A uniform solid cylinder of mass ' m ' and radius ' r ' rolls along an inclined rough plane of inclination 45°. If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder's axis will be

Solution

Physics sir · 4 steps

Rolling down a slope, sir says: take g sin θ and divide it by 1 + k²/r² for the shape.

1. Rolling acceleration. Friction provides the torque that spins the cylinder, so not all of g sin θ goes into speeding up the axis. For a body rolling without slipping, I = mk² gives the result below.

   a = (g sin θ)/(1 + k²/r²)

2. Solid cylinder. For a uniform solid cylinder I = 1/2mr², so k²/r² = 1/2. Then a = (g sin θ)/(3/2) = 2/3g sin θ.
3. Put in 45°. With sin 45° = 1/√2, a = 2/3 × g/√2 = 2g/3√2 = (√2 g)/3.
4. So. The axis accelerates down the plane at (√2 g)/3, which is option (d).

The idea to keep. For anything rolling down an incline, a = (g sin θ)/(1 + k²/r²). Remember k²/r²: ring 1, solid cylinder 1/2, hollow sphere 2/3, solid sphere 2/5.

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