Q.Coefficient of static friction between a coin and a gramophone disc is 0.5. Radius of the disc is 8 cm. Initially the centre of the coin is 2 cm away from the centre of the disc. At what minimum frequency will it start slipping from there? By what factor will the answer change if the coin is almost at the rim? (use g = $\pi^2$ m/s^2)

Slipping starts when $\omega^2r=\mu_sg$; at r = 2 cm this gives $n_{min}=2.5$ rev/s, and at r = 8 cm (4x farther) the minimum frequency is reduced by a factor $1/\sqrt4=1/2$, i.e. $n_2/n_1=1/2$. …

The coin starts slipping once the required centripetal force exceeds the maximum available static friction: $m\omega^2r>\mu_smg \Rightarrow \omega^2>\dfrac{\mu_sg}{r}$. At r = 2 cm = 0.02 m, using $g=\pi^2$ m/s^2: $\omega_{min}^2=\frac{\mu_sg}{r}=\frac{0.5\times\pi^2}{0.02}=25\pi^2 \Rightarrow \omega_{min}=5\pi\text{ rad/s}$ $n_{min}=\frac{\omega_{min}}{2\pi}=\frac{5\pi}{2\pi}=2.5\text{ rev/s}$ …

Set the required centripetal force equal to the maximum available static friction ($m\omega^2r=\mu_smg$) to find the minimum angular speed (hence frequency) at which slipping begins, then note the inverse-square …

Assuming the minimum frequency stays the same regardless of the coin's distance from the centre, forgetting that the centripetal force requirement depends on r; or incorrectly assuming a …