Explanation Given information: At point A: the period of oscillation τ A = 0.87 s . At point B: the period of oscillation τ B = 0.78 s . r a + r b = 10 i n . Calculations: Considering general pendulum of centroidal radius of gyration k ¯ Equation of motion: ∑ M 0 = ∑ ( M 0 ) e f f : − m g r ¯ sin θ = ( m r ¯ θ ¨ ) r ¯ + m k ¯ 2 θ ¨ θ ¨ + [ g r ¯ r ¯ 2 + k ¯ 2 ] sin θ = 0 For θ to be small, sin θ ≈ θ θ .. + [ g r ¯ r ¯ 2 + k ¯ 2 ] θ = 0 ω n 2 = g r ¯ r ¯ 2 + k ¯ 2 τ = 2 π ω n = 2 π r ¯ 2 + k ¯ 2 g r ¯ For rod suspended at point A, τ A = 2 π r a 2 + k ¯ 2 g r a g τ A 2 r a = 4 π 2 ( r a 2 + k ¯ 2 ) .............. ( 1 ) For rod suspended at point B, τ B = 2 π r a 2 + k ¯ 2 g r b g τ B 2 r b = 4 π 2 ( r b 2 + k ¯ 2 ) To determine (b) The centroidal radius of gyration k ¯ .

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ISBN: 9781259977237

Author: BEER

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Chapter 19.2, Problem 19.47P

To determine

(a)

The location of the mass centerG.

To determine

(b)

The centroidal radius of gyration k¯.

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Chapter 19.2, Problem 19.46P

Chapter 19.2, Problem 19.48P

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(a) The location of the mass centerG.

Solution Summary: The author explains the position of the mass centerG, the period of oscillation, and the equation of motion for the theta.