Question

A small sphere of mass m and radius R slides down the smooth surface of a large hemispherical bowl of radius R. If the sphere starts sliding from rest, the total kinetic energy of the sphere at the lowest point A of the bowl will be:

• A. mg(R − r)
• B. \(\frac{7}{10}mg(R - r)\)
• C. \(\frac{2}{7}mg(R - r)\)
• D. \(\frac{7}{10}mg(R - r)\)

Hint:

For rolling motion, the total kinetic energy is the sum of translational and rotational kinetic energies. Use energy conservation to determine the total kinetic energy at the lowest point.

Answer 1

The Correct Option is A

To find the total kinetic energy of a small sphere (mass m, radius r) sliding down a smooth hemispherical bowl (radius R), consider these points:

• The sphere begins at rest, so its initial energy is all potential, based on its height.
• The height difference between the bowl's center and the sphere at the bottom (point A) is R - r.
• At point A, all mechanical energy becomes kinetic energy, assuming no energy loss.

By the conservation of energy:
\[ \text{Initial Potential Energy} = \text{Kinetic Energy at A} \] \[ m g (R - r) = \text{Kinetic Energy at A} \]

Therefore, the total kinetic energy at the bottom is:
\[ \boxed{m g (R - r)} \]

Correct Answer: \( \mathbf{mg(R - r)} \)