The frequency of revolution of a particle performing circular motion changes from $60 \mathrm{rpm}$ to $180 \mathrm{rpm}$ in 20 seconds. Calculate the angular acceleration of the particle.
Q 32.10
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Data $: \mathrm{f}_1=60 \mathrm{rpm}=\frac{60}{60} \mathrm{rev} / \mathrm{s}=1 \mathrm{rev} / \mathrm{s}, \mathrm{f}_2=180 \mathrm{rpm}=\frac{180}{60} \mathrm{rev} / \mathrm{s}=3 \mathrm{rev} / \mathrm{s}, \mathrm{t}=20 \mathrm{~s}$ The angular acceleration in SI units,
$
\begin{aligned}
\alpha & =\frac{\omega_2-\omega_1}{t}=\frac{2 \pi f_2-2 \pi f_1}{t}=\frac{2 \pi(3)-2 \pi(1)}{20} \\
& =\frac{4 \pi}{20}=\frac{\pi}{5}=\frac{3.14}{5}=0.628 \mathrm{rad} / \mathrm{s}^2
\end{aligned}
$
OR
Using non SI units, the angular frequencies are $\omega_1=60 \mathrm{rpm}=1 \mathrm{rps}$ and $\omega_2=180 \mathrm{rpm}=3$ rps.
$
\therefore \alpha=\frac{\omega_2-\omega_1}{t}=\frac{3-1}{20}=\frac{1}{10}=0.1 \mathrm{rev} / \mathrm{s}^2 .
$
$
\begin{aligned}
\alpha & =\frac{\omega_2-\omega_1}{t}=\frac{2 \pi f_2-2 \pi f_1}{t}=\frac{2 \pi(3)-2 \pi(1)}{20} \\
& =\frac{4 \pi}{20}=\frac{\pi}{5}=\frac{3.14}{5}=0.628 \mathrm{rad} / \mathrm{s}^2
\end{aligned}
$
OR
Using non SI units, the angular frequencies are $\omega_1=60 \mathrm{rpm}=1 \mathrm{rps}$ and $\omega_2=180 \mathrm{rpm}=3$ rps.
$
\therefore \alpha=\frac{\omega_2-\omega_1}{t}=\frac{3-1}{20}=\frac{1}{10}=0.1 \mathrm{rev} / \mathrm{s}^2 .
$
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