Question

A torque of 50.0 N-m is applied to a grinding wheel (I=20.0kg-m²) for 20 s. If it starts from rest, what is the angular velocity of the grinding wheel after the torque is removed? Through what angle does the wheel move while the torque is applied?

a) 25.0 rad/s, 500.0 rad
b) 30.0 rad/s, 600.0 rad
c) 35.0 rad/s, 700.0 rad
d) 40.0 rad/s, 800.0 rad

Answer 1

The angular velocity of the grinding wheel after the torque is removed is 50.0 rad/s and the wheel moves through an angle of 500.0 rad while the torque is applied.

Step-by-step explanation:

To find the angular velocity of the wheel after the torque is removed, we can use the equation:

torque = moment of inertia x angular acceleration

Given that the torque is 50.0 N-m and the moment of inertia is 20.0 kg-m², we can rearrange the equation to solve for angular acceleration:

angular acceleration = torque / moment of inertia

Substituting the given values, we get:

angular acceleration = 50.0 N-m / 20.0 kg-m² = 2.5 rad/s²

Since the wheel starts from rest, its initial angular velocity is 0 rad/s. To find the final angular velocity, we can use the equation:

final angular velocity = initial angular velocity + (angular acceleration x time)

Substituting the given values, we get:

final angular velocity = 0 rad/s + (2.5 rad/s² x 20 s) = 50.0 rad/s

Therefore, the angular velocity of the grinding wheel after the torque is removed is 50.0 rad/s.

To determine the angle through which the wheel moves while the torque is applied, we can use the equation:

angle = initial angular velocity x time + (0.5 x angular acceleration x time²)

Substituting the given values, we get:

angle = 0 rad/s x 20 s + (0.5 x 2.5 rad/s² x (20 s)²) = 500.0 rad

Therefore, the wheel moves through an angle of 500.0 rad while the torque is applied.