Accounting for Angular Momentum

Chapter 21

Objectives Understand the basic fundamentals behind

angular momentum Be able to define measures of rotary

motion and measures of rotation

Accounting for Angular Momentum Linear momentum deals with objects

moving in a straight line. Angular momentum deals with objects that

rotate or orbit.

Angular MotionConsider a person on a

spinning carnival ride.Position, velocity, and

acceleration are a function of angular, rather than linear measurements.

rq1

q2

Timepasses

Angular Position and Speed Position can be described by q

Angular displacement, Dq = q2 - q1

Average angular speed is

Instantaneous angular speed is

rs

==radius

length arcq

ttt DD

=--

=qqq

12

12

dtd

tt

qq =DD

=D

lim0

Angular Acceleration Average angular acceleration is

Instantaneous angular acceleration isttt D

D=

--

=

12

12

dtd

tt

=DD

=D

lim0

Analogy to Linear Motion Linear

x = xo + vot + ½aot2

dx/dt = v = vo + aotdv/dt = a = ao

Angular q = qo + ot + ½ot2

dq/dt = = o + otd/dt = = o

Pair Exercise #1 A carnival ride is rotating at 2.0 rad/s. An

external torque is applied that slows the ride down at a rate of –0.05 rad/s2. How long does it take the ride to come to

rest? How many revolutions does it make while

coming to rest?

Measures of RotationFrequency, f is the number of cycles (revolutions) per second

Unit: Herzt or Hz = 1 cycle/second

n revolutions sweep out q radians. There are 2p radians per cycle (revolution). Therefore

Frequency is given by

πnn

π 2radians 2revolution 1 q

q==

p

qpp

qdtd

dtd

dtdnf

21

21

2=

=

== fp 2=

Angular frequency, ω, is the amount of radians per second.

Measures of RotationPeriod, T, is the time it takes to complete one

cycle (seconds per revolution)

p21

==f

T

Pairs Exercise #2

A carnival ride completes 2 revolutions per second. What is its frequency, period, and angular velocity?

More Basic Equations Speed (magnitude of velocity) is

Using some basic calculus and algebra you can find (Section 20.1.3):

rdtdsv ==== VSpeed

22

rrv

dtdva ===

Centripetal Forces

Applying Newton’s second law:

This is the inward force required to keep a mass in a circular orbit. If the force stops being applied, the mass will fly off tangentially.

22

mrrvmmaF ===

Centrifugal Force

From the Inertial Frame The ball is seen to rotate. Centripetal force keep the ball rotating

From a co-rotating frame: a rotating object appears to have an outward force acting on it. This fictitious centrifugal force has the same magnitude as the centripetal force, but acts in the opposite direction.

The centripetal (not the centrifugal) force can be used to design a centrifuge, which separates materials based upon density differences.

Angular momentum (particles) Angular momentum (L) is a vector quantity The direction is perpendicular to the plane of the

orbit and follows the right hand rule 2mrrrmrmvL ===

Moment of Inertia

The mass moment of inertia is define as:

Thus angular momentum can be expressed as:

Which is analogous to linear momentum:

2mrI

IL =

mvp =

Angular Momentum for BodiesFinding the angular momentum of a rotating body requires you to integrate over the volume of the body. (see Figure 21.7)

=

==

L

dhddrrdL

dhdrrdrrdLdmrvdL

0

H

0

2

0

R

0

3pq

q

Angular Momentum for Bodies This gets messy for most shapes (even simple ones),

so tables with moments of inertia have be developed for standard shapes (see Table 21.2).

From AutoCAD you can use the moment of inertia from Mass Properties times the density of the material.

Once you have the moment of inertia, the angular momentum is found from the formula:

IL =

Parallel Axis Theorem Because we want properties about an axis

other than the center of mass, we must use the Parallel Axis Theorem, which states that

Iaxis=Icg+MD2

where Iaxis and ICG are parallel and D is the

distance between them. M is the mass of the part.

Using the Theorem With the mass and CG known, and the desired

axis known, we can find the correct moment of inertia.

The axis we want is through the point 55, 25 mm. Because the CG’s X and Y are 35.54, 25 mm, the distance between them is 19.46 mm.

So the the correct moment is 275.46 kg•mm2 + 0.401 kg • (19.46 mm)2

= 427.4 kg•mm2

Solution

smmkg3.893

srad09.2mmkg4.427

srad09.2

s60min

revrad2

minrev20

22

===

==

p

IL

Torque Torque is a twisting force. It is created by

applying a force to an object in an attempt to make the object rotate. (see Figure 21.8)

Like momentum, it is a vector quantity.

rFT =

Conservation of Angular Momentum Angular momentum is a conserved

quantity. Changing angular momentum of a system

Changing mass (see Figure 21.9) Applying an external torque

Torque and Newton When Torque is applied to an object, it changes

the angular momentum of the object in a manner similar to a force applied to a body changes its linear momentum.

madtdvm

dtmvd

dtdpF

IdtdI

dtId

dtdLT

====

====

Pair Exercise #3 The previous AutoCAD part is to be spun

by applying a torque of 0.001N·m to the hole for 10 seconds.

What is its angular velocity (in rpm’s) at the end of the 10-s period?

Systems without Net Momentum Input

Many times you do not have unbalanced torque or mass transfer. In this case the UAE simplifies to:

Final Amount=Initial Amount This is how ice skaters spin faster.

Pair Exercise #4 A space satellite has an electric motor in it

with a flywheel attached (see Figure 21.11). The motor causes the flywheel to rotate at 10 rpm. If the moment of inertia of the flywheel is 10 kg·m2 and the satellite is 10000 kg·m2, how long must the motor run to twist the satellite by 10 degrees?