Q151 / 180 · KCET 2022 · Physics
The angular speed of a motor wheel is increased from $1200 \mathrm{rpm}$ to $3120 \mathrm{rpm}$ in $16 \mathrm{~s}$. The angular acceleration of the motor wheel is
- Correct answerA$4 \pi \mathrm{rad} / \mathrm{s}^2$
- B$6 \pi \mathrm{rad} / \mathrm{s}^2$
- C$8 \pi \mathrm{rad} / \mathrm{s}^2$
- D$2 \pi \mathrm{rad} / \mathrm{s}^2$
Solution
Given, initial angular frequency of wheel,
$$
f_0=1200 \mathrm{rpm}=\frac{1200}{60} \mathrm{rps}=20 \mathrm{rps}
$$
$\therefore$ Initial angular velocity,
$$
\omega_0=2 \pi f_0=2 \pi \times 20=40 \pi \mathrm{rad} / \mathrm{s}
$$
Similarly, final angular speed,
$$
\omega=2 \pi \times\left(\frac{3120}{60}\right) \mathrm{rad} / \mathrm{s}=104 \pi \mathrm{rad} / \mathrm{s}
$$
Time, $\quad t=16 \mathrm{~s}$
If $\alpha$ be the angular acceleration of the wheel, then
$$
\begin{aligned}
\omega & =\omega_0+\alpha t \\
\Rightarrow \quad \alpha & =\frac{\omega-\omega_0}{t}=\frac{104 \pi-40 \pi}{16} \\
& =\frac{64 \pi}{16}=4 \pi \mathrm{rad} / \mathrm{s}^2
\end{aligned}
$$
Hence, the correct answer is option: 'A'.