Question

A small mass 'm' rests at the edge of a horizontal disc of radius 'R' The coefficient of static friction 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 ' the meters below. What was horizontal distance of travel of mass?

• A. $\sqrt{\mu h}$
• B. $\sqrt{\mu (R+h{)}^2}$
• C. $\sqrt{\mu Rh}$
• D. $\sqrt{2\mu Rh}$

Answer 1

The correct option is D $\sqrt{2\mu Rh}$

$\begin{array}{c}t=\sqrt{\frac{2h}{g}}\\ m{w}^{2}R=\mu mg\\ =w=\sqrt{\frac{\mu g}{R}}\\ v=wR=\sqrt{R\mu g}\\ Range=\sqrt{R\mu g}\times \sqrt{\frac{2h}{g}}\\ =\sqrt{2\mu Rh}\end{array}$

Hence,

option $\left(D\right)$ is correct answer.