Rotational Inertia&

Kinetic Energy

Linear & Angular

Linear Angular

Displacement x θ

Velocity v Acceleration a Inertia m I

KE ½ mv2 ½ I2

N2 F = ma = IMomentum P = mv L = I

Rolling MotionRolling Motion If a round object rolls without slipping, there is a fixed relationship between the translational and rotational speeds:

Rolling MotionRolling Motion We may also consider rolling motion to be a combination of pure rotational and pure translational motion:

Rolling MotionRolling Motion We may also consider rolling motion at any given instant to be a pure rotation at rate about the point of contact of the rolling object.

A Rolling Tire A car with tires of radius 32 cm drives on a highway at a speed of 55 mph.

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

(b) What is the linear speed vtop of the top to the tires?

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

(0.320 m)

v

r

2 110 mphtopv v

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

Rotational Kinetic Energy

• Consider a mass M on the end of a string being spun around in a circle with radius r and angular frequency

– Mass has speed v = r– Mass has kinetic energy

• K = ½ M v2

• K = ½ M 2 r2

• Rotational Kinetic Energy is energy due to circular motion of object.

M

24

Rotational Inertia I• Tells how much “work” is required to

get object spinning. Just like mass tells you how much “work” is required to get object moving.– Ktran = ½ m v2 Linear Motion– Krot = ½ I 2 Rotational Motion

• I = miri2 (units kg m2)

• Note! Rotational Inertia (or “Moment of Inertia”) depends on what you are spinning about (basically the ri in the equation). 13

Inertia Rods

Two batons have equal mass and length.

Which will be “easier” to spin?

A) Mass on endsB) SameC) Mass in center

I = m r2 Further mass is from axis of rotation, greater moment of inertia (harder to spin)

A Dumbbell Use the definition of momentof inertia to calculate that of adumbbell-shaped object withtwo point masses m separatedby a distance of 2r and rotatingabout a perpendicular axis throughtheir center of symmetry.

2 2 2 21 1 2 2 2i iI m r m r m r mr

Nose to the Grindstone A 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 stone’s kinetic energy is 13.0 J, what is its moment of inertia I ?

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

1 2 22 22

2 2(13.0 J)4.30 kg m

(2.46 rad/s)

KK I I

Moment of Inertia of a Hoop

2 2 2 2i i i iI m r m R m R MR

All of the mass of a hoop is at the same distance R from the center of rotation, so its moment of inertia is the same as that of a point mass rotated at the same distance.

Moments of Inertia

More Moments

I is Axis Dependent2 2 22mR mR mR

22 20 2 4m m R mR

Rotation Plus Translation

cm ,reli iv v v

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

Rolling Objects

v r

cmv R

cma R

s R1 12 2

cm cm2 2K mv I

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 22 2

(1.20 kg)(1.41 m/s) 1.19 JtK mv 1 1 1 12 2 2 22 2 2 4

( )( / ) (1.20 kg)(1.41 m/s) 0.595 JrK I mr v r

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

Preflight: Rolling Race (Hoop vs Cylinder)

A hoop and a cylinder of equal mass roll down a ramp with height h. Which has greatest KE at bottom?

A) Hoop B) Same C) Cylinder

20% 50% 30%

Preflight: Rolling Race (Hoop vs Cylinder)

A hoop and a cylinder of equal mass roll down a ramp with height h. Which has greatest speed at the bottom of the ramp?

A) Hoop B) Same C) Cylinder 22% 30%

48%

I = MR2 I = ½ MR2

Rolling Down an Incline

i i f fK U K U

1 2 22

1

2mgh mv I

2

1 2 2122

vmgh mv mr

r

v gh

2Hollow Cylinder : ; I mr v gh

1 422 3

Solid Cylinder: ; I mr v gh

2 1025 7

Solid Sphere: ; I mr v gh

0 0

2Hollow Cylinder : I mr

1 2 2122

2

gh v v

gh v

Compare HeightsA ball is released from rest on a no-slip surface, as shown. After reaching the 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 as its 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 lower height.

Spinning WheelA block of mass m is attached to a string that is wrapped around the circumference of a wheel of radius R and moment of inertia I, initially rotating with angular velocity that causes the block to rise with speed v . The wheel rotates freely about its axis and the string does not slip.

To what height h does the block rise?

i fE E1 1 1 1 12 2 2 2 2 22 2 2 2 2

( / ) (1 / )iE mv I mv I v R mv I mR fE mgh

2

21

2

v Ih

g mR

A Bowling Ball A 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 to roll 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 2cm cm2 2

0 0f f i i i iU K U K Mgh Mv I

2

1 1 2 72 2 2cm cm cm 22 2 5 10

ii i

vMgh Mv MR Mv

R

2 2cm

2

7 7(2.0 m/s)0.29 m

10 10(9.8 m/s )iv

hg

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