Question

A uniform rod of mass m and length l is fixed from point A, which is at a distance l/4 from one end as shown in the fig. The rod is free to rotate in a vertical plane. The rod is released from the horizontal position.
What is the reaction at the hinge, when kinetic energy of the rod is maximum?

• A. $\frac{4}{7}$
• B. $\frac{5}{7}mg$
• C. $\frac{13}{7}mg$
• D. $\frac{11}{7}mg$

Answer 1

Answer: (C)

$\frac{13}{7}mg$

The kinetic energy is maximum when the becomes vertical

When the rod become verticle then CM will fall down by $l/4$

Using energy conservation:

$mg\frac{l}{4}=\frac{I{\omega }^2}{2}$

where I is about A :

$I=\frac{m{l}^2}{12}+\frac{m{l}^2}{16}=\frac{7m{l}^2}{48}$

$mg\frac{l}{4}=\frac{7m{l}^2}{96}{\omega }^2$

${\omega }^2\frac{l}{4}=\frac{6g}{7}$

Now the using the FBD of the rod

$R=mg+\frac{m{\omega }^{2}l}{4}⟹R=\frac{13mg}{7}$

Hence Option C is correct