A particle of mass 1 g executes an oscillatory motion on the concave surface of a spherical dish of radius 2 m placed on a horizontal plane. If the motion of the particle begins from a point on the dish at a height of 1 cm from the horizontal plane and the coefficient of friction is 0.01, the total distance covered by the particle before It comes to rest, is approximately

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10.0 m

2.0 m

1.0 m

20.0 m

answer is C.

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Detailed Solution

Since, the particle is released from a small height, $θ$ (angle of radius with vertical) will be very small. Force of friction throughout the journey can be assumed to be $μ m g$ . Particle will finally come to rest when whole of its energy ( $= m g h$ ) is lost in the work done against friction. Let particle stops after travelling a distance d. Then,

$\begin{matrix} μ m g d = m g h \\ o r d = \frac{h}{μ} = \frac{10^{- 2}}{0.01} = 1.0 m \end{matrix}$