Chapter 13.2, Problem 13.77P

To determine

y_B so that the ball will enter the basket.

Answer to Problem 13.77P

The value of y_B=0.488m.

Explanation of Solution

Consider the diagram of the suspended ball by an inextensible cord.

Let us consider at point “A” as position 1 and then the point described by the angle is considered as position 2.

The path of the ball changes from the circular motion to parabolic motion.

So far then the tension in the cord becomes zero.

From the above diagram x-coordinate of the ball at the position 2 is taken as x_2.

$x_2=l\cos \theta$

Then the y-coordinate of the ball at position 2 is takes as y_2.

$y_2=l\sin \theta$

So the free body diagram of the ball at position 2 is calculated by :

Then apply the newton’s second law after that resolve the forces acting on the ball at the position 2

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

So, here the cord becomes slack at point 2.

And substitute Q=0 in above formula, we get the:

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

Here applying the conservation of energy at position 1 and 2 for the ball.

To find the angle swept by the ball is $\theta$.

$T_1+V_1=T_2+V_2----(2)$

Then formulate the kinetic energy of the ball at position 1 as T_1.

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

$v_0=$ Velocity of the ball at point 1

Formulate the potential energy of the ball, we get:

$V_1=-mgh$

Here we have to substitute h=l, we get:

$V_1=-mgl$

Formulate the kinetic energy of the ball at the position 2 to be considered as T_2.

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

Formulate the potential energy of the ball at point 2 is V_2.

$V_2=mgh$

Consider equation 1 and substitute required values.

$V_2=mgl\sin \theta$

Substitute $T_1=\frac{1}{2}m{v}_0^2$ and $V_2=mgl\sin \theta$, $T_2=\frac{1}{2}m{v}_2^2$ in above equation, we get:

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

Finally from equation (1) we get:

$\begin{array}{l}{v}_0^2-gl\sin \theta =2gl\left(1+\sin \theta \right)\\ gl\left(2+3\sin \theta \right)=v_0^2\\ 2+3\sin \theta =\frac{v_0^2}{gl}\\ \sin \theta =\frac{1}{3}\left(\frac{v_0^2}{gl}-2\right)\end{array}$

Substitute required values, we get:

$\begin{array}{l}\sin \theta =\frac{1}{3}\left(\frac{v_0^2}{gl}-2\right)\\ \sin \theta =\frac{1}{3}\left(\frac{{\left(5m/s\right)}^2}{\left(9.81m/s^2\right)\left(0.6m\right)}-2\right)\\ \theta ={\sin }^{-1}\left(0.74912\right)\\ \theta ={48.514}^0\end{array}$

Then, find the velocity of the ball at point 2.

$\begin{array}{l}{v}_2=\sqrt{gl\sin \theta }\\ =\sqrt{\left(9.81m/s^2\right)\left(0.6m\right)\sin {48.514}^0}\\ =\sqrt{4.4093}\\ =2.0998m/s\end{array}$

After that the velocity of the projectile ball after reaches its point 2 is to be v_p.

$v_p={\left(v_p\right)}_{x}i+{\left(v_p\right)}_{y}j$

In this we have horizontal velocity component and vertical velocity component.

Formulate the horizontal velocity component of the projectile ball along with −ve X-axis.

$\begin{array}{l}{v}_p={\left(v_p\right)}_{x}i+{\left(v_p\right)}_{y}j\\ {\left(v_p\right)}_x=-v_2\cos \left(90-\theta \right)\\ =-v_2\sin \theta \end{array}$

Formulate the horizontal distance between the basket and point of suspension of the ball.

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

Substitute $x_2=l\cos \theta$ in above equation, we get:

$\begin{array}{l}{x}_B=-l\cos \theta -v_2\sin \theta \left(t_B-t_2\right)\\ x_B=\left(0.6m\right)\cos {48.51}^0-\left(2.0998m/s\right)\sin {48.514}^0\left(t_B-t_2\right)\\ 0=0.3975-1.573\left(t_B-t_2\right)\\ 1.573\left(t_B-t_2\right)=0.3975\\ t_B-t_2=0.25267s----(3)\end{array}$

Formulate the vertical velocity component of the projectile ball with the negative X-axis.

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

Vertical distance between the basket and the point of suspension is to be y_B..

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

Substitute $y_2=l\sin \theta$ in above, we get:

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

Find the y-coordinate value by substituting all the required values.

$\begin{array}{l}{y}_B=\left(0.6m\right)\sin {48.514}^0+\left(2.0998m/s\right)\cos {48.514}^0\left(0.25267s\right)-\frac{1}{2}\left(9.81m/s^2\right){\left(0.25267s\right)}^2\\ =0.4495+0.35146-0.313145\\ =0.48779m\end{array}$