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

Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259638091

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

See similar textbooks

Concept explainers

Conservation of Energy

In physics, the rule of the conservation of energy states that energy doesn’t disappear from the atmosphere; instead, it changes its form from one type to another. Energy is one of the most important physical quantities in nature, and it cannot be create…

Videos

Question

Chapter 17.3, Problem 17.126P

To determine

Find the velocity of the sphere immediately after impact.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 17.3, Problem 17.125P

Chapter 17.3, Problem 17.127P

Students have asked these similar questions

Find equation of motion and natural frequency for the system shown in fig. by energy method. H.W2// For the system Fig below find 1-F.B.D 2-Eq.of motion 8wn 4-0 (5) m. Jo m

2. Read the following Vernier caliper measurements. (The scales have been enlarged for easier reading.) The Vernier caliper is calibrated in metric units. (a) 0 1 2 3 4 5 سلسلسله (b) 1 2 3 4 5 6 سلسل (c) 1 23456 (d) 1 2 3 4 5 6 سلسلس

Explain why on the interval 0<x<1000 mm and 1000<x<2000mm, Mt is equal to positive 160 Nm, but at x= 0mm and x=1000mm Mt is equal to -160 Nm (negative value!). What is the reason for the sign change of Mt?

Chapter 17 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Find the velocity of the sphere immediately after impact.

Solution Summary: The author calculates the velocity of the sphere immediately after impact.

Concept explainers

Videos

Find the velocity of the sphere immediately after impact.

Want to see the full answer?

Chapter 17 Solutions