Chapter 10, Problem 77P

(a)

To determine

Find the energy of the pendulum.

Answer to Problem 77P

The energy of the pendulum is $6.1\text{mJ}$.

Explanation of Solution

Write the equation for total mechanical energy of a pendulum.

$E=\frac{1}{2}m{\omega }^2{A}^2$ (I)

Here, $m$ is the mass of the pendulum, $\omega$ is the angular frequency, and $A$ is the amplitude of the oscillating pendulum.

Write the equation for angular frequency of the SHM for pendulum oscillation (Refer Equation 10-26a).

$\omega =\sqrt{\frac{g}{L}}$ (II)

Here, $g$ is the gravitational acceleration and $L$ is the length of the pendulum.

Conclusion:

Substitute equation (II) in (I).

$\begin{array}{c}E=\frac{1}{2}m{\left(\sqrt{\frac{g}{L}}\right)}^2{A}^2\\ =\frac{mg{A}^2}{2L}\end{array}$ (III)

Substitute $0.50\text{kg}$ for $m$, $9.80{\text{m/s}}^2$ for $g$, $5.0\text{cm}$ for $A$, and $1.0\text{m}$ for $L$ in the equation (III).

$\begin{array}{c}E=\frac{\left(0.50\text{kg}\right)\left(9.80{\text{m/s}}^2\right){\left[\left(5.0\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)\right]}^2}{2\left(1.0\text{m}\right)}\\ =6.125\times {10}^{-3}\text{J}\left(\frac{1\text{mJ}}{{10}^{-3}\text{J}}\right)\\ =6.125\text{mJ}\\ \simeq \text{6}\text{.1}\text{mJ}\end{array}$

Therefore, the energy of the pendulum is $6.1\text{mJ}$.

(b)

To determine

Find the percentage of the pendulum’s energy lost during one cycle.

Answer to Problem 77P

The percentage of the pendulum’s energy lost during one cycle is $1.1\%$.

Explanation of Solution

Write the equation for time period of the pendulum.

$T=2\pi \sqrt{\frac{L}{g}}$ (IV)

Here, $T$ is the time period of the oscillating pendulum.

Write the equation for potential energy of an object.

$U=m_{2}gh$ (V)

Here, $U$ is the potential energy of an object, $m_2$ is the mass of an object to drop in a week, and $h$ is the distance of the object to drop in 1 week.

The loss of oscillation time period of cycle for 1 week is.

$\text{cycles per week}=\frac{1\text{wk}}{T}$ (VI)

The percentage of pendulum’s energy loss for 1 week is.

$E_{(\%)}=\frac{\left(\frac{U}{\text{cycles}/\text{wk}}\right)}{E}\times 100\%$ (VII)

Here, $E_{(\%)}$ is the percentage of the pendulum’s energy lost during one cycle.

Conclusion:

Substitute equation (IV) in (VI).

$\begin{array}{c}\text{cycles per week}=\frac{1\text{wk}\left(\frac{604,800\text{s}}{1\text{wk}}\right)}{2\pi }\sqrt{\frac{g}{L}}\\ =\frac{604,800\text{s}}{2\pi }\sqrt{\frac{g}{L}}\end{array}$ (VIII)

Substitute the equation (III), (V) and (VIII) in equation (VII).

$\begin{array}{c}{E}_{(\%)}=\left(\frac{m_{2}gh}{\frac{604,800\text{s}}{2\pi }\sqrt{\frac{g}{L}}\left(\frac{mg{A}^2}{2L}\right)}\right)\times 100\%\\ =\frac{4\pi m_{2}h}{\left(604,800\text{s}\right)m{A}^2}\sqrt{\frac{L^3}{g}}\times 100\%\end{array}$

Substitute $0.50\text{kg}$ for $m$, $2.0\text{kg}$ for $m_2$, $9.80{\text{m/s}}^2$ for $g$, $5.0\text{cm}$ for $A$, $1.0\text{m}$ for $L$, and $h$ in above equation.

$\begin{array}{c}{E}_{(\%)}=\frac{4\left(3.14\right)\left(2.0\text{kg}\right)\left(1.0\text{m}\right)}{\left(604,800\text{s}\right)\left(0.50\text{kg}\right){\left[\left(5.0\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)\right]}^2}\sqrt{\frac{{\left(1.0\text{m}\right)}^3}{9.80{\text{m/s}}^2}}\times 100\%\\ =1.06\%\\ \simeq 1.1\%\end{array}$

Therefore, the percentage of the pendulum’s energy lost during one cycle is $1.1\%$.