A stone of mass $100 \mathrm{~g}$ attached to a string of length $50 \mathrm{~cm}$ is whirled in a vertical circle by giving it a velocity of $7 \mathrm{~m} / \mathrm{s}$ at the lowest point. Find the velocity at the highest point.
Q 60.6
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Data : $m=0.1 \mathrm{~kg}, \mathrm{r}=\mathrm{l}=0.5 \mathrm{~m}, \mathrm{v}_2=7 \mathrm{~m} / \mathrm{s}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$
The total energy at the bottom, $E_{\text {bot }}$
$
=\mathrm{KE}+\mathrm{PE}=\frac{1}{2} m v_2^2+0=\frac{1}{2}(0.1)(7)^2=2.45 \mathrm{~J}
$
The total energy at the top, $\mathrm{E}_{\text {top }}=\mathrm{KE}+\mathrm{PE}=\frac{1}{2} m v_1^2+\mathrm{mg}(2 \mathrm{r})$
$
\begin{aligned}
& =\frac{1}{2}(0.1) v_1^2+(0.1)(10)(2 \times 0.5) \\
& =0.05 v_1^2+1
\end{aligned}
$
By the principle of conservation of energy,
$
\begin{aligned}
& \quad E_{\text {top }}=E_{\text {bot }} \\
& \therefore 0.05 v_1^2+1=2.45 \\
& \therefore v_1^2=\frac{2.45-1}{0.05}=\frac{145}{5}=29 \\
& \therefore v_1=\sqrt{29}=5.385 \mathrm{~m} / \mathrm{s}
\end{aligned}
$
The total energy at the bottom, $E_{\text {bot }}$
$
=\mathrm{KE}+\mathrm{PE}=\frac{1}{2} m v_2^2+0=\frac{1}{2}(0.1)(7)^2=2.45 \mathrm{~J}
$
The total energy at the top, $\mathrm{E}_{\text {top }}=\mathrm{KE}+\mathrm{PE}=\frac{1}{2} m v_1^2+\mathrm{mg}(2 \mathrm{r})$
$
\begin{aligned}
& =\frac{1}{2}(0.1) v_1^2+(0.1)(10)(2 \times 0.5) \\
& =0.05 v_1^2+1
\end{aligned}
$
By the principle of conservation of energy,
$
\begin{aligned}
& \quad E_{\text {top }}=E_{\text {bot }} \\
& \therefore 0.05 v_1^2+1=2.45 \\
& \therefore v_1^2=\frac{2.45-1}{0.05}=\frac{145}{5}=29 \\
& \therefore v_1=\sqrt{29}=5.385 \mathrm{~m} / \mathrm{s}
\end{aligned}
$
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