Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
A slender uniform rod of mass $$M$$ and length $$l$$ is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle $$\theta$$ with the vertical is:
Net torque about the pivot on the rod = torque due to gravity
$$τ=Mg\times\ \frac{L}{2}\sin\theta\ $$
Also, $$τ=I\alpha\ $$
Moment of inertia for a rod about an axis passing through one of its ends is $$I=\frac{1}{3}ML^2$$
$$\therefore\ \frac{1}{3}ML^2\alpha\ =MG\frac{L}{2}\sin\theta\ $$
$$\therefore\ \alpha\ =\frac{3g}{2L}\sin\theta\ $$
Create a FREE account and get:
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.