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

Answer

$\alpha \,is\,propotional\,to\,\omega $

$Let = \alpha  = k\omega $                       ($k$ is constant)

$\frac{{d\omega }}{{dt}} = k\omega \,\,\,\,\,\,\,\,\,\left[ {also\,\frac{{d\theta }}{{dt}} = \omega  \Rightarrow dt = \frac{{d\theta }}{\omega }} \right]$

$\therefore \frac{{\omega d\omega }}{{d\theta }} = k\omega  \Rightarrow d\omega  = kd\theta $

$Now\,\int\limits_\omega ^{\omega /2} {d\omega  = k\int {d\theta } } $

$\int\limits_{\omega /2}^0 {d\omega  = k\int\limits_0^\theta  {d\theta  \Rightarrow  - \frac{\omega }{2} = k\theta  \Rightarrow  - \frac{\omega }{2} = K{\theta _1}} } $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\theta _1} = 2\pi n} \right)$

$\therefore \theta  = {\theta _1}\,\,or\,\,2\pi {n_1} = 2\pi n$

${n_1} = n$

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