Question

A block of mass m moves on a horizontal circle against the wall of a cylindrical room of radius R. The floor of the room oil which the block MC/Veil so smooth but the friction coefficient between the wall and the block is p. The block is given an initial speed $V_0$. As a function of the speed v write (a) the normal force by the wall on the block, (b) the frictional force by the wall and (c) the tangential acceleration of the block. (d) Integrate the tangential acceleration $(\frac{dv}{dt}=v\frac{dv}{dt})$ to obtain the speed of the block after one revolution.

Answer 1

A block of mass 'm' moves on a horizontal circle against the wall of a cylindrical room of radius 'R'.

Friction coefficient between wall and the block is m.

(a) Normal reaction by the wall on the block is $=\frac{m{v}^2}{r}$

(b) $\therefore$ Frictional force by wall $=\frac{\mu m{v}^2}{r}$

(c) $=\frac{\mu m{v}^2}{R}=ma$

$\Rightarrow a=-\frac{\mu v^2}{R}$ (Deceleration)

(d) Now $\left(\frac{dv}{dt}\right)=v\left(\frac{dv}{ds}\right)=-\frac{\mu v^2}{R}$

$\Rightarrow ds=-\frac{r}{\mu }\frac{dv}{v}$

$\Rightarrow s=-\frac{R}{\mu }lnv+c$

At s = 0, $v=v_0$

Therefore, $c=\frac{R}{\mu }ln{v}_0$

So, $s=-\frac{R}{\mu }ln\frac{v}{v_0}$

$\Rightarrow \frac{v}{v_0}=e^{-\frac{\mu s}{R}}$

$\Rightarrow v=v_0{e}^{-\frac{\mu s}{R}}$

For, one rotation

$s=2\pi R$

So, $v=v_0{e}^{-2\pi \mu }$