Question

A hollow, thin-walled cylinder and a solid sphere start from rest and roll without slipping down an inclined plane of length 3.0 m. The cylinder arrives at the bottom of the plane 2.8 s after the sphere. Determine the angle between the inclined plane and the horizontal.

Answer 1

Let's consider a solid rigid body rolling without slipping. The Newton's second law: F = ma mgsino - fr=ma (1) The second Newton's law for rotation: t = la Where I is the moment of inertia Then: fưR = la = 1 ) fr = 1 R2 - (2) Using equation (1) and (2), then: mgsino - 12 = ma Solving for a: mgsine R2 + m 1 The travel distance is giving by: Ax = zat? The time is: 2AX t= ✓ a For a hollow cylinder: 1 = mR2, then: mgsino 1 . ac = m + m 29 The time is: 4AX to = (3) For the solid sphere: 1 = MR2 mgsino 5 as = gsine 9sino 5m + m The time is. 144x ts = 5gsino Using (3)44) 10 C = ts 7 Using that: tc=ts + 2.8

Then: ts + 2.8 ts J7 Solving for ts ts = 14.34s Using equation (4) 14(3.0) 14.34 = 5* 9.8sino Solving for a O = 0.239°