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.

Options

A. `3/7sqrt(2gL)`
B. `7/12 sqrt(2gL)`
C. `1/12sqrt(gL)`
D. `2/7sqrt(2gL)`

Answer

B

Explanation

A angular momentum of system of rod and insect, just after collision =initial angular momentum of insect about `O`
`:. {(ML^(2))/12+M(L/4)^(2)}omega=MupsilonL/4`

`omega=(12v)/(7L)`
Let the insect fall on the rod at time `t=0`, then at time `t`, inclination of rod with horizontal, `theta=omegat`
`:.` Angular momentum
`L=Iomega=((ML^(2))/12+Mx^(2)) omega`
Torque produced by weight of insect, `tau=Mg x cos theta=Mg x cos omegat`
`tau =d/(dt), L F=mu N :. mu=0.2`