Question

One end of a uniform rod having mass m and length $l$ is hinged. The rod is placed on a smooth horizontal surface and rotates on it about the hinged end at a uniform angular velocity $\omega$. The force exerted by the hinge on the rod has a horizontal component equal to :

• A. $m{\omega }^{2}l$
• B. Zero
• C. mg
• D. $\frac{1}{2}m{\omega }^{2}l$

Answer 1

The correct option is D $\frac{1}{2}m{\omega }^{2}l$

The center of mass of the rod lies at its geometric center and the center of mass revolves in a circle of radius $\frac{l}{2}$.

Hinge force is equal to the centripetal force acting on the rod.

Hinge force $F_H=F_c=mr{w}^2$

where $r=\frac{l}{2}$

$⟹F_H=\frac{ml{w}^2}{2}$