A coin placed on a rotating table just slips | vedclass

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Question

The coin placed on the rotating table slips when inertial force $\operatorname{mr} \omega^{2}$ is greater than or equal to the static force of frictio

Vedclass · Accepted Answer

$r$

Vedclass · Answer

The coin placed on the rotating table slips when inertial force $\operatorname{mr} \omega^{2}$ is greater than or equal to the static force of friction $\mu m g .$ Mathematically,

$\operatorname{mr} \omega^{2} \geq \mu m g 	ext { or } r \geq \frac{\mu g}{\omega^{2}}$

The coin just slips when $\mathrm{r}=\frac{\mu \mathrm{g}}{\omega^{2}}$

For radii $r_{1}$ and $r_{2},$ let angular velocities be $\omega_{1}$ and $\omega_{2}$ respectively.

$\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}=\frac{\omega_{1}^{2}}{\omega_{2}^{2}}=\left(\frac{\omega_{1}}{\omega_{2}}ight)^{2}$

${\omega_{2}^{2}}=\left(\frac{\omega_{1}}{\omega_{2}}ight)^{2}$

Hence, $r_{1}=4 r,$ when $\omega_{1}=\omega$ and for $r_{2}, \omega_{2}=2 \omega$

$\frac{r_{2}}{4 r}=\left(\frac{\omega}{2 \omega}ight)^{2}=\frac{1}{4}$

$	herefore r_2=r$

A coin placed on a rotating table just slips if it is placed at a distance $4r$ from the centre. On doubling the angular velocity of the table, the coin will just slip when at a distance from the centre equal to

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