Question

A magnesium cylinder and a steel ring roll down a slope without slipping on the surface. Both objects have the same mass and outside diameter. The moment of inertia of a cylinder is Icyl = m r2 and the moment of inertia of a ring is lring = m (r? + r?). Which object will reach the bottom first? A. The magnesium cylinder will reach the bottom first. B. The steel ring will reach the bottom first. C. They will reach the bottom at the same time.

Answer 1

Solution: - Steps: write equation of motion of a object rolling without slipping a surface: of rolling without stùpping on mi I to i in case a Өкови v = wr and xe mgsino a=28 mg j Thet = Id => fxr = I { f fu required friction force, to prevent slipping -> => [f= Ia 0 Ү- ) E Fx= ma mgsino- fema f= mysino - maj- Now ean (1) = 29 (3) Ta mgsino-ma 82 gsing I M 82 a = Here router diameter !+ 3 of object Step2:- given that cylinder j mass steel ring comment m Same. i) outer dia 7 ro 8=80

a ď Now from equation (3) - I I Icy = {mra Iring = fm23² + 1 moi ? Iving - Toy + £mr? Iring> Icylinder, hence during <acylinder. will reaches the ground latter, it lesser velocity, lessez acceleration A) The magnesium cylinder will reach the bottom first .: will have hence ring compare to cylinder. Any.
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