A meter rod, free to spin abt a stationary vertical axis passing through one of its ends is kept on a rough table of coeff. of friction .5, and spun with an initial angular velocity of 1 rad/sec.
Whats the time taken for it to stop rotating???
Now the rod is replaced by the following object:
A sector (pizza-share like) of a circle of radius 1m and of very small angular width..(say .5 degrees, which means you neglect it).
Again find the time....assume axis passes thru the would-be center of the circle..
Finally do the same for a meter rod whose linear mass density varies as the nth power of the distance from the axis.
for the rod one I am getting t=4/(3g)
ReplyDeleteg=acc. due to gravity
bhaiya please check the previos posts also and post their solns like that of the repose theorem.
the general expression is
ReplyDelete(n+2)/(5(n+3))
put n=0,n=1(for second part)
to get the answer
ans of first is t=3/4g
ReplyDeletebhaiya plz tell tht tape vala ques (its really a good one)
for the first
ReplyDelete4/3g
second
3/2g
third
2(n+2)/(n+3)g
someone of you four is right.
ReplyDeleteI am also getting all answers the same as ankush.
ReplyDeletebhaiya i substituted g as 10.
ReplyDeletemine and ankush's answers are the same