Q.A coin kept at a distance of 5 cm from the centre of a turntable of radius 1.5 m just begins to slip when the turntable rotates at a speed of 90 r.p.m. Calculate the coefficient of static friction between the coin and the turntable. [g = 9.8 m/s²].

Friction alone supplies the centripetal force on an unbanked circular track.

On a flat (horizontal, unbanked) circular track, the only horizontal force available to provide the centripetal force needed for circular motion is friction between the tyres and the road. For the vehicle to move safely without skidding, the required centripetal force must not exceed the maximum available static friction:

$\frac{mv^2}{r} \le f_{max} = \mu mg$ …

On a flat (unbanked) curve, friction alone supplies the centripetal force, so the minimum coefficient of friction needed is $v^2/(rg)$.

For a car negotiating a flat (unbanked) curve of radius $r$ at speed $v$, the necessary centripetal force $mv^2/r$ must be supplied entirely by friction between the tyres and the road. The maximum available frictional force is $f_{\max} = \mu mg$. For the car not to skid, friction must be at least equal to the required centripetal force: …

At the point of slipping, the required centripetal force equals the maximum available force of friction; equate the two to find $\mu_s$.

A coin of mass $m$ placed at radius $r=5\text{ cm}=0.05\text{ m}$ on a rotating turntable just begins to slip when the turntable's angular speed is $N=90$ r.p.m.

Angular velocity:

$\omega=\frac{2\pi N}{60}=\frac{2\pi\times 90}{60}=3\pi\text{ rad/s}\approx 9.42\text{ rad/s}$

At the point of slipping, the centripetal force required to keep the coin moving in a circle is provided entirely by static friction, at its maximum (limiting) value:

$f_{s,max}=mr\omega^2=\mu_s mg$

Cancelling $m$:

$\mu_s=\frac{r\omega^2}{g}$

Substituting values ($r=0.05$ m, $\omega=9.42$ rad/s, so $\omega^2\approx 88.83\ \text{rad}^2/\text{s}^2$, $g=9.8\ \text{m/s}^2$): …