Question

A fly wheel rotates about an axis. Due to friction at the axis, it experiences angular retardation proportional to its angular velocity. If its angular velocity falls to half the value while it makes $n$ revolutions, how many more revolutions will it make before coming to rest ?

• A. $2 n$
• B. $n$
• C. $n / 2$
• D. $n / 3$

Answer 1

Answer: (B)

Given that $\alpha=-k \omega$
$\omega \frac{d \omega}{d \theta}=-k \omega$
On integrating we get
$\frac{d \omega}{d \theta} \cdot \frac{d \theta}{d t} =-k \frac{d \theta}{d t} $
$\Rightarrow \omega=-k \theta+C$
Initially $\theta =0, \omega=\omega_{0}$
$\omega_{0} =C$
After $n$ revolution,
$\frac{\omega_{0}}{2}=-k(2 n \pi)+\omega_{0}$
$\Rightarrow \quad \frac{\omega_{0}}{2}=2 k n \pi$
$k=\frac{\omega_{0}}{4 n \pi}\,\,\,...(1)$
For final stop, $\omega=0$
$0=\frac{-\omega_{0}}{4 n \pi}(\theta)+\omega_{0}$
$\theta=4 n \pi$
i.e. $2 n$ revolution,
Number of revolutions that will further occur before flywheel stops
$=2 n-n$
$=n$