Chapter 10, Problem 103P

(a)

To determine

Find the period of the pendulum for each horizontal axis.

Answer to Problem 103P

The period of the pendulum for each horizontal axis is $1.64\text{s}$.

Explanation of Solution

Write the equation for time period of the physical pendulum.

$T=2\pi \sqrt{\frac{I}{mgd}}$ (I)

Here, $T$ is the time period of the physical pendulum, $g$ is the gravitational acceleration, $I$ is the moment of inertia for uniform rod with the axis through its end, $m$ is the mass of the rod, and $d$ is the distance from the axis to the center of mass.

Write the equation of the moment of inertia for uniform rod.

$I=\frac{1}{3}m{L}^2$ (II)

Here, $L$ is the length of the meter stick stretched to the pendulum.

Conclusion:

Substitute equation (II) in equation (I) and replace $L/2$ for $d$.

$\begin{array}{c}T=2\pi \sqrt{\frac{\frac{1}{3}m{L}^2}{mgd}}\\ =2\pi L\sqrt{\frac{1}{3gd}}\\ =2\pi L\sqrt{\frac{1}{3g\left(L/2\right)}}\end{array}$

Substitute $100\text{cm}$ for $L$ and $9.80{\text{m/s}}^2$ for $g$ in the above equation.

$\begin{array}{c}T=2\left(3.14\right)\left(100\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)\sqrt{\frac{1}{3\left(9.80{\text{m/s}}^2\right)\left(\frac{\left(100\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)}{2}\right)}}\\ =2\left(3.14\right)\left(1.0\text{m}\right)\sqrt{\frac{1}{3\left(9.80{\text{m/s}}^2\right)\left(\frac{1.0\text{m}}{2}\right)}}\\ =1.64\text{s}\end{array}$

Therefore, the time period of the pendulum for each horizontal axis is $1.64\text{s}$.

(b)

To determine

Find the period of the pendulum for two rods.

Answer to Problem 103P

The time period of the pendulum for two rods is $1.53\text{s}$.

Explanation of Solution

Consider the meter stick as two rods with lengths $75\text{cm}$ and $25\text{cm}$.

Write the equation of the moment of inertia for ${\left(1/4\right)}^{\text{th}}$ uniform rod.

$I=\frac{1}{3}\left(\frac{m}{4}\right){\left(\frac{L}{4}\right)}^2$ (III)

Here, $L$ is the length of the meter stick stretched to the pendulum.

Write the equation of the moment of inertia for ${\left(3/4\right)}^{\text{th}}$ uniform rod.

$I=\frac{1}{3}\left(\frac{3m}{4}\right){\left(\frac{3L}{4}\right)}^2$ (IV)

Rewrite the equation (III).

$\begin{array}{c}I=\frac{1}{3}\left(\frac{m}{4}\right)\left(\frac{L^2}{16}\right)\\ =\frac{1}{3}m{L}^2\left(\frac{1}{64}\right)\end{array}$

Rewrite the equation (IV)

$\begin{array}{c}I=\frac{1}{3}\left(\frac{3m}{4}\right)\left(\frac{9L^2}{16}\right)\\ =\frac{1}{3}m{L}^2\left(\frac{27}{64}\right)\end{array}$

The total inertia of the two rods is.

$\begin{array}{c}I=\frac{1}{3}m{L}^2\left(\frac{1}{64}+\frac{27}{64}\right)\\ =\frac{1}{3}m{L}^2\left(\frac{28}{64}\right)\\ =\frac{7}{48}m{L}^2\end{array}$

Conclusion:

Substitute $7/48m{L}^2$ for $I$ and $L/4$ for $d$ in equation (I).

$\begin{array}{c}T=2\pi \sqrt{\frac{\frac{7}{48}m{L}^2}{mg\left(\frac{L}{4}\right)}}\\ =2\pi \sqrt{\frac{7m{L}^2}{48mg\left(\frac{L}{4}\right)}}\\ =2\pi \sqrt{\frac{7L}{12g}}\end{array}$

Substitute $100\text{cm}$ for $L$ and $9.80{\text{m/s}}^2$ for $g$ in the above equation.

$\begin{array}{c}T=2\left(3.14\right)\sqrt{\frac{7\left(100\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)}{12\left(9.80{\text{m/s}}^2\right)}}\\ =1.53\text{s}\end{array}$

Therefore, the time period of the pendulum for two rods is $1.53\text{s}$.

(c)

To determine

Find the period of the pendulum for two rods.

Answer to Problem 103P

The time period of the pendulum for two rods is $1.94\text{s}$.

Explanation of Solution

Consider the meter stick as two rods with lengths $60\text{cm}$ and $40\text{cm}$.

Write the equation of the moment of inertia for ${\left(4/10\right)}^{\text{th}}$ uniform rod.

$I=\frac{1}{3}\left(\frac{4m}{10}\right){\left(\frac{4L}{10}\right)}^2$ (V)

Write the equation of the moment of inertia for ${\left(6/10\right)}^{\text{th}}$ uniform rod.

$I=\frac{1}{3}\left(\frac{6m}{10}\right){\left(\frac{6L}{10}\right)}^2$ (VI)

Rewrite the equation (V).

$\begin{array}{c}I=\frac{1}{3}\left(\frac{4m}{10}\right)\left(\frac{16L^2}{100}\right)\\ =\frac{1}{3}m{L}^2\left(\frac{64}{1000}\right)\end{array}$

Rewrite the equation (VI)

$\begin{array}{c}I=\frac{1}{3}\left(\frac{6m}{10}\right)\left(\frac{36L^2}{100}\right)\\ =\frac{1}{3}m{L}^2\left(\frac{216}{1000}\right)\end{array}$

The total inertia of the two rods is.

$\begin{array}{c}I=\frac{1}{3}m{L}^2\left(\frac{64}{1000}+\frac{216}{1000}\right)\\ =\frac{1}{3}m{L}^2\left(\frac{280}{1000}\right)\\ =\frac{7}{75}m{L}^2\end{array}$

Conclusion:

Substitute $7/75m{L}^2$ for $I$ and $L/10$ for $d$ in equation (I).

$\begin{array}{c}T=2\pi \sqrt{\frac{\frac{7}{75}m{L}^2}{mg\left(\frac{L}{10}\right)}}\\ =2\pi \sqrt{\frac{7L^2}{75g\left(\frac{L}{10}\right)}}\\ =2\pi \sqrt{\frac{14L}{15g}}\end{array}$

Substitute $100\text{cm}$ for $L$ and $9.80{\text{m/s}}^2$ for $g$ in the above equation.

$\begin{array}{c}T=2\left(3.14\right)\sqrt{\frac{14\left(100\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)}{15\left(9.80{\text{m/s}}^2\right)}}\\ =1.94\text{s}\end{array}$

Therefore, the time period of the pendulum for two rods is $1.94\text{s}$.