Rotational Dynamics: Torque, Inertia, and Kinetic Energy, Slides of Physics

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Rotational Dynamics

Rotational Dynamics

• Newton’s First Law: a rotating body will continue to rotate at constant angular velocity as long as there is no torque acting on it.
• Picture a grindstone on a smooth axle.
• BUT the axle must be exactly at the center of gravity— otherwise gravity will provide a torque, and the rotation will not be at constant velocity!

• A

Newton’s Second Law for Rotations

• For the special case of a mass m constrained

by a light disk to circle around an axle, the

angular acceleration α is proportional to the

torque τ exactly as in the linear case the

acceleration a is proportional to the force F :

• τ = mr^2 α F = ma

The angular equivalent of inertial mass m is the

moment of inertia mr^2.

More Complicated Rotating Bodies

• Suppose now a light disk has several different masses attached at different places, and various forces act on them. As before, radial components cause no rotation, we have a sum of torques.
• BUT the rigidity of the disk ensures that a force applied to one mass will cause a torque on the others!
• How do we handle that?

• A

F 1

r (^1) m 1

m 2

F 2

Moment of Inertia of a Solid Body

• Consider a flat square plate rotating about a perpendicular axis with angular acceleration α. One small part of it, Δ mi , distance r (^) i from the axle, has equation of motion
• Adding contributions from all parts of the wheel
• I is the Moment of Inertia.

• Z

ext int 2

τ i = τ i + τ i = ∆ m r i i α

Δ mi

ext 2 i i i i i

τ τ m r α I α

Calculating Moments of Inertia

• A thin hoop of radius R (think a bicycle wheel) has all the mass distance R from a perpendicular axle through its center, so its moment of inertia is
• A uniform rod of mass M , length L , has moment of inertia about one end - v

2 2 i i i

I = ∑∆ m r = MR

2 1 2 3 0

( / )

L

I = ∫ x M L dx = ML

R

x

L

Mass of length dx of rod is ( M / L ) dx

dx

Parallel Axis Theorem

• If we already know I CM about some line through the CM (we take it as the z - axis), then I about a parallel line at a distance h is
• I = I CM + Mh^2
• Here’s the proof:^ Moment of inertia I about perpendicular axis through A
• We prove it for a 2D object—the proof in 3D is exactly the same, taking the line through the CM as the z -axis.

CM at O

dmi

y

x

A h



r i  ri ′ 

2 2

2 2

CM^2

2

(Since 0.)

i i^ i^ i i^ i

i i^ i^ i i^ i

i i^ i

I m r m r h

m r h m r Mh

I Mh m r

= = ′+

= ′^ + ⋅ ′+

= + ′ =

 ^ 

 ^  



Clicker Question

We found the moment of inertia of a rod about a perpendicular line through one end was. Use the parallel axis theorem to figure out what it is about a perpendicular line through the center of the rod.

A

B

C

D

E

1 2 3 ML

1 2 3 ML 7 2 12 ML 1 2 2 ML 1 2 4 ML 1 2 12 ML

Clicker Question

Given that the moment of inertia of a disk about its axle is , use the perpendicular axis theorem to find the moment of inertia of a disk about a line through its center and in its plane.

A

B

C

1 2 2 MR

1 2 2 MR 1 2 4 MR MR^2

Rotational Kinetic Energy

• Imagine a rotating body as composed of many

small masses m i at distances r i from the axis of

rotation.

• The mass mi has speed v = ωr (^) i , so KE = ½ mi r (^) i^2 ω^2.
• The total KE of the rotating body (assuming the

axis is at rest) is

( (^12) i i^2 )^2 i

K = (^) ∑ m r ω = I ω