Physics 207: Lecture 16, Pg 1 Physics 207, Lecture 16, Oct. 29 Agenda: Chapter 13 l Center of Mass l Torque l Moment of Inertia l Rotational Energy l.

Physics 207: Lecture 16, Pg 1

Physics 207, Physics 207, Lecture 16, Oct. 29Lecture 16, Oct. 29

Agenda: Chapter 13Agenda: Chapter 13 Center of Mass Center of Mass TorqueTorque Moment of InertiaMoment of Inertia Rotational EnergyRotational Energy Rotational MomentumRotational Momentum

Assignment: Assignment: Wednesday is an exam review session, Exam will be Wednesday is an exam review session, Exam will be

held in rooms B102 &held in rooms B102 & B130 in Van Vleck at 7:15 PM MP Homework 7, Ch. 11, 5 problems, MP Homework 7, Ch. 11, 5 problems,

NOTE: Due Wednesday at 4 PMNOTE: Due Wednesday at 4 PM MP Homework 7A, Ch. 13, 5 problems, available soonMP Homework 7A, Ch. 13, 5 problems, available soon


Chap. 13: Rotational DynamicsChap. 13: Rotational Dynamics

Up until now rotation has been only in terms of circular motion with ac = v2 / R and | aT | = d| v | / dt

Rotation is common in the world around us. Many ideas developed for translational motion are

transferable.


Conservation of angular momentum has consequencesConservation of angular momentum has consequences

How does one describe rotation (magnitude and direction)?


Rotational Dynamics: A child’s toy, a physics Rotational Dynamics: A child’s toy, a physics playground or a student’s nightmareplayground or a student’s nightmare

A merry-go-round is spinning and we run and jump on it. What does it do?

We are standing on the rim and our “friends” spin it faster. What happens to us?

We are standing on the rim a walk towards the center. Does anything change?


Summary Summary (with comparison to 1-D kinematics)(with comparison to 1-D kinematics)

Angular Linear

constant

0 t

0 021

2t t

constanta

at 0vv

200 2

1v attxx

And for a point at a distance R from the rotation axis:

x = R v = Ra = R


System of Particles (Distributed Mass):System of Particles (Distributed Mass): Until now, we have considered the behavior of very simple

systems (one or two masses). But real objects have distributed mass ! For example, consider a simple rotating disk and 2 equal

mass m plugs at distances r and 2r.

Compare the velocities and kinetic energies at these two points.

1 2


System of Particles (Distributed Mass):System of Particles (Distributed Mass): An extended solid object (like a disk) can be thought of as

a collection of parts. The motion of each little part depends on where it is in the

object!

The rotation axis matters too!

1 K= ½ m v2 = ½ m (r)2

2 K= ½ m (2v)2 = ½ m (2r)2


System of Particles: Center of MassSystem of Particles: Center of Mass

If an object is not held then it rotates about the center of mass. Center of mass: Where the system is balanced !

Building a mobile is an exercise in finding

centers of mass.

m1m2

+m1 m2

+

mobile



How do we describe the “position” of a system made up of many parts ?

Define the Center of MassCenter of Mass (average position):For a collection of N individual pointlike particles whose

masses and positions we know:

M

mN

iii

1CM

rR

(In this case, N = 2)

y

x

rr2rr1

m1m2RRCM


Sample calculation:Sample calculation:

Consider the following mass distribution:

(24,0)(0,0)

(12,12)

m

2m

m

RCM = (12,6)

kji CM CM CM1

CM ZYXM

mN

iii

r

R

XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters

YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters

XCM = 12 meters

YCM = 6 meters



For a continuous solid, convert sums to an integral.

y

x

dm

rr

where dm is an infinitesimal

mass element.

Mdmr

dmdmr

CM

R


Rotational Dynamics: What makes it spin?Rotational Dynamics: What makes it spin?

TOT = |r| |FTang| ≡ |r| |F| sin

Torque is the rotational equivalent of forceTorque is the rotational equivalent of forceTorque has units of kg m2/s2 = (kg m/s2) m = N m

A constant torque gives constant angular acceleration iff the mass distribution and the axis of rotation remain constant.

FFTangential aa

r

FFrandial

FF

A force applied at a distance from the rotation axis


Lecture 16, Lecture 16, Exercise 1Exercise 1TorqueTorque

A. Case 1

B. Case 2

C. Same

In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same.

Remember torque requires F, r and sin or the tangential force component times perpendicular distance

L

LF F

axis

case 1 case 2


Lecture 16, Lecture 16, Exercise 1Exercise 1TorqueTorque

In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same.

Remember torque requires F, r and sin or the tangential force component times perpendicular distance

(A)(A) case 1

(B)(B) case 2

(C) (C) same L

LF F

axis

case 1 case 2


Rotational Dynamics: What makes it spin?Rotational Dynamics: What makes it spin?

TOT = |r| |FTang| ≡ |r| |F| sin

Torque is the rotational equivalent of forceTorque is the rotational equivalent of forceTorque has units of kg m2/s2 = (kg m/s2) m = N m

FFTangential aa

r

FFrandial

FF

A force applied at a distance from the rotation axis

TOT = r FTang = r m a = r m r = m r2

For every little part of the wheel


TOT TOT = m r= m r22 and inertiaand inertia

This is the rotational version of FTOT = ma

Moment of inertia, Moment of inertia, I ≡ m rm r2 2 , (here , (here II is just a point is just a point on the wheel) is the rotational equivalent of mass.on the wheel) is the rotational equivalent of mass.

If I is big, more torque is required to achieve a given angular acceleration.

FFTangential aa

r

FFrandial

FF

The further a mass is away from this axis the greater the inertia (resistance) to rotation


Calculating Moment of InertiaCalculating Moment of Inertia

where r is the distance from the mass to the axis of rotation.

N

iiirm

1

2I

Example: Calculate the moment of inertia of four point masses

(m) on the corners of a square whose sides have length L,

about a perpendicular axis through the center of the square:

mm

mm

L


Calculating Moment of Inertia...Calculating Moment of Inertia...

For a single object, I depends on the rotation axis! Example: I1 = 4 m R2 = 4 m (21/2 L / 2)2

L

I = 2mL2I2 = mL2

mm

mm

I1 = 2mL2


Lecture 16, Lecture 16, Exercise 2Exercise 2Moment of InertiaMoment of Inertia

A triangular shape is made from identical balls and identical rigid, massless rods as shown. The moment of inertia about the a, b, and c axes is Ia, Ib, and Ic respectively.

Which of the following is correct:

((A)A) Ia > Ib > Ic

(B)(B) Ia > Ic > Ib

(C)(C) Ib > Ia > Ic

a

b

c

i

iirm 2I


Calculating Moment of Inertia...Calculating Moment of Inertia...

For a discrete collection of point masses we found:

For a continuous solid object we have to add up the mr2 contribution for every infinitesimal mass element dm.

An integral is required to find I :

N

iiirm

1

2I

r

dm

dmr 2I


Moments of Inertia...Moments of Inertia...

Some examples of I for solid objects:

Solid sphere of mass M and radius R,

about an axis through its center.

I = 2/5 M R2

R

See Table 13.3, Moments of Inertia

Thin spherical shell of mass M and radius R, about an axis through its center.

Use the table…

R


Rotation & Kinetic EnergyRotation & Kinetic Energy Consider the simple rotating system shown below.

(Assume the masses are attached to the rotation axis by massless rigid rods).

The kinetic energy of this system will be the sum of the kinetic energy of each piece:

K = ½m1v1+ ½m2v2

+ ½m3v3+ ½m4v4

rr1

rr2rr3

rr4

m4

m1

m2

m3

4

1

221 v

iiimK


Rotation & Kinetic EnergyRotation & Kinetic Energy

Notice that v1 = r1 , v2 = r2 , v3 = r3 , v4 = r4

So we can rewrite the summation:

We recognize the quantity, moment of inertia or I, and

write:

rr1

rr2rr3

rr4

m4

m1

m2

m3

24

1

221

4

1

2221

4

1

221 ][ rrv

iii

iii

iii mmmK

221 IK


Lecture 16, Lecture 16, Exercise 2Exercise 2Rotational Kinetic EnergyRotational Kinetic Energy

A. ¼

B. ½

C. 1

D. 2

E. 4

We have two balls of the same mass. Ball 1 is attached to a 0.1 m long rope. It spins around at 2 revolutions per second. Ball 2 is on a 0.2 m long rope. It spins around at 2 revolutions per second.

What is the ratio of the kinetic energy

of Ball 2 to that of Ball 1 ?

221 IK

i

iirm 2I

Ball 1 Ball 2


Lecture 16, Lecture 16, Exercise 2Exercise 2Rotational Kinetic EnergyRotational Kinetic Energy

K2/K1 = ½ m r22 / ½ m r1

2 = 2 / 2 = 4

What is the ratio of the kinetic energy of Ball 2 to that of Ball 1 ?

(A) 1/4 (B) 1/2 (C) 1 (D) 2 (E) 4

Ball 1 Ball 2


Rotation & Kinetic Energy...Rotation & Kinetic Energy...

The kinetic energy of a rotating system looks similar to that of a point particle:

Point ParticlePoint Particle Rotating System Rotating System

i

iirm 2I

221 IK2

21 vmK

v is “linear” velocity

m is the mass.

is angular velocity

I is the moment of inertia

about the rotation axis.


Moment of Inertia and Rotational EnergyMoment of Inertia and Rotational Energy

Notice that the moment of inertia I depends on the distribution of mass in the system. The further the mass is from the rotation axis, the

bigger the moment of inertia.

For a given object, the moment of inertia depends on where we choose the rotation axis (unlike the center of mass).

In rotational dynamics, the moment of inertia I appears in the same way that mass m does in linear dynamics !

221 IK

iiirm 2I So where


Work & Kinetic Energy:Work & Kinetic Energy:

Recall the Work Kinetic-Energy Theorem: K = WNET

This is true in general, and hence applies to rotational motion as well as linear motion.

So for an object that rotates about a fixed axis:

NET22

21 I WK if


Connection with CM motionConnection with CM motion

If an object of mass M is moving linearly at velocity VCM without rotating then its kinetic energy is

CM21

RK I

2CM2

12CM2

1 VK MI What if the object is both moving linearly and rotating?

If an object of moment of inertia ICM is rotating in place about its center of mass at angular velocity then its kinetic energy is

2CM2

1T VK M


Angular Momentum:Angular Momentum:

0p

dt

d

F

We have shown that for a system of particles,

momentum is conserved if

What is the rotational equivalent of this?

angular momentum

is conserved if

vm

p

IL

0dtLd


Example: Two DisksExample: Two Disks

A disk of mass M and radius R rotates around the z axis with angular velocity 0. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity F.

0

z

F

z


Example: Two DisksExample: Two Disks A disk of mass M and radius R rotates around the z axis

with initial angular velocity 0. A second identical disk, at rest, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity F.

0

z

F

z

No External Torque so Lz is constant

Li = Lf I i i = I f ½ mR2 0 = ½ 2mR2 f


Lecture 16, Oct. 29Lecture 16, Oct. 29

Assignment: Assignment: Wednesday is an exam review session, Exam will be held Wednesday is an exam review session, Exam will be held

in rooms B102 &in rooms B102 & B130 in Van Vleck at 7:15 PM MP Homework 7, Ch. 11, 5 problems, MP Homework 7, Ch. 11, 5 problems,

NOTE: Due Wednesday at 4 PMNOTE: Due Wednesday at 4 PM MP Homework 7A, Ch. 13, 5 problems, available soonMP Homework 7A, Ch. 13, 5 problems, available soon


Angular Momentum:Angular Momentum:Definitions & DerivationsDefinitions & Derivations

We have shown that for a system of particles

Momentum is conserved if

What is the rotational equivalent of this?

dtdp

F

EXT 0EXT F

prL

Fr The rotational analog of force FF is torque

Define the rotational analog of momentum pp to be

angular momentum, L or

p = mv


Recall from Chapter 9: Linear MomentumRecall from Chapter 9: Linear Momentum

Newton’s 2nd Law:

FF = maa

)v(v

mdtd

dtd

m

Units of linear momentum are kg m/s.

pp ≡ mvv (pp is a vector since vv is a vector)

So px = mvx etc.

Definition: Definition: For a single particle, the momentum pp is defined as:

dt

dp

F


Linear Momentum and Angular MomentumLinear Momentum and Angular Momentum

Newton’s 2nd Law:

FF = maa

)v(v

mdtd

dtd

m

Units of angular momentum are kg m2/s.

So from:So from:

dt

dp

F

vp

m

IL

dt

dL

Frdt

d

L


Putting it all togetherPutting it all together

dtdLEXT EXTEXT Fr prL

EXTddt

L0

In the absence of external torquesIn the absence of external torques

Total angular momentum is conservedTotal angular momentum is conserved

dtFrd L

prdtFr

L

Frdtd

L

pr

L


Angular momentum of a rigid bodyAngular momentum of a rigid bodyabout a fixed axis:about a fixed axis:

kv i

iiii

iiiii

i rmm vrprLz

Consider a rigid distribution of point particles rotating in the x-y plane around the z axis, as shown below. The total angular momentum around the origin is the sum of the angular momentum of each particle:

i rr1

rr3

rr2

m2

m3

j

m1

vv2

vv1

vv3

We see that LL is in the z direction.

Using vi = ri , we get

IL

(since ri , vi , are perpendicular)

k L 2

z i

iirm


Example: Two DisksExample: Two Disks

A disk of mass M and radius R rotates around the z axis with angular velocity 0. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity F.

0

z

F

z


Example: Two DisksExample: Two Disks A disk of mass M and radius R rotates around the z axis

with initial angular velocity 0. A second identical disk, at rest, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity F.

0

z

F

z

No External Torque so Lz is constant

Li = Lf I i i = I f ½ mR2 0 = ½ 2mR2 f


Demonstration:Demonstration:Conservation of Angular MomentumConservation of Angular Momentum

Figure Skating :

A

z

B

z

Arm Arm

IA IB

A B

LA = LB

No External Torque so Lz is constant even if internal work done.


Demonstration:Demonstration:Conservation of Angular MomentumConservation of Angular Momentum

Figure Skating :

A

z

B

z

Arm Arm

IA < IB

A > B

½ IAA2

> ½ IB B2 (work needs to be done)

IAA= LA = LB = IBB

No External Torque so Lz is constant even if internal work done.


Angular Momentum ConservationAngular Momentum Conservation

A freely moving particle has a well defined angular momentum about any given axis.

If no torques are acting on the particle, its angular momentum remains constant (i.e., will be conserved).

In the example below, the direction of LL is along the z axis, and its magnitude is given by LZ = pd = mvd.

y

x

vv

dm


Example: Bullet hitting stickExample: Bullet hitting stick

A uniform stick of mass M and length D is pivoted at the center. A bullet of mass m is shot through the stick at a point halfway between the pivot and the end. The initial speed of the bullet is v1, and the final speed is v2.

What is the angular speed F of the stick after the collision? (Ignore gravity)

v1 v2

MF

before after

mD

D/4


Example: Bullet hitting stickExample: Bullet hitting stick

What is the angular speed F of the stick after the collision? (Ignore gravity).

Process: (1) Define system (2) Identify Conditions

(1) System: bullet and stick (No Ext. torque, L is constant)

(2) Momentum is conserved (Istick = I = MD2/12 )

Linit = Lbullet + Lstick = m v1 D/4 + 0 = Lfinal = m v2 D/4 + I f

v1 v2

MF

before after

mD

D/4


Example: Throwing ball from stoolExample: Throwing ball from stool A student sits on a stool, initially at rest, but which is

free to rotate. The moment of inertia of the student plus the stool is I. They throw a heavy ball of mass M with speed v such that its velocity vector moves a distance d from the axis of rotation. What is the angular speed F of the student-stool

system after they throw the ball ?

Top view: before after

d

vM

I I

F


Example: Throwing ball from stoolExample: Throwing ball from stool

What is the angular speed F of the student-stool system after they throw the ball ?

Process: (1) Define system (2) Identify Conditions

(1) System: student, stool and ball (No Ext. torque, L is constant)

(2) Momentum is conserved

Linit = 0 = Lfinal = m v d + I f

Top view: before after

d

vM

I I

F


An example: Neutron Star rotationAn example: Neutron Star rotation

period of pulsar is 1.187911164 s

Neutron star with a mass of 1.5 solar masses has a diameter of ~11 km.

Our sun rotates about once every 37 days

f / i = Ii / If = ri2 / rf

2 = (7x105 km)2/(11 km)2 = 4 x 109

gives millisecond periods!


Angular Momentum as a Fundamental QuantityAngular Momentum as a Fundamental Quantity

The concept of angular momentum is also valid on a submicroscopic scale

Angular momentum has been used in the development of modern theories of atomic, molecular and nuclear physics

In these systems, the angular momentum has been found to be a fundamental quantity Fundamental here means that it is an intrinsic

property of these objects


Fundamental Angular MomentumFundamental Angular Momentum

Angular momentum has discrete values These discrete values are multiples of a fundamental

unit of angular momentum The fundamental unit of angular momentum is h-bar

Where h is called Planck’s constant

s

m kg10054.1

234

h

),...3,2,1( nnL


Intrinsic Angular MomentumIntrinsic Angular Momentum

photon


Angular Momentum of a MoleculeAngular Momentum of a Molecule

Consider the molecule as a rigid rotor, with the two atoms separated by a fixed distance

The rotation occurs about the center of mass in the plane of the page with a speed of


Angular Momentum of a Molecule Angular Momentum of a Molecule (It heats the water in a microwave over)(It heats the water in a microwave over)

L

CM/ IE = h2/(82I) [ J (J+1) ] J = 0, 1, 2, ….


Center of Mass Example: Astronauts & RopeCenter of Mass Example: Astronauts & Rope

Two astronauts are initially at rest in outer space and 20 meters apart. The one on the right has 1.5 times the mass of the other (as shown). The 1.5 m astronaut wants to get back to the ship but his jet pack is broken. There happens to be a rope connected between the two. The heavier astronaut starts pulling in the rope.

(1) Does he/she get back to the ship ?

(2) Does he/she meet the other astronaut ?

M = 1.5mm


Example: Astronauts & RopeExample: Astronauts & Rope(1) There is no external force so if the larger astronaut

pulls on the rope he will create an impulse that accelerates him/her to the left and the small astronaut to the right. The larger one’s velocity will be less than the smaller one’s so he/she doesn’t let go of the rope they will either collide (elastically or inelastically) and thus never make it.

M = 1.5mm


Example: Astronauts & RopeExample: Astronauts & Rope

(2) However if the larger astronaut lets go of the rope he will get to the ship. (Too bad for the smaller astronaut!)

In all cases the center of mass will remain fixed because they are initially at rest and there is no external force.

To find the position where they meet all we need do is find the Center of Mass

M = 1.5mm

Physics 207: Lecture 16, Pg 1 Physics 207, Lecture 16, Oct. 29 Agenda: Chapter 13 l Center of Mass l Torque l Moment of Inertia l Rotational Energy l.

Documents