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A particle moves in a circle of radius $5 \mathrm{~cm}$ with constant speed and time period $0.2 \pi \mathrm{s}$. The acceleration of the particle is
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Verified Answer
The correct answer is:
$5 \mathrm{~m} / \mathrm{s}^2$
Given, $r=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$ and $T=0.2 \pi \mathrm{s}$
We know that acceleration
$$
\begin{aligned}
a & =r \omega^2 \\
& =\frac{4 \pi^2}{T^2} r \\
& =\frac{4 \times \pi^2 \times 5 \times 10^{-2}}{(0.2 \pi)^2}=5 \mathrm{~ms}^{-2}
\end{aligned}
$$
We know that acceleration
$$
\begin{aligned}
a & =r \omega^2 \\
& =\frac{4 \pi^2}{T^2} r \\
& =\frac{4 \times \pi^2 \times 5 \times 10^{-2}}{(0.2 \pi)^2}=5 \mathrm{~ms}^{-2}
\end{aligned}
$$
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