Lecture 13:Lecture 13:Rotational Kinetic Energy Rotational Kinetic Energy

and Rotational Inertiaand Rotational Inertia

Review: Rotational Kinematics Rotational Kinetic Energy Rotational Inertia Torque Rotational Dynamics Equilibrium

Recall: Rotational KinematicsRecall: Rotational Kinematics

Angular Linearconstant

tαωω 0

0 021

2t t

constanta

v v at 0

x x v t at 0 021

2

And for a point at a distance R from the rotation axis:xT = RvT= RaT = R

Δθ2 220

x2a vv 220

Comment on Axes and SignsComment on Axes and Signs(i.e. what is positive and negative)(i.e. what is positive and negative)

Whenever we talk about rotation, it is implied that there is a rotation “axis”.

This is usually called the “z” axis (we usually omit the z subscript for simplicity).

Counter-clockwise (increasing ) is usuallycalled positive.

Clockwise (decreasing ) is usuallycalled negative. z

Rotational Kinetic Energy 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 = RMass has kinetic energy

» K = ½ M v2

» K = ½ M 2 r2

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

M

Kinetic Energy of Rotating DiskKinetic Energy of Rotating Disk Consider a disk with radius R and mass M,

spinning with angular frequency Each “piece” of disk has speed vi=ri

Each “piece” has kinetic energy » Ki = ½ mi v2

» = ½ mi 2 ri2

Combine all the pieces » Ki = ½ mi 2 r2 » = ½ ( mi ri

2) 2 » = ½ I 2

ri

I is called the Rotational Inertia

Rotational Inertia Rotational Inertia II 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 MotionKrot = ½ I 2 Rotational Motion

I miri2 (units kg-m2)

Note! Rotational Inertia depends on what axis you are spinning about (the ri in the equation).

Rotational Inertia TableRotational Inertia Table For objects with finite number of masses, use I = m r2. For “continuous” objects, use table below:

SummarySummary

Rotational Kinetic Energy Krot = ½ I 2

Rotational Inertia I = miri2

Energy is Still Conserved!

Pulley Example Pulley Example A solid disk wheel is fixed at its center and has a

string wrapped around it and then attached to a block. The mass of the pulley is 5 kg, the radius of the pulley is 0.2 m, and the mass of the block is 8 kg. After the block has fallen 1.5 m, how fast is the block moving and how fast is the disk rotating?

We will use conservation of energy:

Remember:» Ktrans = ½ m v2

» Krot= ½ I 2

» U = m g y

5 kg

8 kg

1.5 m

ffii UKUK

fffiii mgyImvmgyImv 2222

21

21

21

21

Pulley Example Pulley Example A solid disk wheel is fixed at its center and has a string

wrapped around it and then attached to a block. The mass of the pulley is 5 kg, the radius of the pulley is 0.2 m, and the mass of the block is 8 kg. After the block has fallen 1.5 m, how fast is the block moving and how fast is the disk rotating?

Initial Krot and Ktran is zero.

Final U is zero.

5 kg

8 kg

1.5 m

22

21

21

ffi Imvmgy

Pulley Example Pulley Example A solid disk wheel is fixed at its center and has a

string wrapped around it and then attached to a block. The mass of the pulley is 5 kg, the radius of the pulley is 0.2 m, and the mass of the block is 8 kg. After the block has fallen 1.5 m, how fast is the block moving and how fast is the disk rotating?

Remember:» I = ½ m R2 for a disk » v = R so = v/R

5 kg

8 kg

1.5 m

222

21

21

21

Rv

Rmvmgym fpfbib

Pulley Example Pulley Example A solid disk wheel is fixed at its center and has a

string wrapped around it and then attached to a block. The mass of the pulley is 5 kg, the radius of the pulley is 0.2 m, and the mass of the block is 8 kg. After the block has fallen 1.5 m, how fast is the block moving and how fast is the disk rotating?

R2 cancels from the last term.

Simplify:

5 kg

8 kg

1.5 m

22

41

21

fpfbib vmvmgym

pb

bf mm

gymv

2

4

Pulley Example Pulley Example A solid disk wheel is fixed at its center and has a

string wrapped around it and then attached to a block. The mass of the pulley is 5 kg, the radius of the pulley is 0.2 m, and the mass of the block is 8 kg. After the block has fallen 1.5 m, how fast is the block moving and how fast is the disk rotating?

Solve:

5 kg

8 kg

1.5 m

vf = 4.73 m/s

TorqueTorque Rotational effect of force. Tells how effective

force is at twisting or rotating an object.

= r F sin r Fperpendicular

Units: N-mSign: CCW rotation is positive

F

r

Work Done by TorqueWork Done by Torque

Recall W = F d cos

For a wheel:W = Ftangential s

= Ftangential r ( in radians)

W = P = W/t = /t

P

EquilibriumEquilibrium Conditions for Equilibrium:

F = 0 Translational Equilibrium (Center of Mass) = 0 Rotational Equilibrium

» May choose any axis of rotation…. But Choose Wisely!

Equilibrium ExampleEquilibrium Example A 50 kg diver stands at the end of a 4.6

m diving board. Neglecting the weight of the board, what is the force each pivot exerts on the diving board?

First, draw a FBD:

F1

F2

Fg

Equilibrium ExampleEquilibrium Example A 50 kg diver stands at the end of a 4.6

m diving board. Neglecting the weight of the board, what is the force each pivot exerts on the diving board?

Write down F = 0:

-F1 + F2 – Fg = 0-F1 + F2 – (50 kg)(9.8 m/s2) =

0Note: there are two unknowns so we need another equation…

Equilibrium ExampleEquilibrium Example A 50 kg diver stands at the end of a 4.6

m diving board. Neglecting the weight of the board, what is the force each pivot exerts on the diving board?

Pick a pivot point:

F1

F2

Fg

Equilibrium ExampleEquilibrium Example A 50 kg diver stands at the end of a 4.6

m diving board. Neglecting the weight of the board, what is the force each pivot exerts on the diving board?

Write down = 0:

F1 (0 m) + F2 (1.2 m) – (50 kg)(9.8 m/s2)(4.6 m) = 0

F2 (1.2 m) – (50 kg)(9.8 m/s2)(4.6 m) = 0

Equilibrium ExampleEquilibrium Example A 50 kg diver stands at the end of a 4.6

m diving board. Neglecting the weight of the board, what is the force each pivot exerts on the diving board?

Solve:

F2 (1.2 m) – (50 kg)(9.8 m/s2)(4.6 m) = 0

F2 = 1878 N

-F1 + F2 – (50 kg)(9.8 m/s2) = 0

-F1 + (1878 N) – (50 kg)(9.8 m/s2) = 0

F1 = 1388 N

SummarySummary

Torque is a Force that causes rotation = F r sin Work done by torque: W =

Equilibrium F = 0 = 0

» May choose any axis.