An object of uniform density rolls up the curved path with the initial velocity vₒ as shown in the figure. If the maximum height attained by an object is 7v₀²/(10 g) ( g= acceleration due to gravity), the object is a ____ .

Solution

Physics sir · 3 steps

All the kinetic energy, translation plus rotation, turns into height. Read k²/R² from the height given.

1. Total kinetic energy. A body rolling without slipping has v = Rω. Write its moment of inertia as I = mk². Its kinetic energy is 1/2mv₀² + 1/2Iω² = 1/2mv₀²(1 + k²/R²).
2. Energy to height. In pure rolling, static friction does no work, so mechanical energy is conserved. At the top the body stops both moving and spinning, so all of this energy becomes mgh.

   h = v₀²/2g(1 + k²/R²) = 7v₀²/10g ⇒ 1 + k²/R² = 7/5 ⇒ k²/R² = 2/5

   Object	k²/R²	Maximum height h
   Solid cylinder	1/2	3v₀²/4g
   Ring	1	v₀²/g
   Disc	1/2	3v₀²/4g
   Solid sphere	2/5	7v₀²/10g

3. So. Only a solid sphere has k²/R² = 2/5, so the object is a solid sphere, option (d).

The idea to keep. For rolling bodies, h = v²/2g(1 + k²/R²). The larger k²/R² is, the more energy is stored in spin and the higher the body climbs.

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