A circular disc reaches from top to bottom of an inclined plane of length l. When it slips down the plane, if takes t s. When it rolls down the plane then it takes (α/2)^(1 / 2) t s, where α is _________.

Solution

Physics sir · 4 steps

Same incline, same length: time goes as one over the square root of acceleration, so compare accelerations.

1. Slipping down. Slipping means no friction acts, so the disc does not rotate. Its acceleration down the incline is aₛ = g sin θ.
2. Rolling down. When a body rolls without slipping, part of the energy goes into rotation. For a disc, I = 1/2MR², so K²/R² = 1/2.

   aᵣ = (g sin θ)/(1 + K²/R²) = (g sin θ)/(1 + 1/2) = 2/3 g sin θ

3. Compare times. Both start from rest and cover the same length l, so l = 1/2at² gives t = √(2l/a). Then tᵣ/t = √(aₛ/aᵣ) = √(3/2).
4. So. tᵣ = (3/2)^(1/2) t. Matching this with (α/2)^(1/2) t gives α = 3.

The common slip: 4.5. Taking time proportional to 1/a instead of 1/√a gives tᵣ = 1.5 t and α = 4.5, not 3.

The idea to keep. For rolling on an incline, a = (g sin θ)/(1 + K²/R²). With equal distances from rest, time scales as 1/√a, so the slower body takes longer by a square-root factor.

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