A 4.00 kg hollow cylinder of radius 5.00 cm starts from rest and rolls without slipping down a 30.0 degree incline. The acceleration of the center of mass of the cylinder is A) 2.45 m/s². B) 2.98 m/s². C) 3.35 m/s². D) 3.98 m/s². E) 4.05 m/s².

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Quizplus · Accepted Answer

The acceleration of the center of mass of the cylinder is given by:$$a = g\sin	heta = (9.8 m/s^2)\sin30.0^\circ = 4.9 m/s^2$$Since the cylinder is rolling without slipping, the acceleration of the center of mass is also equal to:$$a = \frac{I}{MR^2}\alpha$$where I is the moment of inertia of the cylinder, M is its mass, and R is its radius. For a hollow cylinder, the moment of inertia is given by:$$I = MR^2$$Therefore, we have:$$a = \frac{MR^2}{MR^2}\alpha = \alpha$$and$$\alpha = \frac{a}{R} = \frac{4.9 m/s^2}{0.05 m} = 98 rad/s^2$$The acceleration of the center of mass is then given by:$$a = R\alpha = (0.05 m)(98 rad/s^2) = 4.9 m/s^2$$

A 400 Kg Hollow Cylinder of Radius 5