Question

A uniform rod of mass $m$ length $L$ is sliding along its length on a horizontal table whose top is partly smooth and rest roughly with friction coefficient $\mu$. If the rod after moving through the smooth part enters the rough with velocity $v_0$
(i) What will be the magnitude of the friction force when its $x$ length ($<L$) lies in the rough part during sliding.
(ii) Determine the minimum velocity $v_0$ with which it must enter so that it lies completely in a rough region before coming to rest.
(iii) If the velocity is double the minimum velocity as calculated in part (a) then what distance does its front end $A$ would have travelled in a rough region before rod comes to rest.

Answer 1

i)Weight of length $x$ is

$W_1=\frac{x}{2}mg$

frictional force due to length is

$f=\mu W_1=\frac{\mu mg}{L}x$

ii) Now, work done by frictional force with moving distance $dx$ is

$dw=f{a}_x$

$dW=f.dx$

$W=\int f.dx={\int }_0^L\frac{\mu mg}{L}xdx=\frac{\mu mg}{L}[\frac{x^2}{2}{]}_0^L=\frac{\mu mgL}{2}$

Now, Kinetic energy minimum required to enter completely in rough region is ;

$K.W=W=\frac{1}{2}m{v}_0^2$

$\frac{1}{2}m{v}_0^2=\frac{\mu mgL}{2}$

$v_0=\sqrt{\mu gL}$

iii) If $v=2v_0$ then,

K.E = Work done by friction to enter and travel some $y$ distance

$\frac{1}{2}m{v}^2=\mu mgy+W$

$\frac{1}{2}m(2v_0{)}^2=\mu mgy+\frac{\mu mgL}{2}$

$\frac{1}{2}\times 4(\sqrt{\mu gL}{)}^2=\mu mgy+\frac{\mu mgL}{2}$

$\frac{4}{2}\mu gL=\mu mgy+\frac{\mu mgL}{2}$

$2L=y+\frac{L}{2}$

$y=\frac{3L}{2}$

Hence A travel distance $=L+y=L+\frac{3L}{2}=\frac{5L}{2}$