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7.31 Acylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. The co-efficient of static friction μs = 0.25.(a) How much is the force of friction acting on the cylinder ?(b) What is the work done against friction during rolling ?(c) If the inclination θ of the plane is increased, at what value of θ does the cylinder begin to skidand not roll perfectly? |
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Answer» Here, mass of cylinder (m) = 10 Kg Radius (r) = 15 cm = 0.15 m Inclination of plane (∅) = 30° coefficient of friction (u) = 0.25 (a) friction = mgsin∅/(1 + mr²/I) For solid cylinder (I) = mr²/2 Fr = mgsin∅/3 = 10 × 9.8 × 1/2 × 1/3 = 16.3 N (b) force of friction acts perpendicular to the direction of the displacement.So , work done against during rolling .W = Fscos90° = 0 (c) for rolling without slipping use formula, u = tan∅/( 1 + mr²/I)For solid cylinder u = tan∅/3 tan∅ = 3u = 3×0.25 = 0.75 = tan36°54'∅ = 36°54' ≈ 37° please like the solution 👍 ✔️👍dhamplet |
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