Explanation Given information: The mass ( m S ) of the solid sphere is 2 kg. The radius ( r ) of the solid sphere is 40 mm. The height ( h ) of drop is 200 mm. The mass ( m A B ) of the plank AB is 4 kg. The length ( L ) of the plank AB is 500 mm. The distance ( a ) of the sphere from left end is 40 mm. Calculation: Find the moment of inertia ( I ¯ S ) of the sphere using the equation: I ¯ S = 2 5 m S r 2 Substitute 2 kg for m S and 40 mm for r . I ¯ S = 2 5 ( 2 ) ( 40 mm × 1 m 1 , 000 mm ) 2 = 2 5 ( 2 ) ( 0.040 ) 2 = 1.28 × 10 − 3 kg ⋅ m 2 Find the moment of inertia ( I ¯ A B ) of the slender plank using the equation: I ¯ A B = 1 12 m A B L 2 Substitute 4 kg for m A B and 500 mm for L I ¯ A B = 1 12 ( 4 ) ( 500 mm × 1 m 1 , 000 mm ) 2 = 1 12 ( 4 ) ( 0.500 ) 2 = 83.333 × 10 − 3 kg ⋅ m 2 Find the velocity ( v S ) of the sphere at impact using kinematics. v S = 2 g h Substitute 9.81 m / s 2 for g and 200 mm for h in v S v S = 2 ( 9.81 ) ( 200 mm × 1 m 1 , 000 mm ) = 2 ( 9.81 ) ( 0.200 ) = 3.924 = 1.9809 m / s Find the linear momentum ( p ) before impact using the equation: p S = m A B v S Substitute 4 kg for m A B , and 1.9809 m / s for v S . p S = ( 4 ) ( 1.9809 ) = 7.9236 kg ⋅ m / s ↓ Consider C be the instantaneous center using kinematics. Draw line A C ¯ perpendicular to v ′ A and Draw line B C ¯ perpendicular to v ′ B . Sketch the free body diagram of the given system after the impact as shown in Figure (1). Here, G is the mass center of the slender plank AB and S is the sphere. Refer Figure (1). Find the distance CH using the equation: C H = L cos ( 90 ° − 60 ° ) + r = L cos 30 ° + r Substitute 500 mm for L , and 40 mm for r . C H = ( 500 mm × 1 m 1 , 000 mm ) cos 30 ° + ( 40 mm × 1 m 1 , 000 mm ) = ( 0.500 ) cos 30 ° + ( 0.040 ) = 0.4730 m Find the distance HS using the equation: H S = L 2 − a Substitute 500 mm for L , and 40 mm for a . H S = 500 mm × 1 m 1 , 000 mm 2 − 40 mm × 1 m 1 , 000 mm = 0.500 2 − 0.040 = 0.250 − 0.040 = 0. 21 m Find the distance CS using the equation: C S = ( C H ) 2 + ( H S ) 2 Substitute 0.4730 m for CH and 0.21 m for HS . C S = ( 0.4730 m ) 2 + ( 0.21 m ) 2 = 0.2678 = 0.5175 m Find the distance CG using the equation: C G = C H − r Substitute 0.4730 m for CH and 40 mm for r . C G = 0.4730 m − 40 mm = 0.4730 m − 0.040 m = 0.4330 m Find the angle ( β ) using the equation: tan β = H S C H Substitute 0.4730 m for CH and 0.21 m for HS . tan β = 0

VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)

12th Edition

ISBN: 9781260916942

Author: BEER

Publisher: MCG

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Chapter 17.3, Problem 17.126P

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Find the velocity of the sphere immediately after impact.

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Chapter 17.3, Problem 17.125P

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Chapter 17 Solutions

VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)

Ch. 17.1 - A round object of mass m and radius r is released...Ch. 17.1 - Prob. 17.2CQ Ch. 17.1 - Prob. 17.3CQ Ch. 17.1 - Prob. 17.4CQ 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...Ch. 17.1 - The rotor of an electric motor has an angular...Ch. 17.1 - Prob. 17.3P Ch. 17.1 - Two disks of the same material are attached to a...Ch. 17.1 - The flywheel of a punching machine has a weight of...

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