Explanation Given information: Weight of the uniform cylinder = 15-lb. The center of the cylinder is moved 0.4 in. down the incline and released. Calculations: From the free body diagram shown above: A t P o s i t i o n 1 : Kinetic energy : T 1 = 0 Potential energy : V 1 = 1 2 k ( δ s t + r θ m ) 2 A t P o s i t i o n 2 Kinetic energy : T 2 = 1 2 I ¯ θ . m 2 + 1 2 m v ¯ 2 Potential energy : V 2 = m g h + 1 2 k ( δ s t ) 2 Applying the law of conservation of energy . T 1 + V 1 = T 2 + V 2 : 0 + 1 2 k = m ( δ s t + r θ m ) 2 = 1 2 I ¯ θ . m 2 + 1 2 m v m 2 + m g h + 1 2 k ( δ s t ) 2 k δ s t 2 + 2 k δ s t r θ m + k r 2 θ m 2 = I ¯ θ . m 2 + m v m 2 + 2 m g h + k δ s t 2 .......... ( 1 ) For the disk to be in equilibrium, ∑ M C = 0 m g sin β r − k δ s t r = 0 A l s o , h = r sin β θ m ∴ m g h − k δ s t r = 0 To determine (b) Maximum velocity of the center of the cylinder.

12th Edition

ISBN: 9781259977305

Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.

Publisher: Mcgraw-hill Education,

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Chapter 19.3, Problem 19.79P

To determine

(a)

The period of vibration.

To determine

(b)

Maximum velocity of the center of the cylinder.

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Chapter 19.3, Problem 19.78P

Chapter 19.3, Problem 19.80P

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(a) The period of vibration.

Solution Summary: The author explains how the uniform cylinder is moved 0.4 in. down the incline and released.