Question

Suppose you exert a force of 180 N tangential to a 0.280-m-radius, 75.0-kg grindstone (a solid disk). What torque is exerted? What is the angular acceleration assuming negligible opposing friction? What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?

a) 50.4 N·m, 1.80 rad/s², 1.32 rad/s²
b) 62.4 N·m, 2.10 rad/s², 1.44 rad/s²
c) 75.6 N·m, 2.45 rad/s², 1.58 rad/s²
d) 89.2 N·m, 2.80 rad/s², 1.73 rad/s²

Answer 1

The torque exerted is 50.4 N·m. The angular acceleration assuming negligible opposing friction is 1.80 rad/s². The angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis is 1.32 rad/s².

Step-by-step explanation:

To calculate the torque exerted, we can use the formula:
Torque = Force × Radius.
Substituting the given values:
Torque = 180 N × 0.280 m = 50.4 N·m.

The angular acceleration assuming negligible opposing friction can be found using the formula:
Angular Acceleration = Torque / Moment of Inertia.
Given that the moment of inertia of a solid disk is (1/2) × Mass × Radius^2, we can substitute the known values:
Angular Acceleration = 50.4 N·m / ((1/2) × 75.0 kg × (0.280 m)^2) = 1.80 rad/s².

If there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis, the net torque would be:
Net Torque = Torque (due to force) - Torque (due to friction).
Substituting the given values:
Net Torque = (180 N × 0.280 m) - (20.0 N × 0.015 m) = 46.4 N·m.

The angular acceleration considering the opposing frictional force can be calculated using the formula:
Angular Acceleration = Net Torque / Moment of Inertia.
Substituting the known values:
Angular Acceleration = 46.4 N·m / ((1/2) × 75.0 kg × (0.280 m)^2) = 1.32 rad/s².