A sphere of radius 'a' and mass 'm' rolls along a horizontal plane with constant speed v₀. It encounters an inclined plane at angle θ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

Solution

Physics sir · 4 steps

The sphere's spin also pays for the climb, so use its full kinetic energy, 7/10mv₀².

1. Total kinetic energy. A rolling sphere has translational KE 1/2mv₀² and rotational KE 1/2Iω². For a solid sphere I = 2/5ma², and rolling means ω = v₀/a. So the rotational part is 1/5mv₀², and the total is 7/10mv₀².
2. Energy is conserved. In rolling without slipping the contact point does not slide, so static friction does no work. All of the kinetic energy turns into potential energy by the time the sphere stops at height h.

   7/10mv₀² = mgh ⇒ h = 7v₀²/10g

3. Height to distance. The distance s along the incline is related to the height by h = s sin θ. So s = h/(sin θ).
4. So. The sphere travels s = 7v₀²/(10g sin θ) up the incline, which is option (b).

The idea to keep. For rolling bodies, write total KE as 1/2mv²(1 + I/mR²). Friction in pure rolling does no work, so equate this to mgh.

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