First slide

Rolling motion

Question

A solid sphere of mass m and radius r is released from rest from the given position. If it rolls without sliding on the circular track of radius R, its speed when it reaches its lowest position will be

Difficult

A

$\sqrt{\frac{8}{7} g (R - r)}$
B

$\sqrt{\frac{10}{7} g (R - r)}$
C

$\sqrt{\frac{10}{9} g (R - r)}$
D

$\sqrt{\frac{5}{7} g (R - r)}$

Solution

When the sphere reaches its lowest position it loses gravitational potential energy by $∆$ U and gains a kinetic energy by $∆$ K, where $∆$ U = -mg(R - r)
$\begin{array}{l} and Δ K = \frac{1}{2} m v_{c}^{2} (1 + \frac{K^{2}}{R^{2}}) = \frac{1}{2} m v^{2} (1 + \frac{2}{5}) = \frac{7}{10} m v^{2} \\ According to conservation of energy, Δ U + Δ K = 0 \\ or - m g (R - r) + \frac{7}{10} m v^{2} = 0 \\ or v = \sqrt{\frac{10}{7} g (R - r)} \end{array}$

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