Question

A flywheel of mass 182 kg has an effective radius of \(0.62 \mathrm{m}\) (assume the mass is concentrated along a circumference located at the effective radius of the flywheel). (a) How much work is done to bring this wheel from rest to a speed of 120 rpm in a time interval of 30.0 s? (b) What is the applied torque on the flywheel (assumed constant)?

Short answer

(b) What is the constant applied torque on the flywheel? Answer: (a) The work done to bring the wheel from rest to 120 rpm in 30 seconds is approximately 2,192 J. (b) The constant applied torque on the flywheel is approximately 0.98 Nm.

Step-by-step solution

Convert angular speed to radians per second

First, we need to convert the given angular speed from rpm to radians per second. We are given 120 rpm and can use the following conversion factor: 1 rpm = \(\frac{2\pi}{60}\) rad/s Therefore, continuing the calculation: \(\omega = 120 \times \frac{2\pi}{60}= 2\pi\) rad/s

Calculate the flywheel's moment of inertia

Since the mass is concentrated along a circumference located at the effective radius of the flywheel, we can use the formula for the moment of inertia for a circular ring: \(I = m r^2\) Where \(m\) is the mass (182 kg), \(r\) is the effective radius (0.62 m) of the flywheel, and \(I\) is the moment of inertia. Plugging in the values: \(I = (182\,\text{kg}) (0.62\,\text{m})^2 = 69.93\, \text{kg} \cdot\text{m}^2\)

Calculate the angular acceleration

Now we have the moment of inertia and the final angular speed; we can find the angular acceleration (\(\alpha\)) using: \(\omega ^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)\) Since initially the flywheel is at rest, \(\omega_0 = 0\). Also, we are given the time taken to reach the final speed is 30s. So we can find the total angle covered as follows: \(\theta - \theta_0 = \omega_0t + \frac{1}{2}\alpha t^2\) After plugging in the values, we can solve for the angular acceleration: \(\alpha = \frac{2\pi}{(30\, s)^2} \approx 0.014 \, \text{rad} \cdot\text{s}^{-2}\)

Determine the work done

Now that we know the flywheel's angular acceleration, we can use the work-energy theorem to find the work done on the flywheel: \(W = \frac{1}{2} I (\omega^2 - \omega_0^2)\) The flywheel starts from rest, so \(\omega_0 = 0\): \(W = \frac{1}{2} (69.93\, \text{kg} \cdot\text{m}^2) (2\pi\, \text{rad} \cdot\text{s}^{-1})^2 \approx 2,192 \, \text{J}\)

Calculate the applied torque

Finally, we want to find the constant applied torque. Using the torque-angular acceleration relationship: \(\tau = I\alpha\) \(\tau = (69.93 \, \text{kg} \cdot\text{m}^2)(0.014 \, \text{rad} \cdot\text{s}^{-2}) \approx 0.98\, \text{N} \cdot\text{m}\) So, the applied torque on the flywheel is approximately 0.98 Nm. To summarize the answers: (a) The work done to bring the flywheel from rest to 120 rpm in 30 seconds is approximately 2,192 J. (b) The applied torque on the flywheel is approximately 0.98 Nm.