A small mass $'m'$ rests at the edge of a horizontal disc of radius $'R'$ . The coefficient of static friction between 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 $'h'$ meters below. What was its horizontal distance of travel from the point it left the disc?
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$\frac{{m{v^2}}}{R} = \mu mg;\,v = \sqrt {\mu Rg} \,\therefore t = \sqrt {\frac{{2h}}{g}} $
Horizontal distance = $vt$ = $\sqrt {2\mu Rh} $
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