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Q.
A wheel rotating at $12$ revolution per second is brought to rest in $6\,s$. The average angular deceleration in $rad/s^2$ of the wheel during this process is
KEAMKEAM 2017System of Particles and Rotational Motion
Solution:
$W_{1}^{0}=12 \times 2 \pi \frac{ rad }{s} $
$=24 \pi \frac{ rad }{s} $
$W_{f} =0 $
$\Delta t =6 s $
As $W_{f} =W_{i}+\alpha \Delta t $
We have, $\alpha =\frac{-W_{i}}{\Delta t} $
$=-\frac{24 \pi}{6} $
$=-4 \pi \frac{ rad }{ Sec ^{2}} $