Physics 1: Mechanics

Đào Ngọc Hạnh TâmOffice: A1.503,

Email: dnhtam@hcmiu.edu.vnHCMIU, Vietnam National University

Acknowledgment: Slides are supported by Prof. Phan Bao Ngoc

Part A: Dynamics of Mass Point

Chapter 1 Bases of Kinematics

Chapter 2 Force and Motion (Newton’s Laws)

Group Assignment

Part B: Laws of Conservation

Chapter 3 Work and Mechanical Energy

Midterm exam Chapter 4 Linear Momentum and Collisions

Part C: Dynamics and Statics of Rigid BodyChapter 5 Rotation of a Rigid Body About a Fixed Axis

Group Assignment

Chapter 6 Equilibrium and Elasticity

Chapter 7 Gravitation

Final exam

Contents of Physics 1

Part C: Dynamics and Statics of Rigid Body

Chapter 5: Rotation of a Rigid Body About a Fixed Axis

5.1. Rotational Variables

5.2. Rotation with Constant Angular Acceleration

5.3. Kinetic Energy of Rotation, Rotational Inertia

5.4. Torque, and Newton’s Second Law for Rotation

5.5. Work and Rotational Kinetic Energy

5.6. Rolling Motion of a Rigid Body

5.7. Angular Momentum of a Rotating Rigid Body

5.8. Conservation of Angular Momentum

Overview

•We have studied the motion of translation, in which objects move along a straight or curved line.

•In this chapter, we will examine the motion of rotation, in which objects turn about an axis.

5.1. Rotational variables:

We study the rotation of a rigid body about a fixed axis.Rigid bodies: Bodies can rotate with all its parts locked together andwithout any change in its shape. Fixed axis: A fixed axis means the rotational axis does not move.

Angular Position:

Reference line: To determine the angular position, we must define a reference line, which is fixed in the body, perpendicular to the rotation axis and rotating with the body.The angular position of this line is the angle of the line relative to a fixed direction.

(rad) radians :

r

s

1 rev = 3600 = 2 radwhere s is the arc length of a circular path of radius r and angle .

Angular Displacement

12 Convention: • > 0 in the counterclockwise direction.• < 0 in the clockwise direction.

Angular Velocity

Average angular velocity: ttt

12

12avg

Instantaneous angular velocity:dt

d

tt

0lim

Angular Acceleration

Average angular acceleration: ttt

12

12avg

Instantaneous angular acceleration:

dt

d

tt

0lim

Unit: rad/s or rev/s or rpm; (rev: revolution)

Unit: rad/s2 or rev/s2

Note: Angular displacement, velocity, and acceleration can be treatedas vectors (see page 246).

5.2. Rotation with Constant Angular Acceleration

For one dimensional motion:dt

dva

dt

dxv ;

Let’s change variable names: avx ,,

)(2

2

1

020

2

200

0

tt

t

Checkpoint 2 (p. 248):In four situations, a rotating body has angular position (t) given by (a) =3t-4, (b) =-5t3+4t2+6, (c) =2/t2-4/t, and (d) =5t2-3.

To which situations do the angular equations above apply?

)(2

2

1

0

2

0

2

2

00

0

xxa

attxx

at

Ans: a) and d)

5.3. Kinetic Energy of Rotation

a. Linear and Angular Variable Relationship

•The position: rs where angle measured in rad; s: distance along a circular arc; r: radius of the circle

•The speed:

rv (rad)

rdt

d

dt

ds

The period of revolution:

22

v

rT

•The Acceleration: rdt

d

dt

dv

Tangential acceleration: rat

Radial acceleration:r

r

var

22

b. Kinetic Energy of Rotation:

The KE of a rotating rigid body is calculated by adding up the kinetic energies of all the particles:

2233

222

211

2

1...

2

1

2

1

2

1iivmvmvmvmK

222

2

1)(

2

1iiii rmrmK

2

iirmI

2

2

1IK

Unit for I: kg m2

Rotation Inertia (or moment of inertia)

(radian)K (J)

c. Calculating the Rotational Inertia:

2

iirmI

• For continuous bodies:

dmrI 2

• If the rigid body consists of a few particles:

• Parallel-Axis Theorem: calculate I of a body of mass M about a given axis if we already know Icom:

2MhIIcomO

h: the perpendicular distance between the given axis at 0 and the axis through the COM of the body.

Some Rotational Inertias Table 10-2

Sample problem:

In 1985,Test Devices, Inc. (www.testdevices.com) was spin testing asample of a solid steel rotor (a disk) of mass M= 272 kg and radiusR = 38.0 cm. When the sample reached an angular speed of14000 rev/min, the test engineers heard a dull thump from thetest system, which was located one floor down and one room overfrom them.Investigating, they found that lead bricks had been thrown out inthe hallway leading to the test room, a door to the room had beenhurled into the adjacent parking lot, one lead brick had shot fromthe test site through the wall of a kitchen, the structural beams ofthe test building had been damaged, the concrete floor beneath thespin chamber had been shoved downward by about 0.5 cm, and the900 kg lid had been blown upward through the ceiling and had thencrashed back onto the test equipment. The exploding pieces had notpenetrated the room of the test engineers only by luck.

How much energy was released in the explosion of the rotor?

2

2

1IK

Energy was released in the explosion of the rotor

Rotational Inertia of the disk

(1)

222 .64.19)38.0)(272(2

1

2

1mkgmkgMRI

Angular speed =14000 rev/min = 14000. 2/60= 1.466×103 rad/s

2322 )/10.466.1)(.64.19(2

1

2

1sradmkgIK

= 2.1 × 107 J

(1)

5.4. Torque, and Newton’s Second Law for Rotation

a. Torque (“to twist”):•A force applied at point P of a body that is free to rotate about an axis through O. The force has no component parallel to the rotation axis.

F

F

F

Rotation axis

f

The ability of to rotate the body is defined by torque F

F

trFFr )sin(

F of armmoment the:

r

Fr

(Unit: N.m)

FrFr )sin(

Resolve into 2 components: Ft (tangential) and Fr (radial).

F

f

F

Direction of torque: Use the right hand rule to determine

Important Note: Torque is a vector quantity, however, because we consider only rotation around a single axis, we therefore do not need vector notation. Instead, a torque is a positive value if it would produce a counterclockwise rotation and a negative value for a clockwise rotation.

f1counterclockwise

clockwise

f2

tnet =t1 -t2 = F1r1sinf1 -F2r2 sinf2

b. Newton’s Second Law for Rotation

•We consider a simple problem, the rotation of a rigid body consisting of a particle of mass m on one end of a massless rod.

tt maF

The torque acting on the particle:

)()( 2mrrrmrmarF tt

(at: tangential acceleration)

: angular accelerationmeasure)(radian I

if more than one force applied to the particle:

Inet

This is valid for any rigid body rotating about a fixed axis.

53/270. The figure below shows a uniform disk that can rotatearound its center like a merry-go-round. The disk has a radiusof 2.0 cm and a mass of 20.0 grams and is initially at rest.Starting at time t=0, two forces are to be applied tangentiallyto the rim as indicated, so that at time t=1.25 s the disk hasan angular velocity of 250 rad/s counterclockwise. Force F1 hasa magnitude of 0.1 N. What is magnitude F2?

Inet IRFRFnet 1221

For a uniform disk: 2

2

1MRI

12 FR

IF

For rotation:t

tt

0

(N)14.01.025.12

)250)(100.2)(100.20(

2

23

12

Ft

MRF

+

5.5. Work and Rotational Kinetic Energy

The work-kinetic theorem applied for rotation of a rigid body:

WIIKKK ifif 22

2

1

2

1

•We consider a torque of a force F accelerates a rigid body in rotation about a fixed axis:

The work done by the torque:

f

i

dK

If is a constant:

)( ifW

The power:

dt

dWP

Problem 61 (p. 271): A 32.0 kg wheel, essentially a thin hoop with radius 1.2 m, is rotating at 280 rev/min. It must be brought to a stop in 15.0 s. (a) How much work must be done to stop it? (b) What is the required average power?

•The rotational inertia of a wheel (a thin hoop) about central axis:

)m (kg 1.462.10.32 222 MRI

•To stop the wheel, =0:

(rad/s) 3.29(s) 60

(rad) 2 280 rev/min 2800

(a) The work is needed to stop the wheel:

(kJ) 19.8or (J)197883.291.462

1

2

10 22

0 IKKW if

(W<0: energy transferred from the wheel)

(b) The average power: kW)(32.1or (W) 131915

19788

t

WP

5.6. Rolling Motion of a Rigid Body

We consider a rigid body smoothly rolling along a surface. Its motionconsists of two motions: translation of the center of mass androtation of the rest of the body around that center.

Example: A bike wheel is rollingalong a street. During a time intervalt, both O (the center of mass) and P(the contact point between thewheel and the street) move by adistance s:

Rs where R is the radius of the wheel.

Rvcom

•The speed of the center of the wheel:

•The linear acceleration of the center of the wheel:

Racom

The Kinetic Energy of Rolling

22

2

1

2

1comcom MvIK

Rotational KE Translational KE

Examples: 1. Rolling on a horizontal surface: A 2.0 kg wheel, rolling smoothly on a horizontal surface, has a rotational inertia about its axis I = MR2/2, where M is its mass and R is its radius. A horizontal force is applied to the axle so that the center of mass has an acceleration of 4.0 m/s2. What is the magnitude of the frictional force of the surface acting on the wheel?

Applying Newton’s second law for rotational motion:

(1) comsnet IRf

)(clockwise 0

axis) x theofdirection positive (in the 0,

xcoma

(2) , Ra xcom (1), (2):

appliedF

sf

(for )

+

+

(N)0.42

0.42

22

MR

R

,

2

,2

xcomxcom

coms

Ma

R

aIf

(for acom,x)

-

Examples: 2. Rolling Down a Ramp: We consider a rigid body smoothly rolling down a ramp.Force analysis: Fg, FN, and fs (opposing the sliding of the body, so the force is up the ramp)

(1) sin ,xcoms MaMgf

•Applying Newton’s second law for translational motion:

•Applying Newton’s second law for rotational motion: (2) comsnet IRf

ockwise)(countercl 0

axis) x theofdirection negative (in the 0,

xcoma

2

,

1

sin

MR

I

ga

comxcom

(3) , Ra xcom

(1), (2), (3):

Examples: 3. The Yo-Yo:

y

For translation motion:

(1) ,ycomMaMgT

(2) 0 comITR

For rotational motion:

(3) 0, Ra ycom

20

,

1MR

I

ga

comycom

Examples: 4. Pulley: A 10.0 kg block hangs from a cord which is wrapped around the rim of a frictionless pulley. Calculate the acceleration, a, of the block as it moves down? (The rotational inertia of the pulley is 0.50 kg m2 and its radius is 0.10 m).

• For translation motion:(1) maTmg

(2) pulley0 ITR

• For rotational motion:

(3) 0Ra

)m/s(63.1

1.0

5.010

8.910 2

22

pulley

R

Im

mga

a

T

T

mg

+

(Note: a is the acceleration of the block, is the angular acceleration of the pulley)

Example: (The Kinetic Energy of Rolling) A 10 kg cylinder rolls without slipping. When its translational speed is 10 m/s, What is its translational kinetic energy, its rotational kinetic energy, and its total kinetic energy?

•The rotational inertia of a cylinder about central axis:

2

2

1MRI

•Translational kinetic energy:

(J)50010102

1

2

1 22 MvKk

•Rotational kinetic energy:

22

22

4

1

2

1

2

1

2

1Mv

R

vMRIKr

(J) 250rK•Total kinetic energy:

(J) 750 rk KKK

The Kinetic Energy of Rolling

5.7. Angular Momentum of a Rotating Rigid Body

a. Angular Momentum of a Particle:

)( vrmprl

:r

the position vector of the particle with respect to O

:p

the linear momentum of the particle

•The direction of determined by the right-hand rulel

l

sinrmvl

or

rpprl

(Unit: kg m2 s-1)

l

)( vrmprl

dt

pdrFr net

dt

pdFnet

Newton’s Second Law in Angular Form for a Particle:

For translational motions:

dt

ld

dt

vrdm

vdt

rd

dt

vdrmnet

)(

)(

dt

ldnet

So, Newton’s Second Law in Angular Form:

Note: The torques and the angular momentum must be defined with respect to the same origin O.

l

dt

Ldnet

b. Angular Momentum of a System of Particles:

n

iin lllllL

1321 ...

Newton’s Second Law in Angular Form:

c. Angular Momentum of a Rotating Rigid Body:

Method: To calculate the angularmomentum of a body rotating about afixed axis (here the z axis), we evaluatethe angular momentum of a system ofparticles (mass elements) that formthe body.

iiiiii vmrprl 090sin

il

izlil

is the angular momentum of element i of mass mi:

iiiiii vmrprl 090sin

iiiiiiizi vmrvmrll ))(sin(sin,

i

n

iiii

n

iii

n

iziz rrmrvmlL

111

, )(

n

iiiz rmL

1

2

zz IL

We drop the subscript z: IL

I z : rotational inertia of

the body about the z axis

Note: L and I are the angular momentum and the rotational inertia of a body rotating about the same axis.

Example: At the instant of the figure below, a 2.0 kg particle P has aposition vector of magnitude 5.0 m and angle 1=45

0 and a velocityvector of magnitude 4.0 m/s and angle 2=30

0. Force of magnitude2.0 N and angle 3=30

0, acts on P. All three vectors lie in the xyplane. About the origin, what are the (a) magnitude and (b) directionof the angular momentum of P and the (c) magnitude and (d) directionof the torque acting on P?

r

v

F

prl

(a)

)30sin(0.40.20.5sin 2 rmvl

(b) Using the right-hand rule, points out of thepage and it is perpendicular to the figure plane.

l

l

Fr

(c)

m) (N 5)30sin(0.20.5sin 3 rF

(d) points out of the page and it is perpendicular to the figure plane.

)/m (kg 20 2 s

5.8. Conservation of Angular Momentum

If no net external torque acts on the system:

constantL

fi LL

ffii II

dt

Ldnet

tnet = 0

Example: You stand on a frictionless platform that is rotating atan angular speed of 1.5 rev/s. Your arms are outstretched, andyou hold a heavy weight in each hand. The moment of inertia ofyou, the extended weights, and the platform is 6.0 kg.m2. Whenyou pull the weights in toward your body, the moment of inertiadecreases to 1.8 kg.m2. (a) What is the resulting angular speedof the platform? (b) What is the change in kinetic energy of thesystem? (c) Where did this increase in energy come from?

(a) No net torque acting on the rotating system (platform + you + weights):

(rev/s)58.1

5.10.6

I

IIIconstant

2

1122211

I

(b) The change in kinetic energy:

(J)62125.10.6258.12

1

2

1

2

1 22211

222 IIK

(c) Because no external agent does work on the system, so the increase in kinetic energy comes from your internal energy (biochemical energy in your muscles).

More Corresponding Variables and Relations for Translational and Rotational

1, 3, 7, 14, 20, 26, 39, 43, 53, 56, 62 (p. 267- 271)

2, 5, 9, 17, 35, 38, 41, 43, 54, 60 (p. 297-302)

Homework:

Fr

Inet

•Torque:

Newton’s second law for rotation:

Work – KE Theorem: WIIKKK ifif 22

2

1

2

1

•For a point mass:2mrI

•For a rigid body: I depends on the object shape (table 10-2)

(kg.m2)

• Rotational KE:

2

2

1IK rv

rat

rr

var

22

Angular position: (rad)Angular displacement: (rad)Angular velocity: (rad/s)Angular acceleration: (rad/s2)

1 rev = 3600 = 2 rad

Constant

rs Linear variables

I: Rotation Inertia(or moment of inertia)

)(2

2

1

020

2

200

0

tt

tSummary:

•Rolling motion:Rvcom

Racom 22

2

1

2

1comcom MvIK

Rotational KETranslational KE

•Angular Momentum: prl

sinrmvl

dt

Ldnet

•Newton’s Second Law in Angular Form:

•Angular Momentum of a Rigid Body:

IL •Conservation of Angular Momentum:

If no net external torque acts on the system:

constantL

fi LL

ffii II