When a disc is rotating with angular velocity $\omega$, a particle situated at a distance of 4 cm just begins to slip. If the angular velocity is doubled, at what distance will the particle start to slip?

Question

Dr. Abubacker · Accepted Answer

(a) Angular velocity $= \omega$
Centripetal force $F = mr\omega^{2}$
or $r \propto \frac{1}{\omega^{2}}$
$	herefore$ $\frac{r_{1}}{r_{2}} = \frac{\omega_{2}^{2}}{\omega_{1}^{2}}$
or $\frac{4}{r_{2}} = \frac{4\omega^{2}}{\omega^{2}}$
or $r_{2} = 1$ cm

Q. When a disc is rotating with angular velocity $\omega$, a particle situated at a distance of 4 cm just begins to slip. If the angular velocity is doubled, at what distance will the particle start to slip?

Explanation

Reviewed by Dr. Abubacker

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