Chapter 6, Problem 45SP

Consider the simple pendulum shown in Fig. 6-7. (a) If it is released from point-A, what will be the speed of the ball as it passes through point-C? (b) What is the ball’s speed at point-B? [Hint: How far has it fallen upon arriving at point-B?]

Fig. 6-7

To determine

The speed of the ball at point C attached to a string of length $75\text{ cm}$ as shown in fig 6.7.

Answer to Problem 45SP

Solution:

$3.8\text{ m/s}$

Explanation of Solution

Given data:

Refer to Fig. 6-7.

The length of the string is $75\text{ cm}$.

Formula used:

Understand that, for a simple pendulum at its lowest point, the kinetic energy is maximum and the potential energy is zero, while at its highest point, the potential energy is maximum and the kinetic energy is zero.

The expression for change in kinetic energy is written as:

$KE=\frac{1}{2}m\left(v_f^2-v_i^2\right)$

Here, $m$ is the mass, $v_f$ is the final velocity, $v_i$ is the initial velocity, and $KE$ is the change in kinetic energy.

The expression for change in potential energy is written as:

$P{E}_G=mgh$

Here, $g$ is the acceleration due to gravity, $h$ is the length, and $P{E}_G$ is the change in potential energy.

Explanation:

There is no work done on the ball. Also, the ball’s total energy remains constant, that is, it loses a part of its potential energy but gains an equal amount ofkinetic energy. Therefore,

$\begin{array}{c}\text{Change in kinetic energy}=\text{change in potential energy}\\ KE=P{E}_G\end{array}$

Substitute $\frac{1}{2}m\left(v_f^2-v_i^2\right)$ for $KE$ and $mgh$ for $P{E}_G$

$\begin{array}{c}\frac{1}{2}m\left(v_f^2-v_i^2\right)=mgh\\ \frac{1}{2}\left(v_f^2-v_i^2\right)=gh\end{array}$

The velocity at the highest point is zero. Therefore,

$v_i=0\text{ m/s}$

Recall the expression for the work energy theorem.

$\frac{1}{2}\left(v_f^2-{\left(0\right)}^2\right)=gh$

Substitute $75\text{ cm}$ for $h$ and $9.81{\text{ m/s}}^2$ for $g$.

$\begin{array}{c}\frac{1}{2}\left(v_f^2\right)=\left(9.81{\text{ m/s}}^2\right)\left(75\text{ cm}\left(\frac{0.01\text{ m}}{1\text{ cm}}\right)\right)\\ \frac{1}{2}{v}_f^2=7.3575\\ v_f=\sqrt{\left(2\right)\left(7.3575\right)}\\ v_f\simeq 3.8\text{ m/s}\end{array}$

Conclusion:

The speed of the ball at point C is $3.8\text{ m/s}$.

To determine

The speed of the ball at point B attached to a string of length $75\text{ cm}$ as shown in fig 6.7.

Answer to Problem 45SP

Solution:

$3.4\text{ m/s}$

Explanation of Solution

Given data:

Refer to Fig. 6-7.

The length of the string is $75\text{ cm}$.

Formula used:

Understand that, for a simple pendulum at its lowest point, the kinetic energy is maximum and the potential energy is zero, while at its highest point, the potential energy is maximum and the kinetic energy is zero.

The expression for change in kinetic energy is written as:

$KE=\frac{1}{2}m\left(v_f^2-v_i^2\right)$

Here, $m$ is the mass, $v_f$ is the final velocity, $v_i$ is the initial velocity, and $KE$ is the change in kinetic energy.

The expression for change in potential energy is written as:

$P{E}_G=mgh$

Here, $g$ is the acceleration due to gravity, $h$ is the length, and $P{E}_G$ is the change in potential energy.

Explanation:

Draw the diagram of a simple pendulum.

There is no work done on the ball. Also, the ball’s total energy remains constant, that is, it loses a part of its kinetic energy but gains an equal amount of potential energy. Therefore,

$\begin{array}{c}\text{Change in kinetic energy+change in potential energy = 0}\\ KE+P{E}_G=0\end{array}$

Substitute $\frac{1}{2}m\left(v_f^2-v_i^2\right)$ for $KE$ and $mgh$ for $P{E}_G$

$\begin{array}{c}\frac{1}{2}m\left(v_f^2-v_i^2\right)+mg\left(h_f-h_i\right)=0\\ \frac{1}{2}\left(v_f^2-v_i^2\right)+g\left(h_f-h_i\right)=0\end{array}$

Consider the diagram and determine the distance between points $C^{\prime }$ and $C$ using trigonometric relations.

$\begin{array}{c}C{C}^{\prime }=75\text{cm-}\left(75\text{ cm}\cos 37°\right)\\ =75\text{ cm}\left(1-\cos 37°\right)\\ =15.10\text{ cm}\end{array}$

From previous part, the velocity at point C is $3.8\text{m/s}$. Therefore,

$v_C=3.8\text{m/s}$

Recall the expression for the work energy theorem.

$\begin{array}{l}\frac{1}{2}\left(v_B^2-v_C^2\right)+g\left(h_B-h_C\right)=0\\ \frac{1}{2}\left(v_B^2-v_C^2\right)+g\left(CC\text{'}-0\right)=0\end{array}$

Substitute $15.10\text{ cm}$ for $C{C}^{\prime }$, $9.81{\text{ m/s}}^2$ for $g$, and $3.8\text{ m/s}$ for $v_c$

$\begin{array}{c}\frac{1}{2}\left(v_B{}^2-{\left(3.8\right)}^2\right)+\left(9.81{\text{ m/s}}^2\right)\left(15.10\text{ cm}\left(\frac{0.01\text{ m}}{1\text{ cm}}\right)\right)=0\\ \frac{1}{2}{v}_B^2-7.22+1.481=0\\ v_B=\sqrt{\left(2\right)\left(5.739\right)}\\ v_B\simeq 3.4\text{ m/s}\end{array}$

Conclusion:

The speed of ball at point B is $3.4\text{ m/s}$.