~It Came In Like a Wrecking Ball~

a) The sum of torques must have the equation of a harmonic oscillator for the period and the frequency can be easily found. Remember to consider the moment of inertia of the complete pendulum.

b) The moment of inertia is a different one, considering the chain. Then, calculate again the frequency and the period.

a) The torque is calculated as:

\begin{equation*}
\vec{\tau}_g=\vec{r}\times \vec{W},
\end{equation*}

where in this case we have:

\begin{equation*}
-L_{CM}mg\sin(\theta)=I\alpha.
\end{equation*}

Rearranging the terms, and for small oscillations (\(\sin \theta \approx \theta\)) we can get:

\begin{equation*}
\alpha+\frac{L_{CM}mg}{I}\theta=0,
\end{equation*}

where \(\alpha = \frac{d^2\theta}{dt} \). The angular frequency squared is the coefficient of the \(\theta\) variable. Then:

\begin{equation*}
\omega=\sqrt{\frac{L_{CM}mg}{I}}.
\end{equation*}

The moment of inertia is for a solid sphere and the pivot at a distance \(L+R\). So:

\begin{equation*}
I = \frac{2}{5} MR^2 + M(L+R)^2.
\end{equation*}

So, since the period is \(T = \frac{2 \pi}{\omega}\) we get:

\begin{equation*}
T=2\pi\sqrt{\frac{\frac{2}{5}MR^2+M(L+R)^2}{(L+R)Mg}}.
\end{equation*}

With numerical values:

\begin{equation*}
T\approx 5.695\,\text{s}.
\end{equation*}

The frequency is then \(f=1/T\), hence:

\begin{equation*}
f \approx 0.176\,\text{Hz}.
\end{equation*}

b) With the chain, the lenght of the center of mass is:

\begin{equation*}
L_{CM}=\frac{M(L+R)+m_c(L/2)}{M+m_c}.
\end{equation*}

The moment of inertia is then:

\begin{equation*}
I=\frac{2}{5}MR^2+M(L+R)^2+\frac{m_cL^2}{3}.
\end{equation*}

In this case, the period is:

\begin{equation*}
T=2\pi\sqrt{\frac{\frac{2}{5}MR^2+M(L+R)^2+\frac{m_cL^2}{3}}{\left(\frac{M(L+R)+m_c(L/2)}{M+m_c}\right)(m_c+M)g}},
\end{equation*}

which numerically is:

\begin{equation*}
T\approx 5.659\,\text{s}.
\end{equation*}

The frequency is \(f=1/T\), then

\begin{equation*}
f \approx 0.177\,\text{Hz},
\end{equation*}

almost the same when the effect were not considered.

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

(a) We need to find the frequency and period of the massless chain. To approach the solution we will deduce the expression for the period of oscillation of an arbitrary solid body. We start by drawing a body of arbitrary shape which can rotate with respect to an axis as seen in figure 1.

\begin{figure}[H] \centering
\includegraphics{Laco y Luis A/images/Ondas/physical pendulum.png}
\caption{Physical Pendulum}
\label{fig:physicalpendulum}
\end{figure}

Now we will take the sum of torques with respect to this pivot point as the line that joins the pivot point and center of mass of the body makes an angle \(\theta\) with the vertical. The only force that exerts some torque on the body is its weight, exerted on its center of mass. This torque is calculated as

\begin{equation}
\vec{\tau}_g=\vec{r}\times \vec{W},
\end{equation}

where \(\vec{r}\) is the vector joining the pivot point and the center of mass and \(\vec{W}=-mg\,\hat{\textbf{j}}\) is the weight of the body with \(m\) its mass and \(g\) the gravitational acceleration of Earth. Using the right hand rule, we can deduce that this torque is negative, that is, is entering the screen. The magnitude of this torque is

\begin{equation}
|\vec{\tau}_g|=|\vec{r}||mg|\sin(\theta),
\end{equation}

where \(\theta\) is the angle between \(\vec{r}\) and \(\vec{W}\). Calling the magnitude \(|\vec{r}|=L_{CM}\) we obtain

\begin{equation}
\vec{\tau}_g=-L_{CM}mg\sin(\theta)\,\hat{\textbf{k}},
\end{equation}

where we take the positive direction of \(\hat{\textbf{k}}\) as coming out of the screen. Now, according to Newton's second law, the sum of torques must be equal to the product of moment of inertia \(I\) and angular acceleration \(\vec{\alpha}\), namely

\begin{equation}
\sum \vec{\tau}=I\vec{\alpha}.
\end{equation}

The left-hand side of the equation above can be replaced by the expression for the torque exerted by the gravitational acceleration. We will assume that since the torque is exerted only along the Z-axis, then the angular acceleration will be along this direction also. Thus, Newton's second law can be simplified to

\begin{equation}
-L_{CM}mg\sin(\theta)\,\hat{\textbf{k}}=I\alpha\,\hat{\textbf{k}},
\end{equation}

which after dropping the vector notation is

\begin{equation}
-L_{CM}mg\sin(\theta)=I\alpha.
\end{equation}

Dividing all terms by \(I\) we can rewrite the expression above as

\begin{equation}
\alpha+\frac{L_{CM}mg}{I}\sin(\theta)=0
\end{equation}

This equation can be simplified for small amplitude oscillations by using the approximation \(\sin(\theta)\approx\theta\), thus obtaining

\begin{equation}
\alpha+\frac{L_{CM}mg}{I}\theta=0.
\end{equation}

Because the angular acceleration is defined as \(\alpha=\frac{d^2\theta}{dt^2}\) the equation above is a differential equation which is characteristic of harmonic motion with angular frequency

\begin{equation}
\omega=\sqrt{\frac{L_{CM}mg}{I}}.
\end{equation}

Therefore the frequency of the oscillation is \(f=\omega/2\pi\), explicitly

\begin{equation}
\label{freq}
f=\frac{1}{2\pi} \sqrt{\frac{L_{CM}mg}{I}}.
\end{equation}

The period \(T\) of one oscillation is the inverse of the frequency \(T=1/f\), then

\begin{equation}
\label{period}
T=2\pi\sqrt{\frac{I}{L_{CM}mg}}.
\end{equation}

We will use equations \eqref{freq} and \eqref{period} to determine the frequency and period of the oscillation in both cases.

For this first case, the mass of the chain is negligible with respect to that of the ball. Then, the center of mass is located at the center of the sphere. The distance from the pivot to the center of the sphere will be the length of the rope \(L\) plus the radius of the sphere \(R\)

\begin{equation}
L_{CM}=L+R.
\end{equation}

The moment of inertia \(I\) is taken with respect to the pivot. Since the rope's mass is negligible we will not include it in the moment of inertia calculus. Therefore, the moment of inertia \(I\) will be that of the sphere. To calculate this moment of inertia we use the parallel axis theorem which indicates

\begin{equation}
\label{parallel}
I=I_{CM}+Md^2,
\end{equation}

where \(I_{CM}\) is the moment of inertia with respect to an axis passing through the center of mass of the object, \(M\) is the mass of the object, and \(d\) the distance between the center of mass and a parallel axis in which the inertia \(I\) is calculated. This parallel axis in our case is the pivot, then \(d=L+R\). The moment of inertia with respect to the center of mass for a solid sphere is

\begin{equation}
I_{CM}=\frac{2}{5}MR^2.
\end{equation}

Then the moment of inertia is, according to \eqref{parallel}

\begin{equation}
\label{isphere}
I=\frac{2}{5}MR^2+M(L+R)^2.
\end{equation}

Using the result for \(I\) and for \(L_{CM}\) in equation \eqref{period} we get

\begin{equation}
T=2\pi\sqrt{\frac{\frac{2}{5}MR^2+M(L+R)^2}{(L+R)Mg}}.
\end{equation}

Using the numerical values we obtain

\begin{equation}
T=2\pi\sqrt{\frac{\frac{2}{5}(1000\,\text{kg})(1\,\text{m})^2+(1000\,\text{kg})(7\,\text{m}+1\,\text{m})^2}{(7\,\text{m}+1\,\text{m})(1000\,\text{kg})(9.8\,\text{m/s}^2)}},
\end{equation}

\begin{equation}
T\approx 5.695\,\text{s}.
\end{equation}

The frequency is then \(f=1/T\), hence

\begin{equation}
f\approx \frac{1}{5.695\,\text{s}}\approx 0.176\,\text{Hz}.
\end{equation}

(b) Take now the chain to have a mass of 70 kg. Let us now calculate the frequency and oscillation period of the chain. If it has a mass of \(m_c=70\,\text{kg}\) we must now find the new expressions for \(L_{CM}\) and \(I\) in order to calculate the period and frequency of oscillation. Let us start by calculating the distance between the pivot and the center of mass \(L_{CM}\). To calculate the position of the center of mass we will use the expression

\begin{equation}
L_{CM}=\frac{Mr_{cm}^{\text{sphere}}+m_cr_{cm}^{rope}}{M+m_c},
\end{equation}

where \(r_{cm}^{\text{sphere}}\) is the position of the center of mass for the sphere and \(r_{cm}^{\text{rope}}\) is the position of the center of mass of the rope. As we have seen earlier, the position of the center of mass of the sphere is in the center of the sphere, then \(r_{cm}^{\text{sphere}}=L+R\). The center of mass of the rope will we at half its length, thus \(r_{cm}^{\text{rope}}=L/2\). Hence we can write for \(L_{CM}\) using the expressions above as

\begin{equation}
\label{lcm}
L_{CM}=\frac{M(L+R)+m_c(L/2)}{M+m_c}.
\end{equation}

Now let us calculate the join moment of inertia \(I\). From the superposition principle the moment of inertia of the whole chain-ball system with respect to the pivot will be the sum of the moment of inertia of the ball and that of the chain, that is

\begin{equation}
\label{inertialtotal}
I=I^{\text{sphere}}+I^{\text{chain}}.
\end{equation}

We already calculated \(I^{\text{sphere}}\) and is given by equation \eqref{isphere}. To calculate \(I^{\text{chain}}\) we consider it to be a rigid bar of length \(L\) whose moment of inertia with respect to one of its ends is

\begin{equation}
I^{\text{chain}}=\frac{m_cL^2}{3}.
\end{equation}

Putting everything together in equation \eqref{inertialtotal} we obtain

\begin{equation}
\label{defi}
I=\frac{2}{5}MR^2+M(L+R)^2+\frac{m_cL^2}{3}.
\end{equation}

Using the results of equations \eqref{lcm} and \eqref{defi} into the expression for the period given in equation \eqref{period} we get

\begin{equation}
T=2\pi\sqrt{\frac{\frac{2}{5}MR^2+M(L+R)^2+\frac{m_cL^2}{3}}{\left(\frac{M(L+R)+m_c(L/2)}{M+m_c}\right)(m_c+M)g}},
\end{equation}

where notice that we have used the total mass of the system for \(m=m_c+M\). Numerically we get

\begin{equation}
T=2\pi\sqrt{\frac{\frac{2}{5}(1000\,\text{kg})(1\,\text{m})^2+(1000\,\text{kg})(7\,\text{m}+1\,\text{m})^2+\frac{(70\,\text{kg})(7\,\text{m})^2}{3}}{\left(\frac{(1000\,\text{kg})(7\,\text{m}+1\,\text{m})+(70\,\text{kg})(7\,\text{m}/2)}{1000\,\text{kg}+70\,\text{kg}}\right)(70\,\text{kg}+1000\,\text{kg})(9.8\,\text{m/s}^2)}},
\end{equation}

\begin{equation}
T\approx 5.659\,\text{s}.
\end{equation}

The frequency is \(f=1/T\), then

\begin{equation}
f\approx \frac{1}{5.659\,\text{s}}\approx 0.177\,\text{Hz}.
\end{equation}

Thus we see that the mass of the chain is very small compared to the mass of the sphere, thus we obtain a result very similar to the one where we did not consider its effects. Notice also that the period considering the chain decreases. Although the moment of inertia increases, the distance from the pivot to the center of mass decreases, and the total mass increases, the combination of all these effects make the period decrease and the frequency increase clearly.