Explanation Given information: The mass of the ball is m . The initial horizontal velocity ( v 0 ) is 5 m / s . The length of the cord ( l ) is 0.6 m . The horizontal distance ( x B ) is 0. Calculation: Show the diagram of the suspended ball by an inextensible cord as in Figure (1). Assume that position 1 is at A and position 2 is at the point described by the angle where the path of the ball changes from circular to parabolic. At position 2, the tension ( Q ) in the cord is zero due to cord becomes slack. The expression for the x -coordinate of the ball ( x 2 ) at position 2 as follows; x 2 = l cos θ The expression for the y -coordinate of the ball ( y 2 ) at position 2 as follows; y 2 = l sin θ The expression for calculating the centripetal acceleration component ( a n ) as follows; a n = v 2 2 l Show the free body diagram of the ball at position 2 as in Figure (2). Calculate the velocity of the ball at position 2 by resolving the forces acting by considering the Newton’s second law. Q + m g sin θ = m a n Substitute v 2 2 l for a n . Q + m g sin θ = m ( v 2 2 l ) Here, m is the mass of the ball and g acceleration due to gravity. Substitute 0 for Q . 0 + m g sin θ = m ( v 2 2 l ) g sin θ = ( v 2 2 l ) v 2 2 = g l sin θ v 2 = g l sin θ (1) Consider the position 1. The expression for calculating the kinetic energy of the ball at position 1 ( T 1 ) as follows; T 1 = 1 2 m v 0 2 Here, v 0 is the velocity of the ball at position 1. Calculate the potential energy of the ball at position 1 ( V 1 ) using the formula: V 1 = − m g h Here, negative sign indicates that the ball is located below the datum level, and h is the vertical distance of the ball from the datum level. Substitute l for h . V 1 = − m g l Consider the position 2. The expression for calculating the kinetic energy of the ball at position 2 ( T 2 ) as follows; T 2 = 1 2 m v 2 2 The expression for the vertical distance of the ball ( h ) from datum in position 2 from the figure as follows; h = l sin θ Calculate the potential energy of the ball at position 2 ( V 2 ) using the formual: V 2 = m g h Substitute l sin θ for h . V 2 = m g l sin θ The expression for principle of conservation of energy at position 1 and position 2 for the ball as follows; T 1 + V 1 = T 2 + V 2 Substitute 1 2 m v 0 2 for T 1 , − m g l for V 1 , 1 2 m v 2 2 for T 2 , and m g l sin θ for V 2 . T 1 + V 1 = T 2 + V 2 1 2 m v 0 2 − m g l = 1 2 m v 2 2 + m g l sin θ m ( v 0 2 2 − g l ) = m ( v 2 2 2 + g l sin θ ) v 0 2 2 − v 2 2 2 = g l sin θ + g l 1 2 ( v 0 2 − v 2 2 ) = g l + g l sin θ v 0 2 − v 2 2 = 2 g l ( 1 + sin θ ) Substitute g l sin θ for v 2 2 . v 0 2 − g l sin θ = 2 g l ( 1 + sin θ ) v 0 2 = 2 g l ( 1 + sin θ ) + g l sin θ v 0 2 = 2 g l + 2 g l sin θ + g l sin θ v 0 2 = 2 g l + 3 g l sin θ v 0 2 = g l ( 2 + 3 sin θ ) v 0 2 g l = 2 + 3 sin θ v 0 2 g l − 2 = 3 sin θ sin θ = 1 3 ( v 0 2 g l − 2 ) Substitute 5 m / s for v 0 , 9.81 m / s 2 for g , and 0.6 m for l . sin θ = 1 3 ( ( 5 m / s ) 2 ( 9.81 m / s 2 ) ( 0

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9781259639272

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

Publisher: McGraw-Hill Education

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Chapter 13.2, Problem 13.77P

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Find the vertical distance (yB) so that the ball will enter the basket.

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6/86 The connecting rod AB of a certain internal-combustion engine weighs 1.2 lb with mass center at G and has a radius of gyration about G of 1.12 in. The piston and piston pin A together weigh 1.80 lb. The engine is running at a constant speed of 3000 rev/min, so that the angular velocity of the crank is 3000(2)/60 = 100л rad/sec. Neglect the weights of the components and the force exerted by the gas in the cylinder compared with the dynamic forces generated and calculate the magnitude of the force on the piston pin A for the crank angle 0 = 90°. (Suggestion: Use the alternative moment relation, Eq. 6/3, with B as the moment center.) Answer A = 347 lb 3" 1.3" B 1.7" PROBLEM 6/86

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Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

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Find the vertical distance ( y B ) so that the ball will enter the basket.

Solution Summary: The author calculates the velocity of the ball at position 2 by resolving the forces acting by considering the Newton’s second law.

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Find the vertical distance (yB) so that the ball will enter the basket.

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