Question

A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of 0.0790 m, an angular speed of 80.0 rad/s, and a moment of inertia of 0.596 kg·m2. A brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of 5 during a time of 4.00 s. (a) Find the magnitude of the angular deceleration of the cylinder. (b) Find the magnitude of the force of friction applied by the brake shoe.

Answer 1

final angular velocity = initial angular velocity plus the product of angular acceleration and time

w = wo + at

( 1/2 ) wo = wo + at
- ( 1/2 ) wo = at
- ( 1/2 ) (80.0 rad / s ) = a ( 4.0s )
a = -10 rad /s2

b) Newton's Second Law, rotational form: Torque (force perpendicular to radius) is equal to the product of moment of inertia and angluar acceleration

Fr = I a
F ( .0790 m ) = ( .596kg m^2 ) ( -10 rad / s )
F = 75.4N