Question

A block of mass $m$ moves on a horizontal circular path against the wall of a cylindrical room of radius $R$. The floor of the room on which the block moves is smooth, but the friction coefficient between the wall and the block is $\mu$. Take ${\mu }_s={\mu }_k=\mu$. The block is given an initial speed $v_{\circ }$. Obtain the speed of the block '$v$' after one revolution.

• A. $v_{\circ }{e}^{-\pi \mu }$
• B. $v_{\circ }{e}^{\frac{-\pi \mu }{2}}$
• C. $v_{\circ }{e}^{\frac{-3\pi \mu }{2}}$
• D. $v_{\circ }{e}^{-2\pi \mu }$

Answer 1

The correct option is D $v_{\circ }{e}^{-2\pi \mu }$

In this case the block is moving on a horizontal circular path of radius $R$, against a cylindrical wall.


From equation of dynamics towards centre :
$N=\frac{m{v}^2}{R}$ ......($1$)


$a_t=$ tangential acceleration
$v_{\circ }$= initial velocity of block
$v=$ velocity at the end of $1$ revolution
Sliding is happening along cylindrical wall
$\therefore f_{max}=\mu N$
$\Rightarrow f_{max}=\frac{\mu m{v}^2}{R}$ (from (1))

and tangential acceleration ($a_t$) will be opposite to that of velocity direction.
$\therefore a_t=\frac{-f_{max}}{m}=$$\frac{-\mu v^2}{R}$

Expressing acceleration in the form of $v\frac{dv}{dx}$ and integrating with suitable limits;
$v\frac{dv}{dx}$=$\frac{-\mu v^2}{R}$
$\Rightarrow {\int }_{v_o}^v\frac{1}{v}dv=\frac{-\mu }{R}{\int }_0^{2\pi R}dx$
($\because$ at the end of one revolution, distance covered is $x=2\pi R$)
$\Rightarrow \ln v-\ln v_o=-2\mu \pi$

On solving:
$\ln \frac{v}{v_{\circ }}$= $-2\mu \pi$
or $v=v_{\circ }{e}^{-2\mu \pi }$