When a ceiling fan is switched off its angular velocity reduces to $50\%$ while it makes $ 36$ rotations. How many more rotation will it make before coming to rest (Assume uniform angular retardation)
$18$
$12$
$36$
$48$
When a ceiling fan is switched off its angular velocity reduces to $50\%$ while it makes $ 36$ rotations. How many more rotation will it make before coming to rest (Assume uniform angular retardation)
By using equation ${\omega ^2} = \omega _0^2 - 2\alpha \theta $
${\left( {\frac{{{\omega _0}}}{2}} \right)^2} = \omega _0^2 - 2\alpha (2\pi n)$
$⇒$ $\alpha = \frac{3}{4}\frac{{\omega _0^2}}{{4\pi \times 36}}$, $(n = 36)$..$(i)$
Now let fan completes total $n'$ revolution from the starting to come to rest
$0 = \omega _0^2 - 2\alpha (2\pi n')$ $⇒$ $n' = \frac{{\omega _0^2}}{{4\alpha \pi }}$
substituting the value of from equation $(i)$
$n' = \frac{{\omega _0^2}}{{4\pi }}\frac{{4 \times 4\pi \times 36}}{{3\omega _0^2}} = 48$ revolution
Number of rotation $= 48 -36 = 12$
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