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Question

Physics Question on work, energy and power

A thin uniform rod of mass m and length is hinged at the lower end to a level floor and strands vertically. It is now allowed to fall, then its upper end will strike the floor with the velocity

A

2gl \sqrt{ 2gl }

B

5gl \sqrt{ 5gl }

C

3gl \sqrt{ 3gl }

D

mgl \sqrt{ mgl }

Answer

3gl \sqrt{ 3gl }

Explanation

Solution

The kinetic energy at the point Q is given by = 121ω2=12ml23v2l2\frac{1}{2} 1 \omega^2 = \frac{1}{2} \frac{ ml^2 }{ 3 } \frac{ v^2 }{ l^2 } = 12×13mv2\frac{1}{2} \times \frac{1}{3 } mv^2 \hspace20mm ..(i) The potential energy at G =12mgl= \frac{1}{2} mgl...(ii) From Eqs. (i) and (ii), we get 12mv23=12mgl\frac{1}{2} \frac{mv^2 }{ 3} = \frac{1}{2} mgl v = 3gl \sqrt{ 3gl }