A 6 kg grinding wheel of radius 0.2 m rotates at a constant rotational frequency of 4 rad/s when an object makes contact with the outer edge of the wheel. Friction causes the wheel to stop in 2 seconds. What is the average torque on the wheel by the frictional force?
Torque = (Moment of Inertia)*(Angular deceleration)
The moment of inertia is
I = (1/2)*M*R^2 = 3*(0.2)^2
= 0.12 kg*m^2
Angular deceleration = (4 rad/s)/2s = 2 rad/s^2
Torque = 0.12 * 2 = 0.24 Newton-meter
Is actualy A 6 kg grinding wheel of radius 0.2 m rotates at a constant rotational frequency of 4(pi) rad/s when an object makes contact with the outer edge of the wheel. Friction causes the wheel to stop in 2 seconds. What is the average torque on the wheel by the frictional force?
ANS = 0.754
.754
To calculate the average torque on the wheel by the frictional force, we need to use the equation:
Torque = Moment of Inertia x Angular Acceleration
First, let's find the moment of inertia of the grinding wheel. The moment of inertia of a cylindrical object rotating about its central axis is given by the equation:
Moment of Inertia = (1/2) x mass x radius^2
Plugging in the values, we have:
Moment of Inertia = (1/2) x 6 kg x (0.2 m)^2
Moment of Inertia = 0.12 kg·m^2
Next, we need to find the angular acceleration of the wheel. The equation for angular acceleration is:
Angular Acceleration = (Final Angular Velocity - Initial Angular Velocity) / Time
Given that the final angular velocity is 0 rad/s (as the wheel comes to a stop) and the initial angular velocity is 4 rad/s, and the time is 2 seconds, we have:
Angular Acceleration = (0 rad/s - 4 rad/s) / 2 s
Angular Acceleration = -2 rad/s^2
Now we can calculate the torque:
Torque = 0.12 kg·m^2 x -2 rad/s^2
Torque = -0.24 N·m
The negative sign indicates that the torque is acting in the opposite direction of the initial angular velocity. Therefore, the average torque on the wheel by the frictional force is 0.24 N·m.