rotational

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"

10 Answers

1
niteen9128 ·

what shud b the approach for this problem??

23
qwerty ·

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

23
qwerty ·

vinitial = vo

vfinal = v=rwfinal

winitial = 0

wfinal= v/r

just use work energy theorem now

1
niteen9128 ·

the answer is
(1) 2Vo/3
(2) Vo/3μg
W=-μmg[ Vot-0.5μgt2 ]

23
qwerty ·

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

23
qwerty ·

\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]

1
niteen9128 ·

thanks.....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

23
qwerty ·

hmmm..so wich one u didnt get??

1
niteen9128 ·

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...

23
qwerty ·

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

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