A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of +49 N·m is applied to the wheel for 16 s, giving the wheel an angular velocity of +685 rev/min. The external torque is then removed, and the wheel comes to rest 120 s later. (Include the sign in your answers.)
a. Find the moment of inertia of the wheel.
b. Find the frictional torque, which is assumed to be constant.
To find the moment of inertia of the wheel, we can use the equation:
τ = Iα
where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
In this scenario, we know that a constant external torque of +49 N·m is applied to the wheel, and it reaches an angular velocity of +685 rev/min after 16 s. We need to convert the angular velocity to radians per second since the SI unit for torque is in N·m.
1 revolution = 2π radians. Therefore, +685 rev/min = +685 * 2π radians/60 s = +716.35 radians/s.
The equation for angular acceleration is:
α = (ωf - ωi) / t
where ωf is the final angular velocity, ωi is the initial angular velocity, and t is the time.
Substituting the values, we have:
α = (+716.35 radians/s - 0 radians/s) / 16 s = +44.77 radians/s^2
Now, we can use the equation τ = Iα to find the moment of inertia:
49 N·m = I * 44.77 radians/s^2
Solving for I, we get:
I = 49 N·m / (44.77 radians/s^2) = 1.093 kg·m^2
Therefore, the moment of inertia of the wheel is 1.093 kg·m^2.
Now let's move on to finding the frictional torque.
To find the frictional torque, we need to use the equation:
τ = Iα
At the moment when the external torque is removed, the only torque acting on the wheel is the frictional torque. Since the wheel comes to rest after 120 s, the angular acceleration becomes zero, and therefore the torque is also zero.
0 N·m = I * 0 radians/s^2
Solving for I, we find that the frictional torque is also zero.
Therefore, the frictional torque is 0 N·m.