A solid cylinder (I = MR²/2)has a string wrapped around it many times.When I release the cylinder,holding on to the string,the cylinder falls and spins as the string unwinds.What is the downward acceleration of the cylinder as it falls? A)0 B)4.9 m/s² C)6.5 m/s² D)9.8 m/s²

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Quizplus · Accepted Answer

The downward acceleration of the cylinder is 2/3 of g (9.8 m/s²) because the tension in the string provides a torque that causes angular acceleration. The net force is mg - T = ma, and the torque is T*R = I*α. Solving these equations gives a = 2/3 * g = 6.5 m/s².

A Solid Cylinder (I = MR2/2)has a String Wrapped Around

A Solid Cylinder (I = MR²/2)has a String Wrapped Around