Question

a uniform rod of length 1m and mass 100g is pivoted at one end and hanging vertically.it is displaced through 60° from the vertical.the increase in its potential energy?
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Answer 1

Given:

A uniform rod of length 1 m and mass 100 g is pivoted at one end and hanging vertically.it is displaced through 60° from the vertical.

To Find:

Calculate the increase in its potential energy?

Solution:

Mass of rod = M = 100 g = 0.1 kg

Length of rod = L = 1 m

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

Initial position of centre gravity of rod from pivoted end in vertical direction is given:

$\mathbf{Initial\ position\ of\ centre \ of\ gravity\ = R = \dfrac{L}{2}}$

Final position of centre gravity of rod from pivoted end in vertical direction is given as:

$\mathbf{Final\ position\ of\ centre \ of\ gravity\ = R_{o} = \dfrac{L}{2}\cos 60 }$

Rise in vertical height of centre of gravity is given:

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

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

From formula of potential energy:

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

On putting respective value in above equation:

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

Potential energy = U = 0.25 J

So increase in potential energy is 0.25 Joule.

Answer 2

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Increase in potential energy is 0.25 Joule.