Explanation Given information: The weight ( W P ) of wooden panel is 16 lb. The weight ( W s ) of metal sphere is 4 lb. Calculation: Find the mass ( m p ) of the panel using the equation: m p = W p g Here, W p is the weight of the panel, and g is the acceleration due to acceleration due to gravity. Substitute 16 lb for W p and 32.2 ft / s 2 for g . m p = 16 32.2 = 0.4969 slug Find the mass ( m s ) of the sphere using the equation: m s = W s g Substitute 4 lb for W p and 32.2 ft / s 2 for g . m s = 4 32.2 = 0.1242 slug Find the centroidal moment of inertia ( I ¯ ) of the panel using the equation: I ¯ = 1 6 m p L 2 Here, L is the length of the panel. Substitute 0.4969 slug for m p and 18 in. for L . I ¯ = 1 6 ( 0.4969 ) ( 18 ) 2 = 1 6 ( 0.4969 ) ( 18 12 ) 2 = 1 6 ( 0.4969 ) ( 2 .25 ) = 0.1863 slug ⋅ ft 2 Sketch the free body diagram of the system after the impact as shown in Figure (1). Refer Figure (1). Find the initial velocity ( v C ′ ) 1 of the sphere at point C using kinematics. ( v C ′ ) 1 = 2 g y Here, y is the distance of the sphere from point B to point C’ . Substitute 32.2 ft / s 2 for g and 18 in. for y . ( v C ′ ) 1 = 2 ( 32.2 ) ( 18 in . × 1 ft 12 in . ) = 2 ( 32.2 ) ( 1.5 ) = 9.8285 ft / s From Figure (1), Apply Pythagoras theorem. Find the distance A C ′ . A C ′ = A ′ A 2 + A ′ C 2 Substitute 7 in. for A ′ A and 9 in. for A ′ C . A C ′ = ( 7 in . × 1 ft 12 in . ) 2 + ( 9 in . × 1 ft 12 in . ) 2 = ( 0 .5833 ) 2 + ( 0.75 ) 2 = 0.902777 = 0.9502 ft Find the equation of the velocity ( v ¯ 2 ) of the mass center G of the panel after impact: v ¯ 2 = ( 9 in . × 1 ft 12 in . ) ( ω 2 ) = 0.75 ω 2 (1) Find the equation of the velocity [ ( v C ′ ) 2 ] of the sphere at point C’ after impact

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ISBN: 9781260265453

Author: BEER

Publisher: MCG

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Chapter 17.3, Problem 17.100P

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Find the velocity of the mass center G of the panel immediately after the impact.

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Chapter 17.3, Problem 17.99P

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Find the velocity of the mass center G of the panel immediately after the impact.

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