A wheel is rotating at 480 rpm. Find the magnitude of angular acceleration required to stop the wheel in 8 s.
A wheel is rotating at 480 rpm. Find the magnitude of angular acceleration required to stop the wheel in 8 s.
- $2 \pi \mathrm{rad} \mathrm{s}^{-2}$
- $2.5 \pi \mathrm{rad} \mathrm{s}^{-2}$
- $2 \mathrm{rad} \mathrm{s}^{-2}$
- $3.5 \mathrm{rad} \mathrm{s}^{-2}$
Solution
Given, initial angular speed, $\omega_i=480 \mathrm{rpm}$
$=\frac{480}{60}=8 \mathrm{rps}=8 \times 2 \pi=16 \pi \mathrm{rad} \mathrm{s}^{-1}$
Time, t = 8s
Final angular speed, $\omega_f=0$
From first equation of motion,
$
\begin{array}{rlrl}
\omega_f & =\omega_i+\alpha t \\
\Rightarrow & \frac{\omega_f-\omega_i}{t} & =\alpha \Rightarrow \alpha=\frac{0-16 \pi}{8} \\
\therefore & \alpha & =-2 \pi \mathrm{rad} \mathrm{s}^{-2}
\end{array}
$
Therefore, magnitude of angular acceleration,
$
|\alpha|=2 \pi \mathrm{rads}^{-2}
$
Asked in: AP EAMCET 2021 (25 Aug Shift 2)
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