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Question

The object will slip if centripetal force $\geq$ force of

friction

${\operatorname{mr} \omega^{2} \geq \mu m g} $

${r \omega^{2} \geq \mu g}$

Vedclass · Accepted Answer

$1\,cm$

Vedclass · Answer

The object will slip if centripetal force $\geq$ force of

friction

${\operatorname{mr} \omega^{2} \geq \mu m g} $

${r \omega^{2} \geq \mu g}$

$\mathrm{r} \omega^{2} \geq \mathrm{constant}, \mathrm{or}\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}ight)=\left(\frac{\omega_{2}}{\omega_{1}}ight)^{2}$

$\frac{4 \mathrm{cm}}{\mathrm{r}_{2}}=\left(\frac{2 \omega}{\omega}ight)^{2} \quad 	herefore \mathrm{r}_{2}=1 \mathrm{cm}$

A small object placed on a rotating horizontal turn table just slips when it is placed at a distance $4\, cm$ from the axis of rotation. If the angular velocity of the turn-table is doubled, the object slips when its distance from the axis of rotation is

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