rotational

what shud b the approach for this problem??

basically 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

v_initial = v_o

v_final = v=rw_final

w_initial = 0

w_final= v/r

just use work energy theorem now

bymistakely i hav used wrong symbols, corrected it , see #4

\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]

hmmm..so wich one u didnt get??

while 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...

m preparing for jee 2010 [1]....