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If KE of the particle of mass $m$ performing UCM in a circle of radius $r$ is $E$. Find the acceleration of the particle
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2903 Upvotes
Verified Answer
The correct answer is:
$\frac{2 E}{m r}$
Kinetic energy
$$
\begin{aligned}
\frac{1}{2} m v^{2}=E \\
\frac{1}{2} m r \frac{v^{2}}{r} &=E \\
\frac{1}{2} m r a &=E \\
a &=\frac{2 E}{m r}
\end{aligned}
$$
$$
\begin{aligned}
\frac{1}{2} m v^{2}=E \\
\frac{1}{2} m r \frac{v^{2}}{r} &=E \\
\frac{1}{2} m r a &=E \\
a &=\frac{2 E}{m r}
\end{aligned}
$$
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