Question

A flywheel of a motor has mass 100 kg and radius 1.5 m. The motor develops a constant torque of 2000 Nm. The flywheel starts from rest. Calculate the work done during the first 4 revolutions.

Hint:

This is the rotational version of Work = Force \(\times\) distance. Here, Work = Torque \(\times\) angular distance. Remember to always convert revolutions to radians (\(1 \text{ rev} = 2\pi \text{ rad}\)) before calculating.

Answer 1

The work done (\(W\)) by a constant torque (\(\tau\)) is given by the formula: \[ W = \tau \cdot \theta \] where \(\theta\) is the angular displacement in radians.
Given:
Torque, \(\tau = 2000 \, Nm\)
Number of revolutions = 4
First, convert the number of revolutions into radians for the angular displacement \(\theta\).
One revolution corresponds to an angle of \(2\pi\) radians. \[ \theta = 4 \, \text{revolutions} \times 2\pi \, \frac{\text{rad}}{\text{revolution}} = 8\pi \, \text{rad} \] Now, calculate the work done: \[ W = 2000 \, Nm \times 8\pi \, \text{rad} = 16000\pi \, J \] Using \(\pi \approx 3.14159\): \[ W \approx 16000 \times 3.14159 \approx 50265.48 \, J \] The work done during the first 4 revolutions is \(16000\pi\) Joules, or approximately 50265 J. (Note: The mass and radius of the flywheel are not needed for this calculation as the torque is given directly).