Question

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

• A. 0 m/s
• B. 1 m/s
• C. 2 m/s
• D. 4 m/s

Answer 1

The correct option is C

2 m/s

Take the bob + earth as the system. The external force acting on the system is that 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 total mechanical energy will remain constant. As is clear from figure, the height ascended by the bob at an angular displacement $\theta$ is l - l cos$\theta$ = l (1 - cos$\theta$). This should be equal to the decrease in the kinetic energy of the system. Again, as the earth does not move in the lab frame, this is the decrease in the kinetic energy of the bob. If the speed at an angular displacement $\theta$ is v1, the decrease in kinetic energy is

Where $v_0$ is the speed of the blocks at the lowest position.

Thus, $\frac{1}{2}m{v}_0^2-\frac{1}{2}m{v}_1^2=mgl(1-cos\theta )$

Or, $v_1=\sqrt{v_0^2-2gl(1-cos\theta )}$

= $\sqrt{\left(9m^2/s^2\right)-2x\left(10m/s^2\right)\times \left(0.5m\right)\times (1-\frac{1}{2})}$

= 2 m/s