Explanation Point C is the instantaneous centre. ν = r ω , ω = ν r Position 1, at rest: We know that when the body is at rest, its velocity is zero. So it do not have any kinetic energy. Therefore, ⇒ T 1. = 0 Position 2,Cylinder has fallen through distance s. T 2 = Kinetic energy due to linear velocity ν 2 +Kinetic energy due to angular velocity ω 2 . T 2 = 1 2 m ν 2 2 + 1 2 I ω 2 2 Here, I = Moment of inertia of the cylinder = 1 2 m r 2 Put the value of ω = ν r and moment of inertia I = 1 2 m r 2 in above equation, we get: T 2 = 1 2 m v 2 + 1 2 ( 1 2 m r 2 ) ( v r )

Vector Mechanics for Engineers: Dynamics

11th Edition

ISBN: 9780077687342

Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 17.1, Problem 17.25P

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Velocity of the centre of the cylinder after it has moved downward a distance s.

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Chapter 17 Solutions

Vector Mechanics for Engineers: Dynamics

Ch. 17.1 - A round object of mass m and radius r is released...Ch. 17.1 - Prob. 17.CQ2P Ch. 17.1 - Prob. 17.CQ3P Ch. 17.1 - Prob. 17.CQ4P Ch. 17.1 - Slender bar A is rigidly connected to a massless...Ch. 17.1 - A 200-kg flywheel is at rest when a constant 300 N...Ch. 17.1 - The rotor of an electric motor has an angular...Ch. 17.1 - Two uniform disks of the same material are...Ch. 17.1 - Two disks of the same material are attached to a...Ch. 17.1 - Prob. 17.5P

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Velocity of the centre of the cylinder after it has moved downward a distance s.

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