PHYS 1441 – Section 002 Lecture #20

Monday, Apr. 27, 2009 PHYS 1441-002, Spring 2009 Dr. Jaehoon Yu

PHYS 1441 – Section 002Lecture #20

Monday, Apr. 27, 2009Dr. Jaehoon Yu

• Torque and Angular Acceleration• Work, Power and Energy in Rotation• Rotational Kinetic Energy• Angular Momentum & Its Conservation • Conditions for Equilibrium & Mechanical

Equilibrium• A Few Examples of Mechanical Equilibrium

1

Today’s homework is HW #11, due 9pm, Wednesday, May 6!!


2

Announcements• Term exam resolution

– There will be another exam for those of you who wants to take in the class 1 – 2:20pm this Wednesday, Apr. 25

• You are welcome to take it again but– If you take this exam despite the fact you took last Wednesday, the grade from this exam

will replace the one from last Wednesday’s

– It will cover the same chapters and will be at the same level of difficulties

– The policy of one better of the two term exams will be used for final grading will still be valid

• The final exam– Date and time: 11am – 12:30pm, Monday, May 11– Comprehensive exam– Covers: Ch 1.1 – What we finish next Monday, May 4– There will be a help session Wednesday, May 6, during the class


3

Torque & Angular AccelerationLet’s consider a point object with mass m rotating on a circle.

What does this mean?

The tangential force Ft and the radial force Fr

The tangential force Ft is

What do you see from the above relationship?

mr

Ft

Fr

What forces do you see in this motion?

t tF maThe torque due to tangential force Ft is

tF r I

Torque acting on a particle is proportional to the angular acceleration.

What law do you see from this relationship? Analogs to Newton’s 2nd law of motion in rotation.

How about a rigid object?

r

Ft

m

O

The external tangential force Ft is tF

The torque due to tangential force Ft isThe total torque is

What is the contribution due to radial force and why?

Contribution from radial force is 0, because its line of action passes through the pivoting point, making the moment arm 0.

mr

tma r 2mr I

tma mr

tF r 2r m 2r m I


4

The combined moment of inertia of the dual pulley is 50.0 kg·m2. The crate weighs 4420 N. A tension of 2150 N is maintained in the cable attached to the motor. Find the angular acceleration of the dual pulley.

Ex. 12 Hosting a Crate

yF 2

'T

2ya

1 1 2yT mg ma

1 1T

T2′ −mg yma

ymg ma

2

'1 1 2T T

Isince 2 2mg m I

Solve for 1 1 2T mg 22

I m

2

222

2150 N 0.600 m 451 kg 9.80m s 0.200 m6.3rad s

46.0 kg m 451 kg 0.200 m


5

Work, Power, and Energy in RotationLet’s consider the motion of a rigid body with a single external force F exerting tangentially, moving the object by s.The rotational work done by the force F as the object rotates through the distance s=r is

W

Since the magnitude of torque is rF, W

The rate of work, or power, of the constant torque becomes

P How was the power defined in linear motion?

Fs Fr

W

t

t

FrWhat is the unit of the rotational work? J (Joules)

What is the unit of the rotational power? J/s or W (watts)


6

Rotational Kinetic EnergyWhat do you think the kinetic energy of a rigid object that is undergoing a circular motion is?

Since a rigid body is a collection of masslets, the total kinetic energy of the rigid object is

Since moment of Inertia, I, is defined as

Kinetic energy of a masslet, mi, moving at a tangential speed, vi, is

ri

mi

O x

y vi

iK

RKE

2i i

i

I m r

RKE The above expression is simplified as

21

2 i Tim v 21

2 i im r

ii

K 21

2 i ii

m r 21

2 i ii

m r

1

2I Unit? J


7

A thin-walled hollow cylinder (mass = mh, radius = rh) and a solid cylinder (mass = ms, radius = rs) start from rest at the top of an incline. Determine which cylinder has the greatest translational speed upon reaching the bottom.

Ex. 13 Rolling Cylinders

E 2 21 1

2 2f f fmv I mgh From Energy Conservation21

2 fmvf

212 fmv

2

2 of

mghv

m I r

The cylinder with the smaller moment of inertia will have a greater final translational speed.

Total Mechanical Energy = 21

2 mv 212 I mgh

2 21 102 2i imv I mgh

212 fI imgh since fv

r

2

12 2

fvI

r 0mgh

Solve for vf

What does this tell you?

The final speeds of the cylinders are

hfv

sfv

2

2 o

h

mgh

m I r

2 2

2 o

h h

mgh

m mr r

2

2omgh

m ogh

2

2 o

h

mgh

m I r

2 2

212

o

h h

mgh

m mr r

223

omgh

m

3

2 ogh 3

2hfv 1.15 h

fv

KE+ KER+ PE


8

Kinetic Energy of a Rolling Sphere

Since vCM=R

Let’s consider a sphere with radius R rolling down the hill without slipping.

K

R

xh

vCM

21

2CM

CM

vI

R

Since the kinetic energy at the bottom of the hill must be equal to the potential energy at the top of the hill

What is the speed of the CM in terms of known quantities and how do you find this out?

K

222

1CM

CM vMR

I

222

1CM

CM vMR

I

Mgh

2/1

2

MRI

ghv

CMCM

21

2 CMI 2 21

2MR

21

2 CMMv


9

Example for Rolling Kinetic EnergyFor solid sphere as shown in the figure, calculate the linear speed of the CM at the bottom of the hill and the magnitude of linear acceleration of the CM. Solve this problem using Newton’s second law, the dynamic method.

xF

Gravitational Force,

Since the forces Mg and n go through the CM, their moment arm is 0 and do not contribute to torque, while the static friction f causes torque

Mxh

RaCM

CM

We know that

What are the forces involved in this motion?

Mg

f

Newton’s second law applied to the CM givesFrictional Force, Normal Force

n

x

y

2

5

2MRICM

We obtain

f

Substituting f in dynamic equations CMMaMg

5

7sin

fMg sin CMMa

yF cosMgn 0

fR CMI

R

ICM

R

a

R

MRCM

2

52

CMMa5

2

sin7

5gaCM


10

sinL r p ur r ur

Angular Momentum of a ParticleIf you grab onto a pole while running, your body will rotate about the pole, gaining angular momentum. We’ve used the linear momentum to solve physical problems with linear motions, the angular momentum will do the same for rotational motions.

L

Let’s consider a point-like object ( particle) with mass m located at the vector location r and moving with linear velocity vThe angular momentum L of this particle relative to the origin O is

What do you learn from this? If the direction of linear velocity points to the origin of rotation, the particle does not have any angular momentum.

What is the unit and dimension of angular momentum? 2 /kg m s

Note that L depends on origin O. Why? Because r changesThe direction of L is +zWhat else do you learn?

Since p is mv, the magnitude of L becomes

If the linear velocity is perpendicular to position vector, the particle moves exactly the same way as a point on a rim.

2 1[ ]ML T

mvr 2mr I


11

Conservation of Angular MomentumRemember under what condition the linear momentum is conserved?

Linear momentum is conserved when the net external force is 0.

i fL Lr r

Three important conservation laws for isolated system that does not get affected by external forces

Angular momentum of the system before and after a certain change is the same.

By the same token, the angular momentum of a system is constant in both magnitude and direction, if the resultant external torque acting on the system is 0.

iLr

p constur

ext r

What does this mean?

Mechanical Energy

Linear Momentum

Angular Momentum

L ur

0p

Ft

ur

ur

L

t

ur

0

fLr

constant

i fp pr r

i i f fK U K U

const


12

Example for Angular Momentum Conservation A star rotates with a period of 30 days about an axis through its center. After the star undergoes a supernova explosion, the stellar core, which had a radius of 1.0x104km, collapses into a neutron star of radius 3.0km. Determine the period of rotation of the neutron star.

What is your guess about the answer? The period will be significantly shorter, because its radius got smaller.

fi LL

Let’s make some assumptions: 1. There is no external torque acting on it

2. The shape remains spherical3. Its mass remains constant

The angular speed of the star with the period T is

Using angular momentum conservation

Thus f

ffi II

f

i

I

I

if

i

Tmr

mr 22

2

fTf2

i

i

f Tr

r

2

2

days30100.1

0.32

4

days6107.2 s23.0

2 T


13

An ice skater is spinning with both arms and a leg outstretched. She pulls her arms and leg inward and her spinning motion changes dramatically. Use the principle of conservation of angular momentum to explain how and why her spinning motion changes.

Ex. 14 A Spinning Skater

The system of the ice skater does not have any net external torque applied to her. Therefore the angular momentum is conserved for her system. By pulling her arm inward, she reduces the moment of inertia (mr2) and thus in order to keep the angular momentum the same, her angular speed has to increase.


14

A satellite is placed in an elliptical orbit about the earth. Its point of closest approach is 8.37x106m from the center of the earth, and its point of greatest distance is 25.1x106m from the center of the earth. The speed of the satellite at the perigee is 8450 m/s. Find the speed at the apogee.

Ex. 15 A Satellite in an Elliptical Orbit

L From angular momentum conservation A AI

2I mr2 AA

A

vmr

r A A P Pr v r v

Av

Angular momentum is IP PI

since

Solve for vA

v r and2 P

PP

vmr

r

P P

A

r v

r

6

6

8.37 10 m 8450m s2820m s

25.1 10 m


15

Similarity Between Linear and Rotational MotionsAll physical quantities in linear and rotational motions show striking similarity.

Quantities Linear Rotational

Mass Mass Moment of Inertia

Length of motion Distance Angle (Radian)Speed

AccelerationForce Force TorqueWork Work WorkPower

MomentumKinetic Energy Kinetic Rotational

2I mr

rv

t

t

v

at

t

F maur r

I r ur

W F d rr

P F v ur r P

2

2

1mvK 2

2

1 IKR

L

M

W

p mvur r

L Iur ur


16

A thought problem• Consider two cylinders – one hollow

(mass mh and radius rh) and the other solid (mass ms and radius rs) – on top of an inclined surface of height h0 as shown in the figure. Show mathematically how their final speeds at the bottom of the hill compare in the following cases:

1. Totally frictionless surface2. With some friction but no energy

loss due to the friction3. With energy loss due to kinetic

friction

• Moment of Inertia– Hollow cylinder:

– Solid Cylinder:

2h hI mr

21

2s sI mr

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PHYS 1441 – Section 002 Lecture #20