise and is negative if the turning tendency is clockwise

EXAMPLE 2EXAMPLE 2A one-piece cylinder is shaped with a core section

protrudingfrom the larger drum. The cylinder is free to rotate

around thecentral axis shown in the drawing. A rope wrapped

around thedrum, which has radius R1 = 1.0 m , exerts a force F1 =

5.0 N tothe right on the cylinder. A rope wrapped around the

core, whichhas radius R2 = 0.50 m, exerts a force F2 = 15.0 N

downwardon the cylinder.What is the net torque about the rotation axis, and which

waydoes the cylinder rotate from rest?The net torque is positive, if the cylinder

startsfrom rest, it will commence rotatingcounterclockwise with increasing angular

velocity.

b. The rotational analog of Newton’s second law

Consider a particle of mass m rotating in a circle of radius r under the influence of the force F

FTFR

F

RF FT FR

Torque due to F :

0T TRF RF

Newton’s second law :T TF ma ( )m R

. ( ) ;F R m R

2( )F mR

m

● A rigid object of arbitrary shape rotating about a fixed axis

2( )i i ii i

mR

The net torque on a rigid body :

We put :

I

m1

R1

m2

R22

i ii

mR

2i i

i

I mR

I : the moment of inertia of the rotating particle about the axis

( kg.m2 )

F ma

(rotational analog of Newton's second law for a rigid body)

c. Moments of inertia

CAUTION : Moment of inertia depends on the choice of axis

2i i

i

I mR

Moment of inertia of discrete mass points :

Moment of inertia of continuous mass distribution :

m

2i i

i

I mR 2

0lim

ii im

i

I mR

2

M

I R dm 2

V

I R dV

( : mass density ; V : volume of the object)

EXAMPLE 3EXAMPLE 3Four very tiny spheres are fastened to the corners of a

frame ofnegligible mass lying in the xy plane. (a) If the system rotates about the y axis with an angular

speed find the moment of inertia about this axis.(b) Suppose the system rotates in the xy plane about an

axis through O (the z axis). Calculate the moment of inertia about this axis.

(a)

(b)

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

2 2

M M

I R dm R dm 2I MR

The parallel-axis theorem

2CMI I MD

Suppose the moment of inertia about an axis through the center of mass of an object is ICM.The moment of inertia about any axis parallel to and a distance D away from this axis is

PROBLEM 2

SOLUTION

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 is the initial angular acceleration of the rod and the initial linear acceleration of its right end?

PROBLEM 3

A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axle. A light cord wrapped around the wheel supports an object of mass m. Calculate the angular acceleration of the wheel, the linear acceleration of the object, and the tension in the cord.

SOLUTION

;a

I TRR

2

aT I

R

2;

amg I ma

R

PROBLEM 4

Two blocks having masses m1 and m2 are connected to each other by a light cord that passes over one frictionless pulley, having a moment of inertia I and radius R.Find the acceleration of each block and the tensions T1 , T2 in the cord. (Assume no slipping between cord and pulleys.)

SOLUTION

AB

Rm

mB

mA

AB

++

PA =mAg

TA

TAPB =mBg

TB

TB

Rm

For A : A A Am g T m a

For B : B B BT m g m a

(1)

(2)

For the pulley :

A BT R T R I (3)

For A : A A Am g T m a

For B : B B BT m g m a

(1)

(2)

For the pulley :

A BT R T R I (3)

Ta aR R

(3) ( )A B

aT T R I

R

2A B

IT T a

R(3’)

(1) + (2) ( ) ( )A B A B A Bm g m g T T m m a

2 ( )A B A B

Im g m g a m m a

R(3’)

2 ( )A A A B

Im g m g a m m a

R

2

A B

A B

m g m ga

Im m

R

Acceleration =

Acting force

System’s Inertia

AB

Rm

mB

mA

mAgmBg

With : 212

I mR

1

2

A B

A B

m g m ga

m m m

Notes :

2

A B

A B

m g m ga

Im m

R

PROBLEM 5

Two blocks having masses mA and mB = 5.5 kg are connected to each other by a light cord that passes over one frictionless pulley, which is a thin-walled hollow cylinder and has a mass of 1.0 kg. The system begins to move from rest. After 2.0 s, the speed of A and B is 10 m/sFind mA and the tensions TA , TB in the cord.

A

B

SOLUTION

A

B

mAg

NATA

TA

mBg

TB

TB

For A : A AT m a

For B : B B Bm g T m a+

+ For the pulley :

B AT R T R I

/ /Ta R a R

( ) ;B A

aT T R I

R 2B A

IT T a

R

2( ) ;B B A

Im g m a m a a

R

2

B

A B

m ga

Im m

R2

2 22

B B

A BA B

m g m ga

mmR m mm mR

SOLUTION

2

2

;

22

B B

A BA B

m g m ga

mmR m mm mR

0v at v ;at v

at

210 /5,0 /

2,0m s

m ss

12A B

g mm m

a

10 1,05,5 1 ;

5,0 2Am kg

5,0Am kg

A AT m a 25,0 5,0 / ;kg m s 25AT N

;B B Bm g T m a ( )B BT m g a 25,5 (10 5,0) /kg m s

27,5BT N

3. Rotational Kinetic Energy

212i i i

i i

K K mv

3.1 The total kinetic energy of the rotating rigid object

With the moment of inertia :

m1R1

m2

R2 21( )

2 i ii

m R

2i i

i

I mR

The rotational kinetic energy of a object :

2 21

2 i ii

mR

212

K I

To compare with the linear motion :

212

K mv

dW Fds

3.2 Work– kinetic energy theorem

The work done by external forces ( sin )F rd

dW d

The Newton’s law :

sinr d

I

d : lever arm d

Idt

d dI

d dt

d

Id

;d I d dW d I d ;

f

i

W dW I d

2 21 12 2fiW I I

To compare with the linear motion :

2 21 12 2fiW mv mv

dW Fds

3.3 Work and Power

The work done by external forces ( sin )F rd

dW d

sinr d

d : lever arm

dW ddt dt

P

The power by external forces : PAngular velocity :

PROBLEM 6

A uniform rod of length L and mass M is free to rotate on africtionless pin passing through one end. The rod is released from rest in the horizontal position.(a) What is its angular speed when it reaches its lowest position?

SOLUTION

2 21 1;

2 2fiW I I 21 10

2 2 fMgL I

2 21 12 3 fML

G

G

3f

gL

(a)

PROBLEM 6

A uniform rod of length L and mass M is free to rotate on africtionless pin passing through one end. The rod is released from rest in the horizontal position.(b) Determine the linear speed of the center of mass andthe linear speed of the lowest point on the rod when it is inthe vertical position.SOLUTION

G

G

3f

gL

(a)

CMv R(b)CMv

3;

2L g

L

13

2CMv gL

'v R

3;

gL

L 3v gL v

Two blocks having masses mA and mB

are connected to each other by a light cord that passes over one frictionless pulley, which is a thin-walled hollow cylinder and has a mass m. The system begins to move from rest. Find the acceleration of each block

SYSTEM

A

B

mBg

x

SOLUTION

2 2 21 1 10

2 2 2A B BI m v m v m gx

2

2 22

1 1 10

2 2 2A B B

vI m v m v m gx

R

22 2A B B

Im m v m gx

RDerive with respect to the time :

2 2 ' 2A B B

Im m v v m gv

R

2;A B B

Im m a m g

R

2

B B

A BA B

m g m ga

I m m mm mR

PROBLEM 7

A block with mass m = 2.00 kg slides down a surface inclined 300 to the horizontal. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel is a hollow cylinder and has mass m = 2.00 kg. The string pulls without slipping.(a) What is the acceleration of the block down the plane?(b) What is the tension in the string?

SOLUTION

2 21 10 sin

2 2I mv mgx

PROBLEM 8

M

22 21 1

( ) 0 sin2 2

vmR mv mgx

R

2 sin ;v gx 2 ' sin ;v v gv

1

sin2

a g 2 0 21(9.81 / ).sin30 2.45 /

2a m s m s

(a)

A block with mass m = 2.00 kg slides down a surface inclined 300 to the horizontal. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel is a hollow cylinder and has mass m = 2.00 kg. The string pulls without slipping.(a) What is the acceleration of the block down the plane?(b) What is the tension in the string?

SOLUTION

PROBLEM 8

1

sin2

a g(a)

(b) sinmg T ma

1sin

2m g

1

sin2

T mg 2 012 (9.81 / ) sin30

2m s

4.90T N

M

T

mg

mgsin

4 Rolling Motion of a Rigid ObjectSuppose a cylinder is rolling on a straight path : The center of mass CM moves in a straight line.Each point of the cylinder moves about this CM path called a cycloid

The total kinetic energy of a rigid object rolling on a rough surface without slipping equals the rotational kinetic energy about its center of mass, plus the translational kinetic energy of the center of mass :

2 21 12 2CM CMK I Mv

Combined Translation and Rotation

Relationships between and vCM ?

If cylinder or sphere rolls without slipping (pure rolling motion) :

2 21 12 2CM CMK I Mv

CM

ds dv R

dt dt

CMv R

A primitive yo-yo is made by wrapping a string several times around a solid cylinder with mass M and radius R. You hold the end of the string stationary while releasing the cylinder with no initial motion. The string unwinds but does not slip or stretch as the cylinder drops and rotates. Use energy considerations to find the speed of the center of mass of the solid cylinder after it has dropped a distance h.

SOLUTION

PROBLEM 9

The kinetic energy at point 2 :

Conservation of energy :

For the solid sphere shown in the figure, calculate the linear speed of the center of mass at the bottom of the incline and the magnitude of the linear acceleration of the center of mass.

SOLUTION

PROBLEM 10

Work–kinetic energy theorem :

2 21 12 2CM CMK I Mv CMv R

5 Angular Momentum of a Rotating Rigid Object

5.1 Angular momentum of a particleFor a particle with constant mass m, velocity v, momentum p = mv, and position vector r relative to the origin 0 of an inertial frame, the angular momentum is defined as

r

O

p

m

L

L r p

sin sinL rp mrv

is the cross product of and

r

p

L

is perpendicular to the rp-plane. its direction is upward, and its magnitude is

L

2. /kg m s

( )dL d dr dp

r p p rdt dt dt dt

0dp dp

v p r rdt dt

With : r F dp

rdt

dLdt

To compare with linear motion : dp

Fdt

5.2 Angular momentum of a system of particlesThe net external torque acting on the system :

ii

i i

dL d dLL

dt dt dt

with :i

i

L L

being the total angular momentum of the system

A car of mass 1 500 kg moves with a linear

speed of 40 m/s on a circular race track of radius 50 m.(a) What is the magnitude of its angular momentum

relative to the center O of the track?(b) Find the moment of inertia of the car about O.

SOLUTION

PROBLEM 11

sin sinL rp mrv

is perpendicular to the rp-plane. Its direction is upward, and its magnitude is

L

0sin90mrv mrv rp

1500 50 40 /L kg m m s 6 23.0 10 . /kg m s

(a)

(b) 2( ) ( ) ;L mrv mr r mr I L

I

6 23.0 10 . /

(40 / ) /(50 )kg m s

m s m

23.75 .kg m

5.3 Angular Momentum of a Rotating Rigid Object

Consider a rigid object rotating about a fixed axis that coincides with the z axis of a coordinate system

0 2sin90 ( )i i i i i i i i iL m rv m r r m r

The angular momentum of each particle:

2 2z i i i i

I I

L m r m r

The angular momentum of the whole object:

zL I

zdL dI

dt dt

Differentiate with respect to time :

I dLdt

A sphere of mass m1 and a block of mass m2 are connected by a light cord that passes over a pulley. The radius of the pulley is R, and the moment of inertia about its axle is I. The block slides on a frictionless, horizontal surface. Find an expression for the linear acceleration of the two objects, using the concepts of angular momentum and torque.SOLUTION

PROBLEM 12

1 1 ;L mvR 2 2L mvR

The angular momentum of the sphere and of

the block about the axle of the pulley:

3L I vI

R

The angular momentum of the pulley:

The total angular momentum of the system:

1 2

vL mvR mvR I

R

A sphere of mass m1 and a block of mass m2 are connected by a light cord that passes over a pulley. The radius of the pulley is R, and the moment of inertia about its axle is I. The block slides on a frictionless, horizontal surface. Find an expression for the linear acceleration of the two objects, using the concepts of angular momentum and torque.

PROBLEM 12

1m gR The total external torque acting on thesystem about the pulley axle:

1 2

vL mvR mvR I

R

dLdt

1 1 2

d vm gR mvR mvR I

dt R

1 2( )I

m m Ra aR

1

1 2 2

m ga

Im m

R

PROBLEM 13

P

+

sOM

T

F

By using the concepts of angular momentum

and torque, find the equation of motion for a pendulum.

gSOLUTION

L I The angular momentum of the object

aboutthe axis of rotation:

mgd The total external torque acting on the system:

Line of force

dLdt

Lever arm

sinmgl

sind

mgl Idt

d

Idt

I

With small : '' ;mgl I '' 0mglI

2I mlSimple pendulum: '' 0gl

6 Conservation of Angular Momentum

From :

dLdt

If: 0

The total angular momentum of a system is constant in both magnitude and direction if the resultant external torque acting on the system is zero.

0 ;dLdt

L const

NNNNNNNNNNNNNN

i fL L

i i ffI I

Three Conservation Laws for an Isolated System

Conservation of energy :i i ffK U K U

i fL L

Conservation of linear momentum :

i fp p

Conservation of angular momentum :

A horizontal platform in the shape of a circular

disk rotates in a horizontal plane about a frictionless vertical axle.

The platform has a mass M = 100 kg and a radius R = 2.0 m. A

student whose mass is m = 60 kg walks slowly from the rim of the

disk toward its center. If the angular speed of the system is 2.0 rad/s when the student is at the rim, what is the angular

speedwhen he has reached a point r = 0.50 m from the center?

PROBLEM 14

SOLUTION

A door 1.00 m wide, of mass 15 kg, is hinged at one

side so that it can rotate without friction about a vertical axis. It is

unlatched. A bullet with a mass of 10 g is fired at a speed of400 m/ s into the exact center of the door, in a direction

perpendi- cular to the plane of the door. Find the angular speed of the

door just after the bullet embeds itself in the door. Is kinetic

energy con- served?

PROBLEM 15

SOLUTION

The initial angular momentum of the bullet :

The final angular momentum :( )door bulletI I I

A door 1.00 m wide, of mass 15 kg, is hinged at one

side so that it can rotate without friction about a vertical axis. It is

unlatched. A bullet with a mass of 10 g is fired at a speed of400 m/ s into the exact center of the door, in a direction

perpendi- cular to the plane of the door. Find the angular speed of the

door just after the bullet embeds itself in the door. Is kinetic

energy con- served?

PROBLEM 15

SOLUTION

Conservation of angular momentum requires:

mvl I

We calculate the initial and final kinetic energies:

1 2K K