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Engineering Mechanics - Dynamics Chapter 18 Problem 18-1 At a given instant the body of mass m has an angular velocity Z and its mass center has a velocity v G . Show that its kinetic energy can be represented as T = 1/2 I IC Z 2 , where I IC is the moment of inertia of the body computed about the instantaneous axis of zero velocity, located a distance r GIC from the mass center as shown. Solution: T 1 2 m v G 2 1 2 I G + , Z 2 . where v G = Z r GIC T 1 2 m Z r GIC + , 2 1 2 I G Z 2 . T 1 2 m r GIC 2 I G . + , Z 2 However m ( r GIC ) 2 + I G = I IC T 1 2 I IC Z 2 Problem 18-2 The wheel is made from a thin ring of mass m ring and two slender rods each of mass m rod . If the torsional spring attached to the wheels center has stiffness k , so that the torque on the center of the wheel is M = k T , determine the maximum angular velocity of the wheel if it is rotated two revolutions and then released from rest. Given: m ring 5 kg m rod 2 kg k 2 N m rad r 0.5 m Solution: I O 2 1 12 m rod 2 r ( ) 2 m ring r 2 . I O 1.583 kg m 2 T 1 6 U 12 . T 2 591 Engineering Mechanics - Dynamics Chapter 18 4 S T k T d . 1 2 I O Z 2 Z k I O 4 S Z 14.1 rad s Problem 18-3 At the instant shown, the disk of weight W has counterclockwise angular velocity Z when its center has velocity v . Determine the kinetic energy of the disk at this instant. Given : W 30 lb Z 5 rad s v 20 ft s r 2 ft g 32.2 ft s 2 Solution: T 1 2 1 2 W g r 2 Z 2 1 2 W g v 2 . T 210 ft lb *Problem 18-4 The uniform rectangular plate has weight W . If the plate is pinned at A and has an angular velocity Z , determine the kinetic energy of the plate. Given: W 30 lb Z 3 rad s a 2 ft b 1 ft Solution: T 1 2 m v G 2 1 2 I G Z 2 . 592 Engineering Mechanics - Dynamics Chapter 18 T 1 2 W g Z b 2 a 2 . 2 2 1 2 1 12 W g b 2 a 2 . + , Z 2 . T 6.99 ft lb Problem 18-5 At the instant shown, link AB has angular velocity Z AB . If each link is considered as a uniform slender bar with weight density J , determine the total kinetic energy of the system. Given: Z AB 2 rad s a 3 in J 0.5 lb in b 4 in T 45 deg c 5 in Solution: U J g Guesses Z BC 1 rad s Z CD 1 rad s v Gx 1 in s v Gy 1 in s T 1 lb ft Given Z AB a u Z BC b u . Z CD c cos T + , c sin T + , u . Z AB a u Z BC b 2 u . v Gx v Gy T 1 2 U a 3 3 Z AB 2 1 2 U b 3 12 Z BC 2 .... View Full Document

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