Chapter 10

More on angular momentum and torque

In chapter 9 we described the rotational motion of a rigid body and, based on that, we defined the vector of angular momentum as: L = I

In chapter 10 we will give a more general definition of L and introduce the principle of the conservation of angular momentum. This principle is:

(a) A fundamental principle of Physics, and

(b) A handy tool to solve a certain class of problems on rotational motion

(10-1)

The cross product of two vectors

C = A B We combine A and B to form a new vector C

1. Magnitude of C C = ABsin

2. Direction of C C is perpendicular on the plane

defined by A and B

3. Sense of C

It is given by the right hand rule (RHR)

(10-2)

Right Hand Rule (RHR)

I. Choose the smallest angle by which you have to rotate vector A (in the plane defined by A and B ) so that it coincides with vector B

II. Curl the fingers of the right hand in that direction

III. The thumb of the right hand points along C

(10-3)

In terms of components:z

O

x y

ij

k^

^

^

x y z

x y z

x y z

C A B

A iA jA kA

B iB jB kB

C iC jC kC

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x y z z y

y z x x z

z x y y x

C A B A B

C A B A B

C A B A B

(10-4)

Recipe for those who can handle determinants:

x y z

x y z

i j k

A B A A A

B B B

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P

F

r

O

x

y

z

rod

Example: Find the torque produced by the force F acting at point P on the rod.

Indeed

= rFsin

The cross product gives the correct magnitude and direction for

(10-5)

r F ������������� �

m

New definition of angular momentum L

Consider an object (mass m, position vector r, linear momentum p = mv). According the the definition of chapter 9 its angular momentum L = I

New definition:

We will show that the old and the new definition give the same result

(10-6)

L r p ������������������������������������������

Indeed the two definitions give the same result from L and thus are equivalent

(10-7)

A

m

2

2

, ,

sin

From triangle OAP we have that: sin

Old definition:

New definition:

vL I I md

dv

L md mvdd

L r p L rp

r d

L pd mvd

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(10-8)In chapter 9 we saw that

We will derive the same result using the new definition of L

( )

, (Newton's sec

o

dL

dt

d L d r p dr d pL r p p r

dt dt dt dt

d r d pv F

dt dt

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nd law)

=

The first term is zero.

The second term is

dLv p r F

dt

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dL

dt

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System of particles: here we deviate again from the notion of a rigid body introduced in chapter 9. A system of particles is a collection of N particles of mass m1, m2, … mN but it is not necessarily rigid. For example a group of meteorites traveling in space, although not rigid, is a system of particles

Total angular momentum is the vector sum of the angular momenta of all the particles in the system

L = L1 + L2 + … + LN

Total torque is the vector sum of the torques of all the particles in the system

= 1 + 2 + … + N

(10-9)

The total angular momentum L and total torque of a system obey the equation:

which we have shown to be true for a single particle This equation states that the rate of change of the total angular momentum L of a system is equal to the total torque on the system

Note: When we calculate we only have to take into account the torques generated by external forces. The torques generated by internal forces can be ignored! This is an important simplification

(10-10)

dL

dt

��������������

When we calculate we only have to take into account the torques generated by external forces. This can easily be shown in the simple case of a system that consists of two particles.

Total force on particle 1: F1 = F1

ext + F12

Total force on particle 2: F2 = F2ext + F21

F12 + F21 = 0 (Newton’s third law) F21 = - F12

1 = r1F1ext

+ r1F12

2 = r2F2ext

+ r2F21

2 = r2F2ext

- r2F12

= 1 + 2

= r1F1ext

+ r2F2ext

+ (r1 -r2)F12

the last term vanishes

= r1F1ext

+ r2F2ext (10-11)

Conservation of angular momentum

This principle is true for any system of particles (including rigid bodies)

(10-12)

1

1 22

In a system of particles If 0 0

L is a constant vector. If the angular momentum at t

is L and at time t is is L

dL dL

dt dt

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1 2L L����������������������������

More concisely:

0L ��������������

Example (10-6) page 280 A student sits on a spinning stool with weights in his hands and rotates with an initial angular velocity o The student pulls his hands so that they are close to his torso. Find the final angular velocity o = 2/To = 2.1 rad/s

(10-13)

R = 0.25 m M = 50 kg m = 2 kg d = 0.5 m L = 1 m To = 3s

L+R

m

R

m

m m

Li

Lf

(10-14)

2 22 2

22

22

, ,

2 ( ) , 2 2 2

2 ( ) 2 9 rad/s

22

ii i o f f i f i o f o

f

i f

o

IL I L I L L I I

I

MR MRI m L R I mR

MRm L R

MRmR

Central forces

A central force is one that acts along the line joining the source of the force and the object on which the force acts. For example the force F exerted by the sun on a comet The torque of the force is: = rF = rFcos180º = 0 The rate of change in the angular momentum of the comet is:

(10-15)

r

F

If 0 0

L is a constant vec

tor

dL dL

dt dt

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As a result of the fact that the angular momentum of the comet is constant, the comet sweeps equal areas in equal times (shaded areas in the figure). This is known in astronomy as “Kepler’s second law” .

For that reason the comet has to move faster when it is closer to the sun (between points C and D) and slow down when it is further away (between points A and B)

(10-16)

The kinetic energy of the rolling object is given by:

The first term in the expression for K is the rotational energy about the center of mass. The second term is the kinetic energy due to the translation of the center of mass. We can use energy conservation in problems that involve rotational motion (10-17)

Conservation of rotational energy

Consider a rolling object of radius R and mass m. In chapter 9 we saw that the velocity of the center of mass and the angular velocity are connected via the equation:

vcm = R

2 2

2 2CM CMI mv

K

Example (10-7), page 285. A spool of mass m, moment of inertia about its axis I , and radius R unwinds under the force of gravity. Use conservation of energy to find the speed of the spool’s center of mass after it has fallen by a length h

(10-18)

.

.

vcm

CM

CM

U = 0

before after

h

y

(10-19)

2 2

2 2

0 , ,

2 2

2 2

i f

cm

cmf

E E K U U mgh

mv IK

mv IE mgh

2 2

2 2

cm2

2

0 , Substitute 2 2

0 Solve for v 2 2

2

/

cm cmi f

cm cm

cm

mv vIE E mgh

R

mv Ivmgh

Rmgh

vm I R