A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity 3 m / s (as shown in figure). Maximum height with respect to the initial position covered by it will be __________ cm.

Solution

Physics sir · 4 steps

A rolling ball carries spin energy too, and all of it turns into height when the ball stops on the curve.

1. Total kinetic energy. For a hollow sphere I = 2/3mR², and rolling without slipping means ω = v/R. So the rotational energy 1/2Iω² = 1/3mv² adds to the translational 1/2mv², giving 5/6mv² in total.
2. Energy conservation. Static friction in rolling does no work, so mechanical energy is conserved. At the highest point the ball stops moving and stops spinning, so all of 5/6mv² becomes mgh.

   mgh = 5/6mv² ⇒ h = 5v²/6g

3. Put in numbers. Take g = 10 m s⁻²: h = (5 × 3²)/(6 × 10) = 45/60 = 0.75 m.
4. So. The maximum height is 0.75 m, that is 75 cm.

The common slip: 45. Using only the translational energy 1/2mv² and leaving out the spin gives h = v²/2g = 45 cm, not 75.

The idea to keep. For rolling up a rough surface, use mgh = 1/2mv²(1 + K²/R²). Pick K²/R² for the body: 2/3 hollow sphere, 2/5 solid sphere, 1 ring, 1/2 disc.

Answer from the NTA final answer key. Solution by PrepWiser.