Question

A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of 0.0700 m, an angular speed of 88.0 rad/s, and a moment of inertia of 0.850 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 two during a time of 4.40 s.(a) Find the magnitude of the angular deceleration of the cylinder. rad/s 2 (b) Find the magnitude of the force of friction applied by the brake shoe. N

Answer 1

The magnitude of the angular deceleration of the cylinder is -10.0 rad/s^2, and the magnitude of the force of friction applied by the brake shoe is -8.50 N.

Step-by-step explanation:

(a) To find the magnitude of the angular deceleration of the cylinder, we can use the formula:

angular deceleration = (final angular velocity - initial angular velocity) / time

Given that the initial angular speed is 88.0 rad/s, the final angular speed is half of that, and the time is 4.40 s, we can calculate:

angular deceleration = (44.0 rad/s - 88.0 rad/s) / 4.40 s = -10.0 rad/s^2

(b) To find the magnitude of the force of friction applied by the brake shoe, we can use the formula:

force of friction = moment of inertia * angular deceleration

Given that the moment of inertia is 0.850 kg·m², and the angular deceleration is -10.0 rad/s², we can calculate:

force of friction = 0.850 kg·m² * -10.0 rad/s² = -8.50 N