Question

A homogeneous rod AB of length $L$ and mass $M$ is pivoted at the center O in such a way that it can rotate freely in the vertical plane. The rod is initially in the horizontal position. An insect S of the same mass $M$ falls vertically with speed $V$ on the point C mid-way between O and B and sticks to it. The initial angular velocity $\omega$ in terms of $V$ and $L$ is :

• A. $\frac{6V}{7L}$
• B. $\frac{12V}{7L}$
• C. $\frac{24V}{7L}$
• D. $\frac{V}{7L}$

Answer 1

The correct option is B $\frac{12V}{7L}$

Since there is no external torque angular momentum will be conserved.

$L_i=MV\frac{L}{4}$ where M is the mass of the insect, V is the velocity of the insect and $\frac{L}{4}$ is the distance of the point from center of the rod where the insect falls.

$L_f=I\omega =(\frac{M{L}^2}{12}+M(\frac{L}{4}{)}^2)\omega =\frac{7M{L}^2}{48}\omega$

$MV\frac{L}{4}=\frac{7M{L}^2}{48}\omega$

$\Rightarrow \omega =\frac{12V}{7L}$