Chapter 8Chapter 8

Rotational Equilibrium and

Rotational Dynamics

Wrench DemoWrench Demo

TorqueTorque

Torque, τ , is the tendency of a force to rotate an object about some axis

F is the forced is the lever arm (or moment arm)Units are Newton-m

Fd=τ

Direction of TorqueDirection of Torque

Torque is a vector quantityDirection determined by axis of twistPerpendicular to both r and FIn 2-d, torque points into or out of paperClockwise torques point into paper. Defined as negativeCounter-clockwise torques point out of paper. Defined as positive

NonNon--perpendicular forcesperpendicular forces

Only the y-component, Fsinφ,produces a torque

NonNon--perpendicular forcesperpendicular forces

F is the forcer is distance to point where F is appliedΦ is the angle between F and r

φτ sinFr=

Torque and EquilibriumTorque and Equilibrium

Forces sum to zero (no linear motion)

Torques sum to zero (no rotation)

0and0 =Σ=Σ yx FF

0=Στ

Meter Stick DemoMeter Stick Demo

Axis of RotationAxis of Rotation

Torques require point of referencePoint of reference can be anywhere

Use same point for all torquesPick the point to make problem least difficult

ExampleExampleGiven M = 120 kg.Neglect the mass of the beam.

b) What is the forcebetween the beam andthe wall

a) Find the tensionin the cable

SolutionSolutiona) Given: M = 120 kg, L = 10 m, x = 7 m

Find: T

∑ = 0formula Basic

τ

0=−MgxTL

Solve for T = 824 N

Axis

SolutionSolutionb) Given: M = 120 kg, L = 10 m, x = 7 m, T = 824 N

Find f

∑ = 0formula Basic

yF

0=+− fMgT

Solve for f = 353 N

f

Alternative SolutionAlternative Solutionb) Given: M = 120 kg, L = 10 m, x = 7 m

Find f

0)( =−− fLxLMg

Solve for f = 353 N

fAxis∑ = 0

formula Basic

τ

Another ExampleAnother ExampleGiven: W=50 N, L=0.35 m,

x=0.03 mFind the tension in the muscle

∑ = 0formula Basic

τ0=−WLFx

F = 583 N

xL

W

Center of GravityCenter of Gravity

Gravitational force acts on all points of an extended objectHowever, it can be considered as one net force acting on one point,the center-of-gravity, X.

∑∑

∑∑

=

=

ii

iii

ii

iii

m

xmX

gmXxgm )(

WeightedAverage

ExampleExampleConsider the 400-kgbeam shown below.Find TR

SolutionSolutionFind TR

0formula Basic

=∑τ

to solve for TR, choose axis here

X = 2 mL = 7 m

0=+− LTMgX R

TR = 1 121 N

Fig 8.12, p.228

Slide 17

One Last ExampleOne Last ExampleGiven: x = 1.5 m, L = 5.0 m,

wbeam = 300 N, wman = 600 N

Find: T

x

L

SolutionSolution

Fig 8.12, p.228

Slide 17

Given: x , L , wbeam ,wmanFind: T

x

L

First, find Ty

0formula Basic

=∑τ

02/ =+−− LTLwxw ybeamman

Ty = 330 NNow, find T

)53sin( °= TTy T = 413 N

Baton DemoBaton DemoMomentMoment--of_Inertia Demoof_Inertia Demo

Torque and Angular AccelerationTorque and Angular Acceleration

When τ ≠ 0, rigid body experiences angular accelerationRelation between τ and α is analagous to relation between F and a

ατ ImaF == ,

Moment of Inertia

Moment of InertiaMoment of Inertia

This mass analog is called the moment of inertia, I, of the object

r is defined relative to rotation axisSI units are kg m2

∑≡i

iirmI 2

More About Moment of InertiaMore About Moment of Inertia

I depends on both the mass and its distribution.If an object’s mass is distributed further from the axis of rotation, the moment of inertia will be larger.

Demo: Moment of Inertia OlympicsDemo: Moment of Inertia Olympics

Moment of Inertia of a Uniform RingMoment of Inertia of a Uniform Ring

Divide ring into segmentsThe radius of each segment is R

22 MRrmI ii =Σ=

ExampleExample

What is the moment of inertia of the following point masses arranged in a square?

a) about the x-axis?

b) about the y-axis?

c) about the z-axis?

SolutionSolution

a) Find I about the x-axis?Given: M2=2, M3=3 kg, L=0.6 m

First, find distance to 2-kg masses

r = 0.6·sin(45°)∑= 2formula Basic

iirmI

222

2 rMrMI +=

=0.72 kg·m2

SolutionSolution

b) Find I about the y-axis?

Same as before, except you use the 3-kg masses

r = 0.6·sin(45°)

233

2 rMrMI +=

=1.08 kg·m2

SolutionSolution

c) Find I about the z-axis?

Use all the masses

r = 0.6·sin(45°)

233

222

2

2

rMrMrMrMI

+++=

=1.8 kg·m2

Other Moments of InertiaOther Moments of Inertia

M

ExampleExampleTreat the spindle as a solid cylinder.

a) What is the moment of Inertia of the spindle?

b) If the tension in the rope is 10 N, what is the angular acceleration of the wheel?

c) What is the acceleration of the bucket?

d) What is the mass of the bucket?

SolutionSolutiona) What is the moment of Inertia of the spindle?Given: M = 5 kg, R = 0.6 m

M

middleabout rod,,121

endabout rod,,31

shell spherical,32

sphere solid,52

cylinder solid,21

shell lcylindrica,

2

2

2

2

2

2Inertia of Moments

ML

ML

MR

MR

MRMR

2

21MRI = = 0.9 kgm2

SolutionSolutionb) If the tension in the rope is 10 N, what is α?Given: I = 0.9 kg m2, T = 10 N, r = 0.6 m

M

αττ

IrF

==

formula Basic αIrT =Solve for αα=6.67 rad/s2

c) What is the acceleration of the bucket?Given: r=0.6 m, α = 6.67 rad/s

ra α=formula Basic

a=4 m/s2

SolutionSolutiond) What is the mass of the bucket?Given: T = 10 N, a = 4 m/s2

M

maF =formula Basic

agTM

TMgMa

−=

−=

M = 1.72 kg

ExampleExampleA cylindrical space station of radius R = 12 m and mass M = 3400 kg is designed to rotate around the axis of the cylinder. The space station’s moment of inertia is 0.75 MR2. Retro-rockets are fired tangentially to the surface of space station and provide an impulse of 2.9x104 N·s.

a) What is the angular velocity of the space station after the rockets have finished firing?b) What is the centripetal acceleration at the edge of the space station?

SolutionSolutionGiven: M = 3400, R = 12, I = 0.75 MR2 = 3.67x105

F·t = 2.9x104.a) Find: ω

tII ωατ ∆

==

formula Basic

rF=τformula Basic

ωωτItRF

It=⋅

=⋅

Solve for ω.ω= 0.948 rad/s

b) Find centripetal acceleration

ra 2formula Basic

ω= Ra 2ω= = 10.8 m/s2

Rotational Kinetic EnergyRotational Kinetic Energy

22

21

21 ωImvKE +=

KE of center-of-mass motion

KE due to rotation

Each point of a rigid body rotates with angular velocity ω.

2222

21

21

21 ωω IrmvmKE iiii === ∑∑

Including the linear motion

ExampleExampleWhat is the kinetic energy of the Earth due to the daily rotation?Given: Mearth=5.98 x1024 kg, Rearth = 6.63 x106 m.

Tπω 2

formula Basic

=First, find ω

= 7.27 x10-5 rad/s

2

sphere Solid

52MRI =

2

formula Basic

21 ωIKE =

22

51 ωMRKE = = 2.78 x10-29

3600242⋅

=πω

ExampleExampleA solid sphere rolls down a hill of height 40 m.What is the velocity of the ball when it reaches the bottom? (Note: We don’t know r or m!)

22

formula Basic

21

21 ωImvmgh +=

2

sphere solidFor

52mrI =

m cancels

+= 222

52

21 ωrvgh

rv ω=formula Basic

5/72,

52

21 222 ghvvvgh =

+= v = 23.7 m/s

Suitcase DemoSuitcase Demo

Angular MomentumAngular Momentum

mvrLIL

== ω

Analogy between L and p

Conserved if no net outside forces

Conserved if no netoutside torques

F = ∆p/∆tτ = ∆L/∆t

p = mvL = Iω

Linear momentumAngular Momentum

Rigid body

Point particle

Rotating Chair DemoRotating Chair Demo

Angular Momentum and Angular Momentum and Kepler’s Kepler’s 2nd Law2nd Law

For central forces, e.g. gravity, τ = 0and L is conserved.Change in area in ∆t is:

Lmt

AmrvL

tvrA

21

)(21

=∆∆

=

∆=∆

⊥

⊥

ExampleExampleA 65-kg student sprints at 8.0 m/s and leaps onto a 110-kg merry-go-round of radius 1.6 m. Treating the merry-go-round as a uniform cylinder, find the resulting angular velocity. Assume the student lands on the merry-go-round while moving tangentially.

SolutionSolutionKnown: M, R, m, v0Find: ωF

First, find L0

mvrL =formula Basic

mvRL =0

Next, find Itot

2

Particle

2

cylinder Solid

21

MRI

MRI

=

=

22

21MRmRITot +=

Now, given Itot and L0, find ω

ωIL =formula Basic

TotIL0=ω = 2.71 rad/s

ExampleExampleTwo twin ice skaters separated by 10 meters skate without friction in a circle by holding onto opposite ends of a rope. They move around a circle once every five seconds. By reeling in the rope, they approach each other until they are separated by 2 meters.

a) What is the period of the new motion?

b) If each skater had a mass of 75 kg, what is the work done by the skaters in pulling closer?

SolutionSolutiona) Find: Tf

Given: R0=5 m, Rf=1 m, T0=5 s

2

particlePoint

mrI =

T

ILπω

ω2

formula Basic

=

=

0

200

22T

mRL π=

First, find expression for initial L

Next, apply conservation of L

ff T

mRT

mR ππ 2222 2

0

20 =

20

2

0 RR

TT ff = = T0/25 = 0.2 s

(2 skaters)

SolutionSolutionb) Find: W

Given: m=75 kg, R0=5 m, Rf=1 m, T0=5 s, Tf=0.2 s

2

formula Basic

21 ωIKE =

2200 2,2 ff mRImRI ==

First, find moments of inertia

Next, find values for ω

ff TT

πωπω 2,2

00 ==

Finally, find change in KE

200

2

21

21 ωω IIW ff −= = 7.11x105 J