A ballet dancer spins about a vertical axis at $2.5 \pi \mathrm{rad} / \mathrm{s}$ with his arms outstretched. With the arms folded, the MI about the same axis of rotation changes by $25 \%$. Calculate the new speed of rotation in rpm.
Q 114.3
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Let $I_1, w_1$, and $f_1$, be the moment of inertia, angular velocity and frequency of rotation of the ballet dancer with arms outstretched, and $\mathrm{I}_2, \mathrm{w}_2$ and $\mathrm{f}_2$ be the corresponding quantities with arms folded.
Data : $\omega_1=2.5 \pi \mathrm{rad} / \mathrm{s}$
Since moment of inertia with arms folded is less than that with arms outstretched,
$
\begin{aligned}
I_2 & <I_1 \\
\therefore I_2 & =I_1-0.25 I_1=0.75 I_1=\frac{3}{4} I_1 \\
\omega_1 & =2 \pi f_1=2.5 \pi \\
\therefore f_1 & =\frac{2.5 \pi}{2 \pi}=\frac{5}{4} \mathrm{~Hz}
\end{aligned}
$
According to the principle of conservation of angular momentum, $I_1 \omega_1=I_2 \omega_2$
$
\therefore I_1\left(2 \pi f_1\right)=I_2\left(2 \pi f_2\right)
$
The new frequency of rotation is
$
\begin{aligned}
f_2=\frac{I_1 f_1}{I_2}=\frac{4}{3} \times \frac{5}{4}=\frac{5}{3} \mathrm{~Hz} & =\frac{5}{3} \times 60 \mathrm{rpm} \\
& =100 \mathrm{rpm}
\end{aligned}
$
Data : $\omega_1=2.5 \pi \mathrm{rad} / \mathrm{s}$
Since moment of inertia with arms folded is less than that with arms outstretched,
$
\begin{aligned}
I_2 & <I_1 \\
\therefore I_2 & =I_1-0.25 I_1=0.75 I_1=\frac{3}{4} I_1 \\
\omega_1 & =2 \pi f_1=2.5 \pi \\
\therefore f_1 & =\frac{2.5 \pi}{2 \pi}=\frac{5}{4} \mathrm{~Hz}
\end{aligned}
$
According to the principle of conservation of angular momentum, $I_1 \omega_1=I_2 \omega_2$
$
\therefore I_1\left(2 \pi f_1\right)=I_2\left(2 \pi f_2\right)
$
The new frequency of rotation is
$
\begin{aligned}
f_2=\frac{I_1 f_1}{I_2}=\frac{4}{3} \times \frac{5}{4}=\frac{5}{3} \mathrm{~Hz} & =\frac{5}{3} \times 60 \mathrm{rpm} \\
& =100 \mathrm{rpm}
\end{aligned}
$
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