JEE Main Solved Paper 2017 Question 16
Question: A slender uniform rod of mass M and length $ \ell $ 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 [JEE Main Solved Paper-2017]
Options:
A) $ \frac{3g}{2\ell }\cos \theta $
B) $ \frac{2g}{3\ell }\cos \theta $
C) $ \frac{3g}{2\ell }\sin \theta $
D) $ \frac{2g}{3\ell }\sin \theta $
Show Answer
Answer:
Correct Answer: C
Solution:
- Taking torque about pivot $ \tau =I\alpha $ $ mg\sin \theta \frac{\ell }{2}=\frac{m{{\ell }^{2}}}{3}\alpha $ $ \alpha =\frac{3g}{2\ell }\sin \theta $
