Question

A uniform rod of mass m and length l is hinged at its upper end. It is released from a horizontal position. When it becomes vertical, what force does it exert on the hinge?

• A. 3/2 mg
• B. 2 mg
• C. 5/2 mg
• D. mg

Answer 1

Answer: (C)

5/2 mg

Answer 2

N – mg = $\frac{mv^{2}}{r}$ ………….. (i)

N = mg + $\frac{m(1/2\omega)^{2}}{1/2} = mg + \frac{3}{2}mg = \frac{5}{2}mg$

$\frac{1}{2}I\omega^{2} = mg\frac{1}{2}$

$\frac{1}{2}\frac{ml^{2}}{3}\omega^{2} = mg\frac{1}{2}$

Or $\frac{mv^{2}}{l} = 3mg$ ………… (ii)