Question

A small mass 𝑚 rests at the edge of a horizontal disc of radius $R$. The coefficient of static friction between the mass and the disc is $\mu$. The disc is rotated about its axis at an angular velocity such that the mass slides off the disc and lands on the floor ℎ metres below. The horizontal distance of travel of the mass is$\sqrt{kR\mu h}$. Find $k$.

Answer 1

Formula used: $F=M{\omega }^{2}R$, $T=\sqrt{\frac{2h}{g}}$
According to given situation,
Frictional force is balanced by centrifugal force.
$\therefore \frac{m{v}^2}{R}=\mu mg$
$v=\sqrt{\mu Rg}$
Time taken to reach the particle distance $h$,
$t=\sqrt{\frac{2h}{g}}$
Horizontal distance $=vt=\sqrt{2\mu Rh}$
$\therefore k=2$