VECTOR MECHANICS FOR ENGINEERS W/CON >B
VECTOR MECHANICS FOR ENGINEERS W/CON >B
12th Edition
ISBN: 9781260804638
Author: BEER
Publisher: MCG
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Textbook Question
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Chapter 13, Problem 13.201RP

The 2-lb ball at A is suspended by an inextensible cord and given an initial horizontal velocity of v0. If l = 2 ft, xB = 0.3 ft, and yB = 0.4 ft, determine the initial velocity v0 so that the ball will enter the basket. (Hint: Use a computer to solve the resulting set of equations.)

Fig. P13.201

Chapter 13, Problem 13.201RP, The 2-lb ball at A is suspended by an inextensible cord and given an initial horizontal velocity of

Expert Solution & Answer
Check Mark
To determine

Find the initial velocity (v0) so that the ball will enter the basket.

Answer to Problem 13.201RP

The initial velocity (v0) so that the ball will enter the basket is 15.37m/s_.

Explanation of Solution

Given information:

The weight of the ball (m) is 2lb.

The length of the cord (l) is 2ft.

The horizontal distance between the basket and point of suspension of the ball (xB) is 0.3ft.

The vertical distance between basket and point of suspension of the ball (yB) is 0.4ft.

The acceleration due to gravity (g) is 32.2ft/s2.

Calculation:

Show the diagram of the suspended ball by an inextensible cord as in Figure (1).

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 13, Problem 13.201RP , additional homework tip  1

Assume that position ‘1’ be at A and position ‘2’ be at the point described by the angle where the path of the ball changes from circular to parabolic.

The tension in the cord at position ‘2’, becomes slack (Q) which is zero.

Refer Figure (1),

The expression for the x-coordinate of the ball at position ‘2’ (x2) as follows:

x2=lcosθ

The expression for the y-coordinate of the ball at position ‘2’ (y2) as follows:

y2=lsinθ

Show the free body diagram of the ball at position ‘2’ as in Figure (2).

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 13, Problem 13.201RP , additional homework tip  2

The expression for the normal acceleration of the ball (an) as follows:

an=v22l

Since, the cord becomes slack at position ‘2’, so the tension (Q) will be zero.

Calculate the velocity of the ball by applying Newton’s second law and resolve the forces acting on the ball at position ‘2’ using the relation:

Q+mgsinθ=man

Substitute v22l for an.

Q+mgsinθ=m(v22l)

Substitute 0 for Q.

0+mgsinθ=m(v22l)v22=glsinθv2=glsinθ (1)

The expression for the kinetic energy of the ball at position ‘1’ (T1) as follows:

T1=12mv02

Calculate the potential energy of the ball at position ‘1’ (V1) using the formula:

V1=mgh

Here, negative sign is used as 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.

V1=mgl

The expression for the kinetic energy of the ball at position ‘2’ (T2) as follows:

T2=12mv22

The expression for the vertical distance of the ball by referring the Figure 1 as follows:

h=lsinθ

Calculate the potential energy of the ball at position ‘2’ (V2) using the formula:

V2=mgh

Substitute lsinθ for h.

V2=mglsinθ

The expression for principle of conservation of energy at position ‘1’ and position ‘2’ for the ball, to calculate the angle swept by the ball (θ) as follows:

T1+V1=T2+V2

Substitute 12mv02 for T1, mgl for V1, 12mv22 for T2 and mglsinθ for V2.

T1+V1=T2+V212mv02mgl=12mv22+mglsinθ12(v02v22)=gl+glsinθv02v22=2gl(1+sinθ)v02=v22+2gl(1+sinθ) (2)

Find the velocity at position 2:

Substitute 32.2ft/s2 for g and 2ft for l in the equation (1).

v2=glsinθ=(32.2ft/s2)(2ft)sinθ=64.4sinθ (3)

Show the parabolic motion of the ball after it reaches position ‘2’ as in Figure (3).

VECTOR MECHANICS FOR ENGINEERS W/CON >B, Chapter 13, Problem 13.201RP , additional homework tip  3

The expression for the velocity of the projectile ball after reaching the position ‘2’ (vp) as follows:

vp=(vp)xi+(vp)yj

Here, (vp)x is the horizontal velocity component of the projectile and (vp)y is the vertical velocity component of the projectile.

The expression for the horizontal velocity component of the projectile ball along the negative X-axis as follows:

(vp)x=v2cos(90θ)=v2sinθ

The expression for the horizontal distance between the basket and point of suspension of the ball (xB) as follows:

xB=x2v2sinθ(tBt2)

Here, tB is the time taken by the ball to reach the basket and t2 is the time taken by the ball to reach the position ‘2’.

Substitute lcosθ for x2.

xB=lcosθv2sinθ(tBt2)v2sinθ(tBt2)=lcosθxB(tBt2)=lcosθxBv2sinθ

Substitute 0.3ft for xB and 2ft for l.

(tBt2)=2cosθ0.3v2sinθ (4)

The expression for the vertical velocity component of the projectile ball along the negative X-axis as follows:

(vp)y=v2sin(90θ)=v2cosθ

The expression for the vertical distance between the basket and point of suspension of the ball (yB) as follows:

yB=y2+v2cosθ(tBt2)12g(tBt2)2

Substitute lsinθ for y2.

yB=lsinθ+v2cosθ(tBt2)12g(tBt2)2

Substitute 2ft for l.

yB=2sinθ+v2cosθ(tBt2)12g(tBt2)2 (5)

Use trial and error method to calculate the value of θ.

Case (1):

Try θ=30°:

Find the velocity at position 2:

Substitute 30° for θ in the equation (3).

v2=64.4sinθ=64.4sin30°=32.2=5.6745ft/s

Find the time difference between basket and ball:

Substitute 30° for θ and 5.6745ft/s for v2 in the equation (5).

(tBt2)=2cosθ0.3v2sinθ(tBt2)=2cos30°0.3(5.6745ft/s)sin30°(tBt2)=1.732050.32.83725(tBt2)=0.50473s

Find the vertical distance between basket and ball:

Substitute 30° for θ, 5.6745ft/s for v2, 32.2ft/s2 for g, and 0.50473s for (tBt2) in the equation (5).

yB=2sinθ+v2cosθ(tBt2)12g(tBt2)2=2sin30°+(5.6745ft/s)cos30°(0.50473s)12(32.2ft/s2)(0.50473s)2=1+2.480384.1015=0.62116ft

Case (2):

Try θ=45°.

Find the velocity at position 2:

Substitute 45° for θ in equation (3).

v2=64.4sinθ=64.4sin45°=45.5377=6.7482ft/s

Find the time difference between basket and ball:

Substitute 45° for θ and 6.7482ft/s for v2 in equation (4).

(tBt2)=2cosθ0.3v2sinθ(tBt2)=2cos45°0.3(6.7482ft/s)sin45°(tBt2)=1.4140.34.77169(tBt2)=0.23351s

Find the vertical distance between basket and ball:

Substitute 45° for θ, 6.7482ft/s for v2, 32.2ft/s2 for g and 0.23351s for (tBt2) in equation (5).

yB=2sinθ+v2cosθ(tBt2)12g(tBt2)2=2sin45°+(0.23351s)cos45°(6.7482ft/s)12(32.2ft/s2)(0.23351s)2=1.414+1.114230.87788=1.625060ft

Case (3):

Try θ=37.5°:

Find the velocity at position 2:

Substitute 37.5° for θ in equation (3).

v2=64.4sinθ=64.4sin37.5°=39.204=6.2613ft/s

Find the time difference between basket and ball:

Substitute 37.5° for θ and 6.2613ft/s for v2 in equation (4).

(tBt2)=2cosθ0.3v2sinθ(tBt2)=2cos37.5°0.3(6.2613ft/s)sin37.5°(tBt2)=1.586710.33.81164(tBt2)=0.33757s

Find the vertical distance between basket and ball:

Substitute 37.5° for θ, 6.2613ft/s for v2, 32.2ft/s2 for g and 0.33757s for (tBt2) in equation (5).

yB=2sinθ+v2cosθ(tBt2)12g(tBt2)2=2sin37.5°+(6.2613ft/s)cos37.5°(0.33757s)12(32.2ft/s2)(0.33757s)2=1.21752+1.676851.83465=1.05972ft

The expression for the data point as follows:

u=θ30°.

From above calculations, the following set of data points is obtained.

(u,yB)=(0°,0.62113ft),(7.5°,1.05972ft),(15°,1.65060ft)

Calculate the general form of quadratic Equation.

yB=au2+bu+c (6)

Here, a, b, c are constants.

Substitute 0 for u and 0.62113ft for yB in equation (6).

0.62113ft=a(0)2+b(0)+cc=0.62113

Substitute 7.5 for u, 1.05972ft for yB and 0.62113 for c in equation (6).

1.05972ft=a(7.5)2+b(7.5)0.621131256.25a+7.5b=1.6809 (7)

Substitute 15 for u, 1.65060ft for yB and 0.62113 for c in equation (6).

1.65060ft=a(15)2+b(15)0.62113225a+15b=2.27173 (8)

Solve the equation (7) and equation (8).

a=0.009688711b=0.29678

Substitute 0.009688711 for a, 0.29678 for b and 0.62113 for c in equation (6) to obtain the quadratic curve fit.

yB=0.009688711u2+0.29678u0.62113

Substitute 0.4ft for yB so that the ball enters the basket.

0.4ft=0.009688711u2+0.29678u0.621130.009688711u2+0.29678u1.02114=0

Solve the above equation.

u=3.95°

Calculate the angle (θ) using the relation:

u=θ30°

Substitute 37.5° for θ.

θ30°=3.95°θ=33.95°

Find the velocity at position 2:

Substitute 33.95° for θ in equation (3).

v2=64.4sinθv2=64.4sin33.95°v2=35.96542v2=5.9676m/s

Find the initial velocity (v0):

Substitute 5.9676m/s for v2, 32.2ft/s2 for g, 2 ft for l and 33.95° for θ in equation (2).

v02=v22+2gl(1+sinθ)v02=(5.9676m/s)2+2(32.2ft/s2)(2ft)(1+sin33.95°)v02=35.6122+200.7308v02=236.343

v0=236.343v0=15.37m/s

Therefore, the initial velocity (v0) so that the ball will enter the basket is 15.37m/s_.

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Chapter 13 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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