1soooooo long and still no reply..........
ne one???
if the system is slightly disturbed the rod starts falling down in the direction of the block,
form equations to find velocity of block at the instant it leaves contact with the ball
(i have solved this one....but just posted for others)
(hint-frame all equations possible first)
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7 Answers
24eqns..
At angle @ from vertical. (clockwise)
T+Nsin@+mgcos@=m.ac
&& mgsin@-Ncos@=m.at
24Also constraint eqn.
accos@+a=atsin@
a=acc of block
and one energy eqn too
mgl(cos@-1)+1/2 Iω2+1/2Mv2=0
24so what its massless ????
its a rigid rod anyways...so there will be force exerted by it..net force obviously zero
1my equations(see it only after u have solved it)
let the rod make angle @ with the horizontal when ball leaves contact
and angular acceleration be a
and angular velocity be w
then
0=lasin@-l(w^2)cos@............1
mg(l-lsin@)=(m(l^2))(w^2)/2 + M(v^2)/2..........2
mglcos@=m(l^2)a.........3
v=lwsin@..............4
p.s. dont cheat
1@ eureka
wat u have to understand in the first one is that after it leaves contact with the block the velocity of x direction becomes 0............