Explanation Given: The weight of the pendulum sphere, W s = 30 lb . The weight of the slender rod, W r = 10 lb . The radius of the sphere, r = 1 ft . The length of the rod, l = 2 ft . Show the free body diagram of the pendulum as in Figure (1). Conclusion: Express the mass moment of inertia of the whole body about O using the parallel axis theorem. I O = ( I G + m s d 2 ) + I r = 2 5 [ W s g ] ( r ) 2 + W s g ( l + r ) 2 + 1 3 [ W r g ] ( l ) 2 (I) Here, acceleration due to gravity is g . Substitute 32 .2 ft / s 2 for g , 30 lb for W s , 1 ft for r , 2 ft for l and 10 lb for W r in Equation (I). I O = 2 5 [ 30 32.2 ] ( 1 ) 2 + 30 32.2 ( 2 + 1 ) 2 + 1 3 [ 10 32.2 ] ( 2 ) 2 = 0.3726 + 8.3850 + 0.4140 = 9.171 slug ⋅ ft 2 Express the half of the centroidal distance from x axis. x ¯ = W s ( l + r ) + W r ( r ) W s + W r (II) Substitute 30 lb for W s , 1 ft for r , 2 ft for l and 10 lb for W r in Equation (II). x ¯ = 30 ( 2 + 1 ) + 10 ( 1 ) 30 + 10 = 100 40 = 2.5 ft Resolve forces along normal dircetion. + → ∑ F n = m a n O x = 0 Here, acceleration along normal direction is a n and reaction about O in x axis is O x . Resolve forces along transverse dircetion. + ↓ ∑ F t = m a t ( W s + W r ) − O y = ( W s + W r ) g a G (III) Here, acceleration along transverse direction is a t and reaction about O in y axis is O y

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Author: HIBBELER

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Chapter 17, Problem 6RP

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The reaction at the pin O just before the cord A B is cut.

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