1. R OTATIONAL M OTION III Torque, Angular Acceleration and Rotational Energy

2. N EWTON ’ S 2 ND L AW & R OTATIONAL M OTION The net force on a particle is proportional to its TANGENTIAL acceleration. The net torque on a particle is proportional to its ANGULAR acceleration.

3. R OTATIONAL I NERTIA I = ∑mr 2 Rot. Inertia = ∑ masses of particles x radius 2 Smaller radius  Lower Rotational Inertia Larger radius  Higher Rotational Inertia

4. E XAMPLE The motor of an electric saw brings the circular blade up to the rated angular speed of 80 rev/s in 240 rev. If the rotational inertia of the blade is 1.41 x 10 -3 kg m 2, what net torque must the motor apply to the blade?

5. R OTATIONAL W ORK The rotational work W R done by a constant torque τ in turning an object through an angle θ is W R = τθ Θ must be in radians; Unit – Joule (J) The rotational work done by a net external torque equals the change in rotational kinetic energy Therefor the formulas for rotational work and energy are analogous to the translational formulas

6. R OTATIONAL W ORK AND K INETIC E NERGY RotationalTranslational Work:W =  W = Fd = Fx Kin. Energy:K = ½ I  2 K = ½ mv 2 Power:P =  P = Fv Work-Energy Theorem: W =  K = ½ I  2 - ½ I  o 2 W = ½ mv 2 - ½ mv o 2

7. R OLLING B ODIES The kinetic energy of a rolling body (without slipping) relative to an axis through the contact point is the sum of the rotational kinetic energy about an axis through the center of mass and the translational kinetic energy of the center of mass. K = ½ I CM  2 + ½ mv CM 2 total = rotational + translational KE KE + KE If the rolling body experiences a change in height then potential energy ( mgh ) must also be included

8. C ONCEPT C HECK

9. E XAMPLE A 1 kg cylinder with a rotational inertia of 0.0625 kg m 2 rolls without slipping down a one meter high incline. At the bottom of the incline the cylinder’s translational speed is 3.13 m/s. What is the cylinder’s angular velocity?