The solution is fairly simple..
Assume the impulse delivered at the hinge to be zero. Due to the impulse I, the gong acquires an angular velocity w..
Newton's law abt. the hinge gives:
I(x)=(MK^2+MR^2)w.........(1).
Note that
MK^2 is the moment of inertia abt. the COM.
I(x) is the angular impulse abt. the hinge.
The same abt. the center of the gong gives:
I(x-R)=(MK^2)w..........(2)
Solve these to get x = [1+(K^2/R^2)] times the radius..
Some of you may appreciate the result..i'll post some cool similar questions someday.
wont it be I(x-R) in 2nd eqn?
ReplyDeletemy mistake..this happens when you solve it online!!
ReplyDeleteis there be MK^2w IN SECOND STEP
ReplyDeletesorry people..i'm making too many mistakes..won't type directly now-on!
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteWHERE IS CONDITION OF ZERO IMPULSE AT THE HINGE APPLICATION. PLEASE EXPLAIN SIR?!!!!!!!!!
ReplyDeleterohan
ReplyDeleteI(x-R)=(MK^2)w
agar hinge pe impulse F hoti to ye eqn hoti
I(x-R)-FR=(MK^2)w
why the minus sign comes as the impulse direction on hinge and on the gong is of sane direction. ANKUSH SACHDEVA or any other please tell about the above post.
ReplyDeletealso ankush please see my post a brick lying on an inclined plane problem and please reply it.
ReplyDeleteu can take that + also,if ur answer comes negative it means u took the opp direction.
ReplyDeletei hope i m rite