Rotational Inertia Energy 1151

8/12/2019 Rotational Inertia Energy 1151

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Rotational Inertia

&Kinetic Energy


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Linear & Angular

Linear Angular

Displacement x

Velocity v

Acceleration a

Inertia m I

KE mv2 I 2

N2 F = ma = I

Momentum P = mv L = I


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Rolling MotionIf a round object rolls without slipping, there is a fixed relationship

between the translational and rotational speeds:


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Rolling MotionWe may also consider rolling motion to be a combination of pure rotational and

pure translational motion:


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A Rolling TireA car with tires of radius 32 cm drives on a

highway at a speed of 55 mph.

(a) What is the angular speed wof the tires?

(b) What is the linear speed vtopof the top to the

tires?

(55 mph)(0.447 m/s/mph)77 rad/s

(0.320 m)

v

rw

2 110 mphtopv v

(77 rad/s) / (2 rad/rev) 12.25 rev/s


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Rotational Kinetic Energy

Consider a mass Mon the end of astring being spun around in a circle withradius rand angular frequency w

Mass has speed v = wr

Mass has kinetic energy K = M v2

K = M w2r2

Rotational Kinetic Energyis energy due

to circular motion of object.

M

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Rotational Inertia I Tells how much work is required to get

object spinning. Just like mass tells youhow much work is required to getobject moving. Ktran= m v2 Linear Motion

Krot= I w2 Rotational Motion

I = Smiri2 (units kg m2)

Note!Rotational Inertia (or Moment ofInertia) depends on what you arespinning about (basically the riin theequation).

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Inertia Rods

Two batons have equal mass andlength.

Which will be easier to spin?

A) Mass on ends

B) SameC) Mass in center

I = Sm r2 Further mass is from axis of rotation,

greater moment of inertia (harder to spin)


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A DumbbellUse the definition of moment

of inertia to calculate that of adumbbell-shaped object with

two point masses mseparated

by a distance of 2rand rotating

about a perpendicular axis throughtheir center of symmetry.

2 2 2 2

1 1 2 2 2i iI m r m r m r mr


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Nose to the GrindstoneA grindstone of radius r= 0.610 m is being

used to sharpen an axe.If the linear speed of the stone is 1.50 m/s

and the stones kinetic energy is 13.0 J,

what is its moment of inertiaI ?

/ (1.50 m/s) / (0.610 m) 2.46 rad/sv rw

1 2 2

2 22

2 2(13.0 J)4.30 kg m

(2.46 rad/s)

KK I Iw

w


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Moment of Inertia of a Hoop

2 2 2 2

i i i iI m r m R m R MR

All of the mass of a hoop is at the same distance Rfrom the centerof rotation, so its moment of inertia is the same as that of a pointmass rotated at the same distance.


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Moments of Inertia


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More Moments


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I is Axis Dependent2 2 22mR mR mR

22 20 2 4m m R mR


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Rotation Plus Translation

cm ,reli iv v v

bottom 0v axel cmv v Rw top cm2 2v v Rw


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Rolling Objects

v rw

cmv Rw

cma R

s R1 12 2

cm cm2 2K mv I w


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Like a Rolling Disk

A 1.20 kg disk with a radius 0f 10.0 cm rolls without slipping. The

linear speed of the disk is v = 1.41 m/s.

(a) Find the translational kinetic energy.

(b) Find the rotational kinetic energy.

(c) Find the total kinetic energy.

1 12 2

2 2 (1.20 kg)(1.41 m/s) 1.19 JtK mv 1 1 1 12 2 2 2

2 2 2 4( )( / ) (1.20 kg)(1.41 m/s) 0.595 JrK I mr v rw

(1.19 J) (0.595 J) 1.79 Jt rK K KS

P fli ht R lli R


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Preflight: Rolling Race(Hoop vs Cylinder)

A hoop and a cylinder of equal mass rolldown a ramp with height h. Which hasgreatest KE at bottom?

A) Hoop B) Same C) Cylinder20% 50% 30%

P fli ht R lli R


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Preflight: Rolling Race(Hoop vs Cylinder)

A hoop and a cylinder of equal mass rolldown a ramp with height h. Which hasgreatest speed at the bottom of the

ramp?A) Hoop B) Same C) Cylinder22% 30% 48%I = MR2 I = MR2


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Rolling Down an Incline

i i f f K U K U

1 2 2

2

1

2mgh mv I w

2

1 2 2122

vmgh mv mr

r

v gh

2Hollow Cylinder : ;I mr v gh

1 42

2 3Solid Cylinder: ;I mr v gh

2 102

5 7Solid Sphere: ;I mr v gh

0 0

2

Hollow Cylinder : I mr

1 2 2122

2

gh v v

gh v


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Compare HeightsA ball is released from rest on a no-

slip surface, as shown. After reachingthe lowest point, it begins to rise again

on a frictionless surface.

When the ball reaches its maximum

height on the frictionless surface, it is

higher, lower, or the same height asits release point?

The ball is not spinning when released, but will be spinning when it

reaches maximum height on the other side, so less of its energy will

be in the form of gravitational potential energy. Therefore, it will

reach a lowerheight.


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Spinning WheelA block of mass mis attached to a string

that is wrapped around the circumference of

a wheel of radiusRand moment of inertiaI,initially rotating with angular velocity wthat

causes the block to rise with speed v. The

wheel rotates freely about its axis and the

string does not slip.

To what height hdoes the block rise?

i fE E

1 1 1 1 12 2 2 2 2 22 2 2 2 2

( / ) (1 / )iE mv I mv I v R mv I mRw

fE mgh

2

21

2

v Ih

g mR

B l B ll


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A Bowling BallA bowling ball that has an 11 cm radius and

a 7.2 kg mass is rolling without slipping at 2.0

m/s on a horizontal ball return. It continues toroll without slipping up a hill to a height h

before momentarily coming to rest and then

rolling back down the hill.

Model the bowling ball as a uniform sphere

and calculate h.

ext mech therm mech0 0W E E E

1 12 2

cm cm2 20 0f f i i i iU K U K Mgh Mv I w

2

1 1 2 72 2 2cmcm cm22 2 5 10

ii i

vMgh Mv MR MvR

2 2

cm

2

7 7(2.0 m/s)0.29 m

10 10(9.8 m/s )

iv

hg


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Rotational Inertia Energy 1151