Question

A block of mass $m$ is constrained to move inside a ring of radius $R$ which is fixed on a horizontal smooth table. At time $t=0$, block is moving along the inside of the ring (i.e., in the tangential direction) with velocity $v_0$. The coefficient of friction between the block and the ring is $\mu$. Then the speed of the block at time $t$ is

• A. $\frac{v_0}{R+\mu v_{0}t}$
• B. $\frac{v_0}{(1+\frac{\mu v_{0}t}{R})}$
• C. $\frac{v_0}{1-\frac{\mu v_{0}t}{R}}$
• D. $\frac{v_{0}R}{1+\mu v_{0}t}$

Answer 1

The correct option is B $\frac{v_0}{(1+\frac{\mu v_{0}t}{R})}$

At some instant $t$,

$f_k=-ma$
$\mu N=-ma(N=\frac{m{v}^2}{R})$
$\mu \left(\frac{m{v}^2}{R}\right)=-ma$
$\Rightarrow a=-\frac{\mu v^2}{R}$

$\frac{dv}{dt}=-\frac{\mu v^2}{R}$

$\frac{dv}{v^2}=-\frac{\mu }{R}dt$

Integrating,
${\int }_{v_0}^v\frac{dv}{v^2}=-\frac{\mu }{R}{\int }_0^{t}dt$

$[-\frac{1}{v}{]}_{v_0}^v=-\frac{\mu }{R}t$

$\frac{1}{v}-\frac{1}{v_0}=\frac{\mu t}{R}$

$\frac{1}{v}=\frac{\mu t}{R}+\frac{1}{v_0}$

$\frac{1}{v}=\frac{v_0\mu t+R}{v_{0}R}$

$v=\frac{v_0}{1+\frac{\mu v_{0}t}{R}}$