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ROTATIONAL MOTION III Torque, Angular Acceleration and Rotational Energy NEWTON’S 2ND LAW & ROTATIONAL MOTION The net force on a particle is proportional to its TANGENTIAL acceleration. The net torque on a particle is proportional to its ANGULAR acceleration. ROTATIONAL INERTIA I = ∑mr2 Rot. Inertia = ∑ masses of particles x radius2 Smaller radius Lower Rotational Inertia Larger radius Higher Rotational Inertia EXAMPLE 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 m2, what net torque must the motor apply to the blade? ROTATIONAL WORK The rotational work WR done by a constant torque τ in turning an object through an angle θ is WR = 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 ROTATIONAL WORK AND KINETIC ENERGY Work: Rotational Translational W = W = Fd = Fx Kin. Energy: K = ½ I 2 Power: P = Work-Energy Theorem: W = K = ½ I 2 - ½ I o2 K = ½ mv2 P = Fv W = ½ mv2 - ½ mvo2 ROLLING BODIES 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 ICM 2 + ½ mvCM2 total = rotational + translational KE KE + KE If the rolling body experiences a change in height then potential energy (mgh) must also be included CONCEPT CHECK EXAMPLE A 1 kg cylinder with a rotational inertia of 0.0625 kg m2 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? EXAMPLE A very common problem is to find the velocity of a ball rolling down an inclined plane. It is important to realize that you cannot work out this problem they way you used to. In the past, everything was SLIDING. Now the object is rolling and thus has MORE energy than normal. So let’s assume the ball is like a thin spherical shell and was released from a position 5 m above the ground. Calculate the velocity at the bottom of the Ebefore Eafter incline. U g KT K R Ebefore Eafter If you HAD NOT included the U g KT mgh 1 mv 2 1 I 2 2 2 rotational kinetic 2 2 1 2 mgh mv energy, you see the v R I sphere@ cm mR 3 2 answer is very much 2 2 1 v 2 2 mgh mv different. mgh 1 mv 1 ( 2 mR )( 2 ) 2 2 2 3 R 1 v2 2 2 gh 1 1 gh v v 2 2 3 v 6 gh 6 (9.8)(5) 7.67 m/s 5 5 v 2 gh 2(9.8)(5) 9.90 m / s