Explanation Given information: Radius of caster A=B, r = r 4 Mass of caster A=B, m = m 8 Concept used: By using the principle of conservation of energy for the free vibration of un-damped system the energy is partly kinetic and partly potential. Thus, T 1 + V 1 = T 2 + V 2 Calculation: Since the semi-circle is initially at rest at an angle ?. The kinetic energy of the semi-circle, T 1 = 0 And potential energy, V 1 = m g ( 4 r 3 π − 4 r 3 π cos θ ) V 1 = m g 4 r 3 π ( 1 − cos θ ) For small oscillations: cos θ = 1 − 2 sin 2 ( θ 2 ) 2 sin 2 ( θ 2 ) = 1 − cos θ θ 2 2 = 1 − cos θ Hence, potential energy, V 1 = m g 4 r 3 π ( θ 2 2 ) V 1 = 2 m g r ( θ 2 3 π ) When the cylinder rotate as there is not any slippage occurs then, The kinetic energy of the semi-circle, T 2 = 1 2 I A ω A 2 + 1 2 I B ω B 2 + 1 2 I ω 2 Here, I A = I B = Moment of inertia of casters A and B. I = Moment of inertia of semi-circle. And, ω A = ω B = 4 ω T 2 = 1 2 ( 1 2 m A r A 2 ) ω A 2 + 1 2 ( 1 2 m B r B 2 ) ω B 2 + 1 2 ( 1 2 m r 2 ) ω 2 T 2 = 1 2 ( 1 2 × m 8 × ( r 4 ) 2 ) ( 4 ω ) 2 + 1 2 ( 1 2 × m 8 × ( r 4 ) 2 ) ( 4 ω ) 2 + 1 2 ( 1 2 m r 2 ) ω 2 T 2 = 1 2 ( m r 2

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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Free undamped vibrations

Whenever an object/system displaces in either side of its equilibrium position with the help of an external/internal means, then the object/system moves regularly or randomly on both sides of its equilibrium position. This motion is called vibration. Gen…

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Chapter 19.3, Problem 19.92P

To determine

The frequency of oscillation.

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Chapter 19.3, Problem 19.91P

Chapter 19.3, Problem 19.93P

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The frequency of oscillation.

Solution Summary: The author explains the principle of conservation of energy for the free vibration of un-damped system the energy is partly kinetic and partly potential.

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The frequency of oscillation.

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Chapter 19 Solutions