A) \[\frac{3}{2}mg\]
B) 3 mg
C) 5 mg
D) \[\frac{5}{2}mg\]
Correct Answer: D
Solution :
[d]: As the rod rotates about A, therefore, from conservation of mechanical energy, decrease in potential energy = increase in rotational kinetic energy about A \[mg\left( \frac{l}{2} \right)=\frac{1}{2}{{I}_{A}}{{\omega }^{2}}=\frac{1}{2}\left( \frac{m{{l}^{2}}}{3} \right){{\omega }^{2}}\]or\[{{\omega }^{2}}=\frac{3g}{l}\] Centripetal force of centre of mass of the rod in this position is \[=mr{{\omega }^{2}}=m\frac{l}{2}\frac{3g}{l}=\frac{3mg}{2}\]. If F is the force exerted by the hinge on the rod (upwards), then \[F-mg=\frac{3mg}{2}\] \[F=\frac{3mg}{2}+mg=\frac{5}{2}mg\].
You need to login to perform this action.
You will be redirected in
3 sec
