Question

A meter stick of mass $400$ gm is pivoted at one end and displaced through an angle $60$ from vertical The increase in its angular speed is:-

Answer 1

We know initial potential energy $E_{P1}=mg\frac{h}{2}$ ---------(A)

Given m=400 g=.4 kg and h=1 m and g=10$m/s^2$

Putting the above values in A we get

$E_{P1}=0.4\times 10\frac{1}{2}=2J$

Now when the stick is displaced with an angle ${60}^0$

Potential Energy $E_{P2}=mg\frac{h}{2}\cos {60}^0$

$E_{P2}=0.4\times 10\times \frac{1}{2}\times \frac{1}{2}=1J$

Thus change in potential energy $=2-1$

$=1J$