Question

A particle of mass $1\text{g}$ executes an oscillatory motion on the concave surface of a spherical dish of radius $2\text{m}$ placed on a horizontal plane. If the motion of the particle begins from a point on the dish at a height of $1\text{cm}$ 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

• A. $2.0\text{m}$
• B. $10.0\text{m}$
• C. $1.0\text{m}$
• D. $20.0\text{m}$

Answer 1

The correct option is C $1.0\text{m}$


Given:
$h=1\text{cm};R=2\text{m};\mu =0.01$

Since the particle is releases from a small height, $\theta$ (angle of radius with vertical) will be very small.

Force of friction throughout the journey can be assumed to be $\mu mg$ for small $\theta$. The particle will finally come to rest when the whole of its energy $(=mgh)$ is lost in the work done against friction.

Let the particle stops after travelling a distance $d$. Then,

$W_{friction}=mgh$

$\mu mgd=mgh$

$\therefore d=\frac{h}{\mu }=\frac{{10}^{-2}}{0.01}=1.0\text{m}$

Hence, option $\left(c\right)$ is correct.