rotational kinetic energy and the stone has translational kinetic energy. Let m be the mass of the stone and let M be the mass of the cylinder. For the cylinder The speed of the stone and the angular speed of the cylinder are related by Solve: Conservation of energy says and so The conservation of energy expression becomes so and 9.43. Set Up: The speed of the weight is related to of the cylinder by where Use coordinates where is upward and for the weight. where h is the unknown distance the weight descends. Let and For the cylinder Solve: (a) Conservation of energy says and (b) Reflect: The net work done by the rope that connects the cylinder and weight is zero. The speed of the weight equals the tangential speed at the outer surface of the cylinder, and this gives 9.44. Set Up: I for a point mass M a distance R from the axis is Solve: (a) For For a very short rod all the mass is at the axis. (b)
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.