Question

A bicycle wheel, of radius \(0.30 \mathrm{m}\) and mass \(2 \mathrm{kg}\) (concentrated on the rim), is rotating at 4.00 rev/s. After 50 s the wheel comes to a stop because of friction. What is the magnitude of the average torque due to frictional forces?

Short answer

Answer: The magnitude of the average torque due to frictional forces is approximately 0.36 N·m.

Step-by-step solution

Determine the initial angular velocity in rad/s

First, we need to convert the initial angular velocity from rev/s to rad/s. As 1 revolution is equal to \(2\pi\) radians, we can convert the given initial angular velocity (4.00 rev/s) to rad/s as follows: \(\omega_i = 4.00\,\text{rev/s} \times 2\pi\,\mathrm{rad/rev} = 8\pi\,\mathrm{rad/s}\).

Calculate the final angular velocity

In this step, we will find the final angular velocity (\(\omega_f\)) of the wheel. As the wheel comes to a stop, the final angular velocity would be 0 rad/s. So, \(\omega_f=0\) rad/s.

Determine the angular acceleration

To find the angular acceleration (\(\alpha\)), we can use one of the angular kinematic equations: \(\omega_f = \omega_i + \alpha t\). Now, we plug in the given values and solve for \(\alpha\): \(0 = 8\pi\,\mathrm{rad/s} + \alpha (50\,\mathrm{s})\) \(\alpha = -\frac{8\pi}{50}\,\mathrm{rad/s^2}\).

Calculate the moment of inertia

The mass of the bicycle wheel is concentrated on the rim, so we can use the moment of inertia formula for a hoop which is \(I = m r^2\), where \(m\) is the mass and \(r\) is the radius of the wheel. \(I = (2\,\mathrm{kg})(0.30\,\mathrm{m})^2 = 0.18\,\mathrm{kg\cdot m^2}\).

Calculate the magnitude of the average torque

Now we can use the formula relating torque (\(\tau\)), moment of inertia, and angular acceleration: \(\tau = I \alpha\). Plug in the values for \(I\) and \(\alpha\) and find the magnitude of torque: \(\tau = (0.18\,\mathrm{kg\cdot m^2})(-\frac{8\pi}{50}\,\mathrm{rad/s^2})\). The magnitude of the average torque due to frictional forces is: \(|\tau| = 0.18 \times \frac{8\pi}{50} = \frac{36\pi}{100}\,\mathrm{N\cdot m} \approx 0.36\,\mathrm{N\cdot m}\).