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?
To find the average torque on the wheel by the frictional force, we can use the following formula:
τ = I * α
Where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
First, let's calculate the moment of inertia (I) of the wheel. The moment of inertia of a solid disk rotating about its central axis is given by the formula:
I = (1/2) * m * r^2
Where m is the mass of the wheel and r is its radius.
Substituting the given values into the formula, we get:
I = (1/2) * (6 kg) * (0.2 m)^2 = 0.12 kg·m^2
Next, let's calculate the angular acceleration (α) of the wheel. Since the wheel stops rotating in 2 seconds, the final angular velocity (ωf) is 0 rad/s. We can use the following kinematic equation to find the angular acceleration:
ωf = ωi + α * t
Where ωi is the initial angular velocity, α is the angular acceleration, and t is the time.
Since the initial angular velocity (ωi) is given as 4 rad/s, we can solve for α:
0 rad/s = 4 rad/s + α * 2 s
Rearranging the equation, we get:
α * 2 s = -4 rad/s
α = -2 rad/s^2
Now that we have the mass of inertia (I) and the angular acceleration (α), we can calculate the torque (τ):
τ = I * α
τ = 0.12 kg·m^2 * (-2 rad/s^2) = -0.24 N·m
Therefore, the average torque on the wheel by the frictional force is -0.24 N·m. The negative sign indicates that the torque is in the opposite direction to the initial rotation of the wheel.