A homogeneous rod AB of length L = 1.8 m and mass M is pivoted at the center O in such a way that it can rotate it can rotate freely position. An insect S of the same mass M falls vertically with speed V on the point C, midway between the points O and B. Immediately after falling, the insect moves towards the end B such that the rod rotates with a constant angular velocity omega.
(a) Determine the angular velocity `omega` in terms of V and L.
(b) If the insect reaches the end B when the rod has turned through an angle of `90^@`, determine V.

Answer

A, B, C

Explanation

(a) Let us consider the system of homogeneous rod and insect and apply conservation of angular momentum during
collision about the point O.
Angular momentum of the system before collision = angular
momentum of the system after collision.
` Mvxx(L)/(4) =I omega`
Where I is the moment of inertia of the system just after
collision and `omega` is the angular velocity just after collision.
`rArr Mv = (L)/(4) = [M((L)/(4))^2 +(1)/(12) ML^2]omega`
`rArr Mvxx(L)/(4) = (ML^2)/(4) [(1)/(4) + (1)/(3)]omega = (ML^2)/(4) [(3+4)/(12)]`
`=(MKL^2)/(4)xx(7)/(12)xx omega rArr omega = (12)/(7) (v)/(L).
(b) Note: initially the torque due to mass OB of the rod
(acting in clockwise direction) was balanced by the torque
due to mass OA of the rod (acting in anticlockwise direction).
But after collision there is an extre mass M of the insect which creates a torque in the clockwise direction, which tends to creates a torque in the clockwise direction, which tends to create angular acceleration in the rod. But the is compensated by the movement of insect towards B due
to which moment of inertia I of the system increases.
Let at any instant of time t the insect be at a disctance x from
the centre of the rod and the rod has turned through an
angle `theta ( = omegat)` w.r.i its original position.
Instantaneous torque,

`tau = (dL)/(dt) = (d)/(dt)(I omega)`
`=omega(dl)/(dt)`
`=omega(d)/(dt)[(1)/(12) ML^2 + Mx^2]`
`=2M omega x (dx)/(dt) .....(i)`
This torque is balanced by the torue due to weight of
insect.
`tau` = Force xx perpendicular distance of force with exis fo
rotation = Mg xx (OM)
`=Mg (xcos theta)`
From (i) and (ii)
`2M omegax(dx)/(dt) = Mg (xcos theta) rArr dx = (g)/(2omega) cos omega dt`
On integration, taking limits
`int_(L//4)^(L//2) dx = (g)/(2omega)int_0 ^(pi//2omega) cos omega t dt`
when `x = (L)/(4) omega t = 0`
`[x]_(L//4)^(L//2) = dx = (g)/(2omega^2) [sin omega t]_0 ^(pi//2omega)`
when `x = (L)/(2), omega t = (pi)/(2)`
`rArr ((L)/(2) - (L)/(4)) = (g)/(2omega^2) [ sin(pi)/(2) - sin 0]`
`rAr (L)/(4) = (g)/(2omega^2) rArr omega = sqrt((2g)/(L))`
But `omega = (12)/(7)(v)/(L) rArr (12)/(7)(v)/(L) = sqrt((2g)/(L)) rArr v = (7)/(12) sqrt(2gL)`
`rArr v = (7)/(12) sqrt(2xx10xx1.8 = 3.5 ms^(-1)`