Question

A particle of mass $1\text{g}$ executes an oscillatory motion on the concave surface of a spherical dish of radius $2\text{m}$. 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 $1.0\text{m}$

The particle will keep oscillating on the concave surface till its potential energy $\text{mgh}$ ( at a height $\text{h}$ from horizontal ) is lost in work against frictional force.

$F.B.D$ of mass:


The value of kinetic fricton $f_k$ acting on mass is

$f_k=\mu N=\mu mgcos\theta$

Since the height $\text{h}=1\text{cm}$ is very less compared to radius of the concave surface ( $R=2$ $\text{m}$ ) $h<<<R$

This $\theta$ is very small angle. $\theta \approx 0^0$,

$f_k=\mu mgcos\theta =\mu mg$

From work energy theorem,

$W_{ext}=\Delta K.E$

$W_N+W_g+W_f=(K.E{)}_f-(K.E{)}_i=0-0=0$

or

$0+mgh+[-(f_k\left(d\right)]=0$ ...........(1)

Here $\text{d}$ = distance covered by particle on concave surface, and considering the small portion travelled friction is assumed constant & tangential at all points on the path.

$\therefore W_f=-f_{k}d=-\mu mgd$

From equation $\left(1\right)$, $mgh=-\mu mgd$

$\Rightarrow .d=\frac{h}{\mu }=\frac{1}{0.01}$

or $\text{d}=100\text{cm}=1\text{m}$

Why this question? since $\text{h}<<<\text{L}$, we can assume friction to be constant and the curved length can be visualised as straight path for very small curvature i.e. $\theta \approx 0^0$.