Chapter 10, Problem 75P

(a)

To determine

The period.

Answer to Problem 75P

The period is $2.01\text{s}$.

Explanation of Solution

Write an expression for the period.

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

Here, $T$ is the period, $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.

Conclusion:

Substitute $1.000\text{m}$ for $L$ and $9.80{\text{m/s}}^2$ for $g$ in equation (I) to find $T$.

$\begin{array}{c}T=2\pi \sqrt{\frac{1.000\text{m}}{9.80{\text{m/s}}^2}}\\ =2.01\text{s}\end{array}$

Thus, the period is $2.01\text{s}$.

(b)

To determine

The percentage age of the total mass of the pendulum in the uniform thin rod.

Answer to Problem 75P

The percentage age of the total mass of the pendulum in the uniform thin rod is $11.3\%$.

Explanation of Solution

Refer figure 1.

Figure 1

Write an expression for the net rotational moment of inertia

$\begin{array}{c}I=I_r+I_p\\ =\frac{1}{3}{m}_r{L}^2+m_p{L}^2\\ =\frac{L^2}{3}\left(m_r+3m_p\right)\end{array}$ (II)

Here, $I$ is the net rotational moment of inertia, $I_r$ is the moment of inertia of the uniform thin rod, $I_p$ is the moment of inertia of the point mass, $m_r$ is the mass of the uniform thin rod, $m_p$ is the point mass and $L$ is the length of the rod.

Write an expression to find the distance from the pivot to the center of mass of the compound object.

$\begin{array}{c}d=\frac{m_r\left(\frac{L}{2}\right)+m_{p}L}{m_r+m_p}\\ =\frac{L}{2}\left(\frac{m_r+2m_p}{m_r+m_p}\right)\end{array}$ (III)

Write an expression for the period.

$\begin{array}{c}T=\frac{2\pi }{\omega }\\ =2\pi \sqrt{\frac{I}{mgd}}\\ =2\pi \sqrt{\frac{\left(\frac{L^2}{3}\left(m_r+3m_p\right)\right)}{mg\left(\frac{L}{2}\left(\frac{m_r+2m_p}{m_r+m_p}\right)\right)}}\\ =2\pi \sqrt{\frac{2L\left(m_r+3m_p\right)}{3g\left(m_r+2m_p\right)}}\\ =2\pi \sqrt{\frac{2L\left(1+\frac{3m_p}{m_r}\right)}{3g\left(1+\frac{2m_p}{m_r}\right)}}\end{array}$ (IV)

Rearrange equation (IV).

$\begin{array}{c}{\left(\frac{T}{2\pi }\right)}^2=\frac{2L\left(1+\frac{3m_p}{m_r}\right)}{3g\left(1+\frac{2m_p}{m_r}\right)}\\ \frac{3g}{2L}{\left(\frac{T}{2\pi }\right)}^2=\frac{\left(1+\frac{3m_p}{m_r}\right)}{\left(1+\frac{2m_p}{m_r}\right)}\\ \frac{3T^{2}g}{8{\pi }^{2}L}=\frac{\left(1+\frac{3m_p}{m_r}\right)}{\left(1+\frac{2m_p}{m_r}\right)}\\ 1+\frac{3m_p}{m_r}=\frac{3T^{2}g}{8{\pi }^{2}L}\left(1+\frac{2m_p}{m_r}\right)\\ \frac{m_r+m_p}{m_r}=1+\frac{\frac{3T^{2}g}{8{\pi }^{2}L}-1}{3-\frac{3T^{2}g}{4{\pi }^{2}L}}\\ \left(\frac{m_r+m_p}{m_r}\right)100\%=\left(1+\frac{16{\pi }^{2}L+3T^{2}g}{24{\pi }^{2}L+6T^{2}g}\right)100\%\end{array}$ (V)

Conclusion:

Substitute $1.000\text{m}$ for $L$, $\left(100\%-1\%\right)\left(\frac{1}{100\%}\right)\left(2.01\text{s}\right)$ for $T$ and $9.80{\text{m/s}}^2$ for $g$ in equation (V) to find $\left(\frac{m_r+m_p}{m_r}\right)100\%$.

$\begin{array}{c}\left(\frac{m_r+m_p}{m_r}\right)100\%=\left(1+\frac{16{\pi }^2\left(1.000\text{m}\right)+3{\left(\left(100\%-1\%\right)\left(\frac{1}{100\%}\right)\left(2.01\text{s}\right)\right)}^2\left(9.80{\text{m/s}}^2\right)}{24{\pi }^2\left(1.000\text{m}\right)+6{\left(\left(100\%-1\%\right)\left(\frac{1}{100\%}\right)\left(2.01\text{s}\right)\right)}^2\left(9.80{\text{m/s}}^2\right)}\right)100\%\\ =\left(1+\frac{16{\pi }^2\left(1.000\text{m}\right)+3{\left(\left(0.99\right)\left(2.01\text{s}\right)\right)}^2\left(9.80{\text{m/s}}^2\right)}{24{\pi }^2\left(1.000\text{m}\right)+6{\left(\left(0.99\right)\left(2.01\text{s}\right)\right)}^2\left(9.80{\text{m/s}}^2\right)}\right)100\%\\ =11.3\%\end{array}$

Thus, the percentage age of the total mass of the pendulum in the uniform thin rod is $11.3\%$.