A circular disc reaches from top to bottom of an inclined plane of length ℓ. When it slips down the plane, it takes t s. When it rolls down the plane then it takes (α/2)¹ᐟ² t s, where α i s ______.
Physics · Rotational Motion · rolling on inclined surface
JEE Mains 2024 09 Apr Evening Shift

Question

A circular disc reaches from top to bottom of an inclined plane of length ℓ. When it slips down the plane, it takes t s. When it rolls down the plane then it takes (α/2)¹ᐟ² t s, where α i s ______.

Physics · Rotational Motion · rolling on inclined surface

JEE Mains 2024 09 Apr Evening Shift

G1 Physics · Accepted Answer

For slipping without rotation: a₁ = g sinθ and: ℓ = ½a₁t² For a solid disc rolling without slipping: I = ½MR² Therefore: a₂ = g sinθ/[1 + I/(MR²)] = g sinθ/(1 + 1/2) = 2g sinθ/3 Let rolling time be t'. Then: ℓ = ½a₂t'² Comparing with the slipping case: t'²/t² = a₁/a₂ = (g sinθ)/(2g sinθ/3) = 3/2 Hence: t' = √(3/2)t Given: t' = √(α/2)t Therefore: α = 3