a uniform disc of mass m and radius R is projected horizontally with velocity "Vo" on a rough horizontal surface so that it starts off with a purely sliding motion at t=0.After time "to" sec,it acquires a purely rolling motion
(1)--calculate the velocity of the center of mass of the disc at time "to"
(2)--Assuming coefficient of friction μ calculate "to"
ALSO,calculate the work done by the frictional force as a function of time and the total work done by it over a time "t" much longer than "to"
23basically if there is friction , then it will act in such a way that it tries to make the sliding motion into pure rolling motion , now here initially v>0
and w = 0 i.e omega
so friction will act such that it will decrease v and increase w such that v=rw ,
v=rw since in dis case pure rolling will occur if velocity of the bottom most point is zero . hence v - rw = 0
23vinitial = vo
vfinal = v=rwfinal
winitial = 0
wfinal= v/r
just use work energy theorem now
23\mu mg=ma
so\; a=\mu g
also,\mu mgR=I\alpha
\mu mgR=m\frac{R^{2}}{2}\alpha
\Rightarrow \alpha =\frac{2\mu g}{R}
v=v_{o}-\mu gt_{o}..............(1)
\omega =\alpha t_{o}=\frac{2\mu gt_{o}}{R }
\Rightarrow \omega R = v=2\mu gt_{o}................(2)
from 1 and 2
v=v_{o}-\frac{v}{2}
\Rightarrow v=\frac{2v_{o}}{3}
now try d oder questions [1]
1thanks.....evn i was able to solve it.......its a similar problem which u can solve by conservation of momentum as in H C Verma...similar solved eg is der in HC in the end of rotational mechanics
1while posting the question i din had its solution.....but later i got it....thanks for the help.....
bdw qwerty ..r u a student preparing for jee or wot??thanks again.....
i got two solutions for the same problem now...