Question

A uniform steel rod of mass $m$ and length $l$ is pivoted at one end. If it is inclined with the horizontal at an angle $\theta$, find its potential energy.
A. $\dfrac{1}{2}mgl\cos \theta$
B. $\dfrac{1}{2}mgl\left( {1 - \sin \theta } \right)$
C. $mgl\cos \theta$
D. $mglsin\theta$

Answer 1

Find the position of the centre of the mass of the rod and find the potential energy about this point. Thereafter find the change in the potential energy when the rod is rotated through an angle $\theta .$

Complete step by step answer: It is given that, mass of the rod = $m$, Length of the rod = $l$ and Angle = $\theta$
We know that the centre of mass of a uniform rod of length $l$above the ground is $\dfrac{l}{2}.$
Therefore, the potential energy of the rod is given by $\dfrac{{ - mgl}}{2}.$
When, the rod is displaced through an angle $\theta$, the height of the centre of mass from the ground becomes $\dfrac{1}{2}\sin \theta .$
So, the final potential energy of the rod becomes $\dfrac{{ - mgl}}{2}\sin \theta .$
Change in the potential energy = Final potential energy – Initial potential energy
$= \dfrac{{ - mgl}}{2}\sin \theta - \dfrac{{ - mgl}}{2}$ $= \dfrac{{mgl}}{2}\left( {1 - \sin \theta } \right)$
Hence the correct option is (B).

Note: The centre of mass of a rigid body is defined as a point where the entire mass of the body is supposed to be concentrated; the nature of the motion of the body shall remain unaffected, if all the forces acting on the body were applied directly on the centre of mass of the body. For rigid bodies of regular geometrical shapes and having uniform distribution of mass, the centre of mass is at the geometrical centre. In case of a uniform rod of length $l$, the centre of mass lies at the centre of the rod.