Question

A uniform rod of length 1m and mass 100 g is pivoted at one end and is hanging vertically. It is displaced through 600 from the vertical. The increase in its potential energy is​

Answer 1

Given:

Rod' mass (M) = 100 g = 0.1 kg

Length of rod (L) = 1 m

Angular movement is 60° from the vertical.

To Find:

What is increase in its potential energy due to change in angular position?

Solution:

Mass  = M = 0.1 kg

Length = L = 1 m

$\mathbf{Gravitational \ acceleration\ value= g= 10 \ \ \dfrac{m}{s^{2}}}$

$\mathbf{Vertically\ initial\ position\ of\ centre \ of\ gravity\ = \dfrac{L}{2}}$

$\mathbf{Vertically\ final\ position\ of\ centre \ of\ gravity\ = \dfrac{L}{2}\cos 60 }$

Vertically rise in height of centre of gravity due to change in angular position is given as:

$\mathbf{\Delta H=\dfrac{L}{2}-\dfrac{L}{2}\cos 60}$

$\mathbf{\Delta H=\dfrac{L}{2}(1-\cos 60)=\dfrac{1}{2}(1-\cos 60)}$

We know the formula of potential energy:

$\mathbf{Potential\ Energy = U = Mg\Delta H}$

$\mathbf{Potential\ Energy = U = 0.1\times 10\times \dfrac{1}{2}(1-\cos 60)}$

Potential energy (U) = 0.25 J

So increase in potential energy is 0.25 Joule by displaced through 60° from vertical.