Question

A uniform rod of mass m is hinged at its upper end. it is released from the horizontal position. When it reached the vertical position, determine the force does it exert on the hinge?

• A. $\frac{3mg}{2}$
• B. $2mg$
• C. $\frac{5mg}{2}$
• D. $\frac{7mg}{2}$

Answer 1

Answer: (C)

$\frac{5mg}{2}$

Given: A uniform rod of mass m is hinged at its upper end. it is released from the horizontal position. When it reached the vertical position,

To find the force it exert on the hinge

Solution:

As the rod is released it will start to fall downward due to its weight.

Let the hinge reaction be V along the vertical direction and H along the horizontal direction

By conservation of energy,

Decrease in Potential energy = increase in the rotational kinetic energy

So, $\frac{mgL}{2}=\frac{I{\omega }^2}{2}$

And $I=\frac{m{L}^2}{3}$ about the end i.e., at hinge point

So we have

$\frac{mgL}{2}=\frac{m{L}^2}{3}\times \frac{{\omega }^2}{2}⟹{\omega }^2=\frac{3g}{L}$

This is the angular velocity in the vertical position

As the center of mass of the rod is moving in a circular path of radius $\frac{L}{2}$

So we have,

$V-mg=\frac{m{\omega }^{2}L}{2}⟹V=mg+\frac{m\times \frac{3g}{L}\times L}{2}⟹V=\frac{5mg}{2}$

as there is no horizontal force so net horizontal hinge reaction, H=0