Question

A wheel is rotating at a frequency \(f_0\) Hz about a fixed vertical axis. The wheel stops in \(t_0\) seconds, with constant angular deceleration. The number of turns covered by the wheel before it comes to rest is given by:

• A. \(f_0 t_0\)
• B. \(2f_0 t_0\)
• C. \(\dfrac{f_0 t_0}{2}\)
• D. \(\dfrac{f_0 t_0}{\sqrt{2}}\)

Hint:

For uniform angular deceleration, average angular velocity = half of initial angular velocity, so total revolutions = \(\frac{1}{2} f_0 t_0\).

Answer 1

The Correct Option is C

Step 1: Angular motion relationship. 
The wheel stops under uniform angular deceleration. \[ \omega = \omega_0 - \alpha t \] At rest, \(\omega = 0 \Rightarrow \alpha = \dfrac{\omega_0}{t_0}\). 
 

Step 2: Use rotational kinematic equation. 
\[ \theta = \omega_0 t_0 - \frac{1}{2}\alpha t_0^2 = \frac{1}{2}\omega_0 t_0 \] Since \(\omega_0 = 2\pi f_0\), number of revolutions \(N = \dfrac{\theta}{2\pi} = \dfrac{f_0 t_0}{2}\). 
 

Step 3: Conclusion. 
Hence, the wheel makes \(\dfrac{f_0 t_0}{2}\) revolutions before stopping.