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A particle of mass $m$ is moving in a horizontal circle of radius $r$ under a centripetal force given by $\left(\frac{-K}{r^2}\right)$, where $K$ is a constant. Then
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Verified Answer
The correct answer is:
the total energy of the particle is $\left(\frac{-K}{2 r}\right)$
The potential energy is given by
$$
\begin{aligned}
U & =\int F \cdot d r=\int \frac{K}{r^2} d r=K \int r^{-2} d r \\
& =K\left[\frac{r^{-1}}{-1}\right]=\frac{-K}{r}
\end{aligned}
$$
The kinetic energy is given by
$$
K=\frac{1}{2} m v^2=\frac{1}{2}\left(\frac{K}{r}\right) \quad\left[\because \frac{m v^2}{r}=\frac{K}{r^2}\right]
$$
Total energy, $E=U+K$
$$
=\frac{-K}{r}+\frac{K}{2 r}=\frac{-K}{2 r}
$$
$$
\begin{aligned}
U & =\int F \cdot d r=\int \frac{K}{r^2} d r=K \int r^{-2} d r \\
& =K\left[\frac{r^{-1}}{-1}\right]=\frac{-K}{r}
\end{aligned}
$$
The kinetic energy is given by
$$
K=\frac{1}{2} m v^2=\frac{1}{2}\left(\frac{K}{r}\right) \quad\left[\because \frac{m v^2}{r}=\frac{K}{r^2}\right]
$$
Total energy, $E=U+K$
$$
=\frac{-K}{r}+\frac{K}{2 r}=\frac{-K}{2 r}
$$
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