Chapter 5, Problem 19P

(a)

To determine

Draw the free body diagram for net force acts on the ball.

Answer to Problem 19P

The free body diagram is drawn for the tension force and gravitational force exerted on the ball.

Explanation of Solution

The figure 1 shows the free body diagram for the tension force and gravitational force exerted on the ball.

Here, $T_A$ is the tension acts on the string $A$ attached to the ball, $T_B$ is the tension acts on the string $B$ attached to the ball, $m$ is the mass of the ball, and $g$ is the gravitational acceleration.

(b)

To determine

The magnitude of the tension acts on the two strings.

Answer to Problem 19P

The magnitude of the tension acts on the string $A$ is $T_A=3.64\text{N}$ and the string $B$ is $T_B=1.68\text{N}$

Explanation of Solution

Write the equation for net force exerted on the ball along y axis (vertical) by using Newton’s second law.

$\begin{array}{c}{\displaystyle \sum F_y}=0\\ T_A\sin \theta -T_B\sin \theta -mg=0\end{array}$ (I)

Here, $T_A$ is the tension acts on the string $A$ attached to the ball, $T_B$ is the tension acts on the string $B$ attached to the ball, $m$ is the mass of the ball, and $g$ is the gravitational acceleration.

Write the equation for net force exerted on the ball along x axis (horizontal) by using Newton’s second law.

$\begin{array}{c}{\displaystyle \sum F_x}=m{a}_x\\ T_A\cos \theta +T_B\cos \theta =m{a}_x\end{array}$ (II)

Here, $a_x$ is the acceleration of the ball

Rewrite the equation (I) for difference between the tension.

$\begin{array}{c}\sin \theta \left(T_A-T_B\right)=mg\\ T_A-T_B=\frac{mg}{\sin \theta }\end{array}$ (III)

Write the relation between radial acceleration and angular speed.

$a_x={\omega }^{2}r$ (IV)

Here, $\omega$ is the angular speed of circular motion of ball and $r$ is the radius of the ball.

Conclusion:

Substitute equation (IV) in equation (II).

$\begin{array}{c}{T}_A\cos \theta +T_B\cos \theta =m{\omega }^{2}r\\ \cos \theta \left(T_A+T_B\right)=m{\omega }^{2}r\\ T_A+T_B=\frac{m{\omega }^{2}r}{\cos \theta }\end{array}$ (V)

Solve to find $T_A$ by adding the equation (III) and (V).

$\begin{array}{c}\underset{\_}{\begin{array}{l}{T}_A-T_B=\frac{mg}{\sin \theta }\\ T_A+T_B=\frac{m{\omega }^{2}r}{\cos \theta }\end{array}}\\ 2T_A=m\left(\frac{g}{\sin \theta }+\frac{{\omega }^{2}r}{\cos \theta }\right)\\ T_A=\frac{m}{2}\left[\left(\frac{g}{\sin \theta }+\frac{{\omega }^{2}r}{\cos \theta }\right)\right]\end{array}$ (VI)

The length of the strings make an angle of $30.0°$ so that $r=\left(15.0\text{cm}\right)\cos 30.0°$.

Substitute $0.100\text{kg}$ for $m$, $9.80{\text{m/s}}^2$ for $g$, $30.0°$ for $\theta$,$6.00\pi \text{rad/s}$ for $\omega$, and $\left(15.0\text{cm}\right)\cos 30.0°$ for $r$ in equation (VI).

$\begin{array}{c}{T}_A=\frac{0.100\text{kg}}{2}\left[\left(\frac{9.80{\text{m/s}}^2}{\sin 30.0°}+\frac{{\left(6.00\pi \text{rad/s}\right)}^2\left(\left(15.0\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)\cos 30.0°\right)}{\cos 30.0°}\right)\right]\\ =0.05\text{kg}\left[19.6{\text{m/s}}^2+53.2{\text{m/s}}^2\right]\\ =3.64\text{N}\end{array}$

Thus the magnitude of the tension acts on the string $A$ is $T_A=3.64\text{N}$.

Substitute $3.64\text{N}$ for $T_A$, $0.100\text{kg}$ for $m$, $9.80{\text{m/s}}^2$ for $g$, and $30.0°$ for $\theta$ in equation (III).

$\begin{array}{c}{T}_B=3.64\text{N}-\frac{\left(0.100\text{kg}\right)\left(9.80{\text{m/s}}^2\right)}{\sin 30.0°}\\ =3.64\text{N}-1.96\text{N}\\ =1.68\text{N}\end{array}$

Thus the magnitude of the tension acts on the string $B$ is $T_B=1.68\text{N}$.