Question

In figure, a smooth curved path ABC is shown. Its path after B is vertical circular path of radius r. From what minimum height, a ball should be released so that it may cross the highest point C remaining in contact the plane?

Answer 1

When the ball reaches point B, then potential energy = kinetic energy

or mgh = \(\frac{1}{2}mv_B{}^2 \Rightarrow\ V_B = \sqrt{2gh}\)

For completing vertical circular path i.e., to reach highest point C, the velocity at B should be equal to critical velocity at lowest point of vertical circle.

i.e., v_B = \(\sqrt{5rg}\)

\(\therefore\ \sqrt{2gh} = \sqrt{5rg} = \sqrt{2h} = \sqrt{5r}\)

or h \( = \frac{5}{2}r\)