QUESTION:
A physical pendulum consists of 4\ \rm m long sticks joined together as shown in the figure. What is the pendulum’s period of oscillation about the midpoint of the horizontal stick?
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SOLUTION:
Given:
- The length of the stick is L = 4.0\rm \ m.
Let:
- Mass of the stick is m
So, the total mass of the physical pendulum will be;
M = 2m
Now, let’s find the center of mass of the system of two sticks.
Let’s choose a coordinate system. Consider an x-y coordinate system such that the x-axis is passing through stick 1 and the y-axis is passing through stick 2 as shown in the figure.
Let C_1 be the center of mass of stick 1, C_2 is the center of mass of stick 2, and C is the center of mass of the system.
Since, C_1 and C_2 both lie on the y-axis. So, the x-coordinate of the center of mass of the system is zero Therefore, we need to find only the y-coordinate of the center of mass of the system.
The y-coordinate of the center of mass is given by;
\begin{aligned}y_{cm} &= \dfrac {\sum\limits_{i=1}^{n}m_iy_i}{M}\\&= \dfrac {m_1y_1+m_2y_2}{M}\\&=\dfrac{m\times0+m\times 2}{2m}\\&=1.0\rm \ m\\\end{aligned}
Thus the distance between pivoted point and the center of mass of the physical pendulum is;
l = 1.0\rm \ m
The moment of inertia of the stick 1 about the center of mass of the physical pendulum is given by;
\begin{aligned}\mathrm I_{1} &= \dfrac {mL^2}{12}+ml^2 \\\end{aligned}
The moment of inertia of the stick 2 about the center of mass of the physical pendulum is given by;
\mathrm I_{2} = \dfrac {mL^2}{12}+ml^2
Thus, the moment of inertia of the physical pendulum about the center of mass is given by;
\begin{aligned}\mathrm I_{cm} &=\mathrm I_{1}+\mathrm I_{2}\\&= 2 \times \left \{dfrac {mL^2}{12}+ml^2 \right \}\\&=\dfrac {mL^2}{6}+2ml^2 \\\end{aligned}
The time period of the physical pendulum is given by the expression;
\begin{aligned}T &= 2\pi \sqrt{\dfrac{I_{cm}+Ml^2}{Mgl}}\\&= 2\pi \sqrt{\dfrac{\dfrac{mL^2}{6}+2ml^2+2ml^2}{2mgl}}\\&= 2\pi \sqrt{\dfrac{\dfrac{mL^2}{6}+4ml^2 }{2mgl}}\\&= 2\pi \sqrt{\dfrac{\dfrac {L^2}{6}+4l^2 }{2gl}}\\&= 2\pi \sqrt{\dfrac{\dfrac {4^2}{6}+4\times 1^2 }{2\times 9.8\times 1}}\\&= 3.66\rm \ s\\&\approx 4\rm \ s\\\end{aligned}
This is the final answer of this question.
