Question

A homogeneous Rod xy of length L and mass M is pivoted at the centre C such that it can rotate freely in vertical plane initially the rod is in the horizontal position a blob of wax of same mass M as that of rod falls vertically with the speed V And sticks to the rod midway between the points C and Y if the rod rotates with angular speed womega what will be angular speed in terms of V andL

Answer 1

Answer:

The answer will be 12V/7L

Explanation:

According to the problem the mass of the rod is M and the length of the rod is L

It is rotating with respect to its center.

Now in this problem as there is no external momentum is conserved,

Therefore,

L(i) = MVL/4 [ where V is the velocity of the particle and L/4 is the distance of the point from center to the rod]

L(f) = Ix omega = (ML^2/12 +M(L/4)^2) x omega)  = 7ML^2/48 omega

MVL/4 = 7ML^2/48 omega

=>omega = 12V/7L