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 on which the block moves in smooth but the function coefficient

between the wall and the block is u . If the initial speed of the block is Vo then what will be it's speed after one revolution?​

Answer 1

as v decrease with s, dv/ds is -ve

Answer 2

$\displaystyle\Large\boxed {\sf {\quad(a)\quad\!V_0\cdot e^{-2\pi\mu}\quad}}$

Here the block experiences a retardation due to the friction on the wall, and therefore,

$\displaystyle\longrightarrow\sf{a=-\dfrac {f}{m}}$

where $\displaystyle\sf {f}$ is the frictional force which is known that,

$\displaystyle\longrightarrow\sf{f=\mu\,N}$

where $\displaystyle\sf {N}$ is the reaction acting on the block by the walls, which provides necessary centripetal force to the block.

$\displaystyle\longrightarrow\sf{N=\dfrac {mv^2}{R}}$

Therefore,

$\displaystyle\longrightarrow\sf{f=\dfrac {\mu\,mv^2}{R}}$

And hence the acceleration becomes,

$\displaystyle\longrightarrow\sf{a=-\dfrac {\mu\,v^2}{R}\quad\quad\dots (1)}$

But,

$\displaystyle\longrightarrow\sf{a=\dfrac {dv}{dt}}$

$\displaystyle\longrightarrow\sf{a=\dfrac {dv}{dx}\cdot\dfrac {dx}{dt}}$

$\displaystyle\longrightarrow\sf{a=v\cdot\dfrac {dv}{dx}}$

Then (1) becomes,

$\displaystyle\longrightarrow\sf{v\cdot\dfrac {dv}{dx}=-\dfrac {\mu\,v^2}{R}}$

$\displaystyle\longrightarrow\sf{\dfrac {dv}{dx}=-\dfrac {\mu\,v}{R}}$

$\displaystyle\longrightarrow\sf{\dfrac {dv}{v}=-\dfrac {\mu}{R}\ dx}$

Integrating both sides with,

• velocity from $\displaystyle\sf {v=V_0}$ to $\displaystyle\sf {v=V.}$

• position from $\displaystyle\sf {x=0}$ to $\displaystyle\sf {x=2\pi R}$

$\displaystyle\longrightarrow\sf{\int\limits_{V_0}^{V}\dfrac {dv}{v}=\int\limits_0^{2\pi R}-\dfrac {\mu}{R}\ dx}$

$\displaystyle\longrightarrow\sf{\left [\ln|v|\right]_{V_0}^V=-\dfrac {\mu}{R}\int\limits_0^{2\pi R}dx}$

$\displaystyle\longrightarrow\sf{\ln|V|-\ln\left|V_0\right|=-\dfrac {\mu}{R}\big [x\big]_0^{2\pi R}}$

$\displaystyle\longrightarrow\sf{\ln\left|\dfrac {V}{V_0}\right|=-2\pi\mu}$

$\displaystyle\longrightarrow\sf{\dfrac {V}{V_0}=e^{-2\pi\mu}}$

$\displaystyle\longrightarrow\sf{\underline {\underline {V=V_0\cdot e^{-2\pi\mu}}}}$