A pendulum bob has a speed 3m/s while passing thorugh its lowest position. What is its speed when it makes an angle of ${60}^0$ with the vertical? The length of the pendulum is 0.5m Take $g=10\frac{m}{s^2}$.

Solution: Thake the bob +earth as the system. The exterN/Al force acting on the system is thst due to the string. But this force is always perpendicular to the velocity of the bob and so the work done by this force is zero. Hence, the totla mefchanicla energy will remain constant. As is clear from ure the height ascended by the bob at an angular displacement $\theta$ is $l-l\cos \theta =l\left(1-\cos \theta \right)$. The increase in the potential energy s $mgl\left(1-\cos \theta \right)$. This should be equal to the decrease in the kinetic energy of the system. Again as the earth does not move in teh labframe, this is the decrease in the kinetic energy of the bob. If the speed at an angular displacement $\theta is{v}_1$, the decrese in kinetic energy is

$\frac{1}{2}m{v}_0^2=-\frac{1}{2}m{v}_1^2$
$where{v}_0$ is the speed of the block t the lowest position.
Thus, $\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2=mgl\left(1-\cos \theta \right)$
or $v_1=\sqrt{v_2^2-2gl\left(1-\cos \theta \right)}$
$=\sqrt{\left(9\frac{m^2}{s^2}\right)-2x\left(10\frac{m}{s^2}\right)\times \left(0.5m\right)\left(1-\frac{1}{2}\right)}$
$=2\frac{m}{s}$