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, x_B = 0.3 ft, and y_B = 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

To determine

Find the initial velocity $\left(v_0\right)$ so that the ball will enter the basket.

Answer to Problem 13.201RP

The initial velocity $\left(v_0\right)$ so that the ball will enter the basket is $\underset{\_}{15.37\text{m}/\text{s}}$.

Explanation of Solution

Given information:

The weight of the ball (m) is $\text{2}\text{lb}$.

The length of the cord (l) is $2\text{ft}$.

The horizontal distance between the basket and point of suspension of the ball $\left(x_B\right)$ is $\text{0}\text{.3ft}$.

The vertical distance between basket and point of suspension of the ball $\left(y_B\right)$ is $\text{0}\text{.4ft}$.

The acceleration due to gravity (g) is $32.2{\text{ft/s}}^{\text{2}}$.

Calculation:

Show the diagram of the suspended ball by an inextensible cord as in Figure (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 $\left(Q\right)$ which is zero.

Refer Figure (1),

The expression for the x-coordinate of the ball at position ‘2’ $\left(x_2\right)$ as follows:

$x_2=l\cos \theta$

The expression for the y-coordinate of the ball at position ‘2’ $\left(y_2\right)$ as follows:

$y_2=l\sin \theta$

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

The expression for the normal acceleration of the ball $\left(a_n\right)$ as follows:

$a_n=\frac{v_2^2}{l}$

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:

$Q+mg\sin \theta =m{a}_n$

Substitute $\frac{v_2^2}{l}$ for $a_n$.

$Q+mg\sin \theta =m\left(\frac{v_2^2}{l}\right)$

Substitute 0 for $Q$.

$\begin{array}{l}0+mg\sin \theta =m\left(\frac{v_2^2}{l}\right)\\ v_2^2=gl\sin \theta \\ v_2=\sqrt{gl\sin \theta }\end{array}$ (1)

The expression for the kinetic energy of the ball at position ‘1’ $\left(T_1\right)$ as follows:

$T_1=\frac{1}{2}m{v}_0^2$

Calculate the potential energy of the ball at position ‘1’ $\left(V_1\right)$ using the formula:

$V_1=-mgh$

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 $h$.

$V_1=-mgl$

The expression for the kinetic energy of the ball at position ‘2’ $\left(T_2\right)$ as follows:

$T_2=\frac{1}{2}m{v}_2^2$

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

$h=l\sin \theta$

Calculate the potential energy of the ball at position ‘2’ $\left(V_2\right)$ using the formula:

$V_2=mgh$

Substitute $l\sin \theta$ for h.

$V_2=mgl\sin \theta$

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 $\left(\theta \right)$ as follows:

$T_1+V_1=T_2+V_2$

Substitute $\frac{1}{2}m{v}_0^2$ for $T_1$, $-mgl$ for $V_1$, $\frac{1}{2}m{v}_2^2$ for $T_2$ and $mgl\sin \theta$ for $V_2$.

$\begin{array}{l}{T}_1+V_1=T_2+V_2\\ \frac{1}{2}m{v}_0^2-mgl=\frac{1}{2}m{v}_2^2+mgl\sin \theta \\ \frac{1}{2}\left(v_0^2-v_2^2\right)=gl+gl\sin \theta \\ v_0^2-v_2^2=2gl\left(1+\sin \theta \right)\\ v_0^2=v_2^2+2gl\left(1+\sin \theta \right)\end{array}$ (2)

Find the velocity at position 2:

Substitute $32.2\text{ft}/{\text{s}}^2$ for $g$ and $2\text{ft}$ for $l$ in the equation (1).

$\begin{array}{c}{v}_2=\sqrt{gl\sin \theta }\\ =\sqrt{\left(32.2\text{ft}/{\text{s}}^2\right)\left(2\text{ft}\right)\sin \theta }\\ =\sqrt{64.4\sin \theta }\end{array}$ (3)

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

The expression for the velocity of the projectile ball after reaching the position ‘2’ $\left(v_p\right)$ as follows:

$v_p={\left(v_p\right)}_x\text{i}+{\left(v_p\right)}_y\text{j}$

Here, ${\left(v_p\right)}_x$ is the horizontal velocity component of the projectile and ${\left(v_p\right)}_y$ 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:

$\begin{array}{c}{\left(v_p\right)}_x=-v_2\cos \left(90-\theta \right)\\ =-v_2\sin \theta \end{array}$

The expression for the horizontal distance between the basket and point of suspension of the ball $\left(x_B\right)$ as follows:

$x_B=x_2-v_2\sin \theta \left(t_B-t_2\right)$

Here, $t_B$ is the time taken by the ball to reach the basket and $t_2$ is the time taken by the ball to reach the position ‘2’.

Substitute $l\cos \theta$ for $x_2$.

$\begin{array}{l}{x}_B=l\cos \theta -v_2\sin \theta \left(t_B-t_2\right)\\ v_2\sin \theta \left(t_B-t_2\right)=l\cos \theta -x_B\\ \left(t_B-t_2\right)=\frac{l\cos \theta -x_B}{v_2\sin \theta }\end{array}$

Substitute $0.3\text{ft}$ for $x_B$ and $2\text{ft}$ for $l$.

$\left(t_B-t_2\right)=\frac{2\cos \theta -0.3}{v_2\sin \theta }$ (4)

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

$\begin{array}{c}{\left(v_p\right)}_y=v_2\sin (90-\theta )\\ =v_2\cos \theta \end{array}$

The expression for the vertical distance between the basket and point of suspension of the ball $\left(y_B\right)$ as follows:

$y_B=y_2+v_2\cos \theta \left(t_B-t_2\right)-\frac{1}{2}g{\left(t_B-t_2\right)}^2$

Substitute $l\sin \theta$ for $y_2$.

$y_B=l\sin \theta +v_2\cos \theta \left(t_B-t_2\right)-\frac{1}{2}g{\left(t_B-t_2\right)}^2$

Substitute $2\text{ft}$ for $l$.

$y_B=2\sin \theta +v_2\cos \theta \left(t_B-t_2\right)-\frac{1}{2}g{\left(t_B-t_2\right)}^2$ (5)

Use trial and error method to calculate the value of $\theta$.

Case (1):

Try $\theta =30°$:

Find the velocity at position 2:

Substitute $30°$ for $\theta$ in the equation (3).

$\begin{array}{c}{v}_2=\sqrt{64.4\sin \theta }\\ =\sqrt{64.4\sin 30°}\\ =\sqrt{32.2}\\ =5.6745\text{ft}/\text{s}\end{array}$

Find the time difference between basket and ball:

Substitute $30°$ for $\theta$ and $5.6745\text{ft}/\text{s}$ for $v_2$ in the equation (5).

$\begin{array}{l}\left(t_B-t_2\right)=\frac{2\cos \theta -0.3}{v_2\sin \theta }\\ \left(t_B-t_2\right)=\frac{2\cos 30°-0.3}{\left(5.6745\text{ft}/\text{s}\right)\sin 30°}\\ \left(t_B-t_2\right)=\frac{1.73205-0.3}{2.83725}\\ \left(t_B-t_2\right)=0.50473\text{s}\end{array}$

Find the vertical distance between basket and ball:

Substitute $30°$ for $\theta$, $5.6745\text{ft}/\text{s}$ for $v_2$, $32.2\text{ft}/{\text{s}}^2$ for $g$, and $0.50473\text{s}$ for $\left(t_B-t_2\right)$ in the equation (5).

$\begin{array}{c}{y}_B=2\sin \theta +v_2\cos \theta \left(t_B-t_2\right)-\frac{1}{2}g{\left(t_B-t_2\right)}^2\\ =2\sin 30°+\left(5.6745\text{ft}/\text{s}\right)\cos 30°\left(0.50473\text{s}\right)-\frac{1}{2}\left(32.2\text{ft}/{\text{s}}^2\right){\left(0.50473\text{s}\right)}^2\\ =1+2.48038-4.1015\\ =-0.62116\text{ft}\end{array}$

Case (2):

Try $\theta =45°$.

Find the velocity at position 2:

Substitute $45°$ for $\theta$ in equation (3).

$\begin{array}{c}{v}_2=\sqrt{64.4\sin \theta }\\ =\sqrt{64.4\sin 45°}\\ =\sqrt{45.5377}\\ =6.7482\text{ft}/\text{s}\end{array}$

Find the time difference between basket and ball:

Substitute $45°$ for $\theta$ and $6.7482\text{ft}/\text{s}$ for $v_2$ in equation (4).

$\begin{array}{l}\left(t_B-t_2\right)=\frac{2\cos \theta -0.3}{v_2\sin \theta }\\ \left(t_B-t_2\right)=\frac{2\cos 45°-0.3}{\left(6.7482\text{ft}/\text{s}\right)\sin 45°}\\ \left(t_B-t_2\right)=\frac{1.414-0.3}{4.77169}\\ \left(t_B-t_2\right)=0.23351\text{s}\end{array}$

Find the vertical distance between basket and ball:

Substitute $45°$ for $\theta$, $6.7482\text{ft}/\text{s}$ for $v_2$, $32.2\text{ft}/{\text{s}}^2$ for $g$ and $0.23351\text{s}$ for $\left(t_B-t_2\right)$ in equation (5).

$\begin{array}{c}{y}_B=2\sin \theta +v_2\cos \theta \left(t_B-t_2\right)-\frac{1}{2}g{\left(t_B-t_2\right)}^2\\ =2\sin 45°+\left(0.23351\text{s}\right)\cos 45°\left(6.7482\text{ft}/\text{s}\right)-\frac{1}{2}\left(32.2\text{ft}/{\text{s}}^2\right){\left(0.23351\text{s}\right)}^2\\ =1.414+1.11423-0.87788\\ =1.625060\text{ft}\end{array}$

Case (3):

Try $\theta =37.5°$:

Find the velocity at position 2:

Substitute $37.5°$ for $\theta$ in equation (3).

$\begin{array}{c}{v}_2=\sqrt{64.4\sin \theta }\\ =\sqrt{64.4\sin 37.5°}\\ =\sqrt{39.204}\\ =6.2613\text{ft}/\text{s}\end{array}$

Find the time difference between basket and ball:

Substitute $37.5°$ for $\theta$ and $6.2613\text{ft}/\text{s}$ for $v_2$ in equation (4).

$\begin{array}{l}\left(t_B-t_2\right)=\frac{2\cos \theta -0.3}{v_2\sin \theta }\\ \left(t_B-t_2\right)=\frac{2\cos 37.5°-0.3}{\left(6.2613\text{ft}/\text{s}\right)\sin 37.5°}\\ \left(t_B-t_2\right)=\frac{1.58671-0.3}{3.81164}\\ \left(t_B-t_2\right)=0.33757\text{s}\end{array}$

Find the vertical distance between basket and ball:

Substitute $37.5°$ for $\theta$, $6.2613\text{ft}/\text{s}$ for $v_2$, $32.2\text{ft}/{\text{s}}^2$ for $g$ and $0.33757\text{s}$ for $\left(t_B-t_2\right)$ in equation (5).

$\begin{array}{c}{y}_B=2\sin \theta +v_2\cos \theta \left(t_B-t_2\right)-\frac{1}{2}g{\left(t_B-t_2\right)}^2\\ =2\sin 37.5°+\left(6.2613\text{ft}/\text{s}\right)\cos 37.5°\left(0.33757\text{s}\right)-\frac{1}{2}\left(32.2\text{ft}/{\text{s}}^2\right){\left(0.33757\text{s}\right)}^2\\ =1.21752+1.67685-1.83465\\ =1.05972\text{ft}\end{array}$

The expression for the data point as follows:

$u=\theta -30°$.

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

$\left(u,y_B\right)=\left(0°,-0.62113\text{ft}\right),\left(7.5°,1.05972\text{ft}\right),\left(15°,1.65060\text{ft}\right)$

Calculate the general form of quadratic Equation.

$y_B=a{u}^2+bu+c$ (6)

Here, a, b, c are constants.

Substitute 0 for $u$ and $-0.62113\text{ft}$ for $y_B$ in equation (6).

$\begin{array}{l}-0.62113\text{ft}=a{\left(0\right)}^2+b\left(0\right)+c\\ c=-0.62113\end{array}$

Substitute $7.5$ for $u$, $1.05972\text{ft}$ for $y_B$ and $-0.62113$ for $c$ in equation (6).

$\begin{array}{l}1.05972\text{ft}=a{\left(7.5\right)}^2+b\left(7.5\right)-0.62113\frac{1}{2}\\ 56.25a+7.5b=1.6809\end{array}$ (7)

Substitute $15$ for $u$, $1.65060\text{ft}$ for $y_B$ and $-0.62113$ for $c$ in equation (6).

$\begin{array}{l}1.65060\text{ft}=a{\left(15\right)}^2+b\left(15\right)-0.62113\\ 225a+15b=2.27173\end{array}$ (8)

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

$\begin{array}{l}a=-0.009688711\\ b=0.29678\end{array}$

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

$y_B=-0.009688711u^2+0.29678u-0.62113$

Substitute $0.4\text{ft}$ for $y_B$ so that the ball enters the basket.

$\begin{array}{l}0.4\text{ft}=-0.009688711u^2+0.29678u-0.62113\\ -0.009688711u^2+0.29678u-1.02114=0\end{array}$

Solve the above equation.

$u=3.95°$

Calculate the angle $\left(\theta \right)$ using the relation:

$u=\theta -30°$

Substitute $37.5°$ for $\theta$.

$\begin{array}{l}\theta -30°=3.95°\\ \theta =33.95°\end{array}$

Find the velocity at position 2:

Substitute $33.95°$ for $\theta$ in equation (3).

$\begin{array}{l}{v}_2=\sqrt{64.4\sin \theta }\\ v_2=\sqrt{64.4\sin 33.95°}\\ v_2=\sqrt{35.96542}\\ v_2=5.9676\text{m}/\text{s}\end{array}$

Find the initial velocity $\left(v_0\right)$:

Substitute $5.9676\text{m}/\text{s}$ for $v_2$, $32.2\text{ft}/{\text{s}}^2$ for $g$, 2 ft for $l$ and $33.95°$ for $\theta$ in equation (2).

$\begin{array}{l}{v}_0^2=v_2^2+2gl\left(1+\sin \theta \right)\\ v_0^2={\left(5.9676\text{m}/\text{s}\right)}^2+2\left(32.2\text{ft}/{\text{s}}^2\right)\left(2\text{ft}\right)\left(1+\sin 33.95°\right)\\ v_0^2=35.6122+200.7308\\ v_0^2=236.343\end{array}$

$\begin{array}{l}{v}_0=\sqrt{236.343}\\ v_0=15.37\text{m}/\text{s}\end{array}$

Therefore, the initial velocity $\left(v_0\right)$ so that the ball will enter the basket is $\underset{\_}{15.37\text{m}/\text{s}}$.