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

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

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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Chapter 13.2, Problem 13.76P

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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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Chapter 13 Solutions