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When a disc is rotating with angular velocity $\omega$, a particle situated at a distance of $4 \mathrm{~cm}$ just begins to slip. If the angular velocity is doubled, at what distance will the particle start to slip?
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Verified Answer
The correct answer is:
$1 \mathrm{~cm}$
Angular velocity $=\omega$ Centripetal force $F=m r \omega^{2}$
or $\quad r \propto \frac{1}{\omega^{2}}$
$\therefore$
$$
\frac{r_{1}}{r_{2}}=\frac{\omega_{2}^{2}}{\omega_{1}^{2}}
$$
or
$$
\frac{4}{r_{2}}=\frac{4 \omega^{2}}{\omega^{2}}
$$
or $r_{2}=1 \mathrm{~cm}$
or $\quad r \propto \frac{1}{\omega^{2}}$
$\therefore$
$$
\frac{r_{1}}{r_{2}}=\frac{\omega_{2}^{2}}{\omega_{1}^{2}}
$$
or
$$
\frac{4}{r_{2}}=\frac{4 \omega^{2}}{\omega^{2}}
$$
or $r_{2}=1 \mathrm{~cm}$
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