A solid cylinder is released from rest from the top of an inclined plane of inclination 30° and length 60 cm. If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is __________ ms⁻¹. (Given g=10 ms⁻²)

Solution

Physics sir · 4 steps

Height gives the energy, but a rolling cylinder spends one third of it on spinning, so v is less than √(2gh).

1. Find the height. The cylinder travels 60 cm = 0.6 m along the slope. The vertical drop is h = 0.6 × sin 30° = 0.3 m.
2. Energy while rolling. In rolling without slipping, ω = v/R and friction does no work. For a solid cylinder I = 1/2MR², so the total kinetic energy is 1/2Mv² + 1/4Mv² = 3/4Mv².

   Mgh = 3/4Mv² ⇒ v = √(4gh/3)

3. Put in numbers. v = √((4 × 10 × 0.3)/3) = √4 = 2 m s⁻¹.
4. So. The cylinder reaches the bottom at 2 m s⁻¹.

The common slip: 2.45. Leaving out the rotational kinetic energy gives v = √2gh = √6 ≈ 2.45, not 2. A rolling body also has 1/2Iω².

The idea to keep. For rolling down an incline, use v = √(2gh/(1 + k²/R²)). Use the vertical height, not the slope length, and put in k²/R² for the body.

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