A pendulum is formed by pivoting a long thin rod about a point on the rod. In a series of experiments, the period is measured as a function of the distance xbetween the pivot point and the rod’s center.

1. If the rod’s length is L=2.20 mand its mass is m=22.1 g, what is the minimum period?
2. If xis chosen to minimize the period and then Lis increased, does the period increase, decrease, or remain the same?
3. If, instead, m is increased without Lincreasing, does the period increase, decrease, or remain the same?

1. The mass of the rod is,$\mathrm{m}=22.1\mathrm{g}=0.0221\mathrm{kg}$
2. The length of the rod is,$\mathrm{L}=2.20\mathrm{m}$

The problem deals with the calculation of moment of inertia. It is the quantity expressed by the body resisting angular acceleration which is the sum of the product of the mass of every particle with its square of a distance from the axis of rotation. Using the equation for moment of inertia of rod about its center and the parallel axis theorem, we can calculate the moment of inertia of the rod about the pivot point.

Formulae:

$$\begin{gathered} \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{I}}{\mathrm{mgx}}} \\ {\mathrm{I}}_{\mathrm{c}}=\frac{{\mathrm{mL}}^2}{12} \\ \mathrm{I}=\frac{{\mathrm{mL}}^2}{12}+{\mathrm{mx}}^2 \end{gathered}$$

Where T is time, m is mass L is the length, I is the rotational inertia, and ${\mathrm{I}}_{\mathrm{c}}$is the rotational inertia about the center

The equation for rotational inertia of a rod about its center is

${\mathrm{I}}_{\mathrm{c}}=\frac{{\mathrm{mL}}^2}{12}$

Now, according to the parallel axis theorem, the rotational inertia of a rod about a point distance ‘d’ from its center is

$\mathrm{I}={\mathrm{I}}_{\mathrm{c}}+{\mathrm{mx}}^2$

So,

$\mathrm{I}=\frac{{\mathrm{mL}}^2}{12}+{\mathrm{mx}}^2$

Now, the equation for period of the pendulum is

$\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{I}}{\mathrm{mgx}}}$

By substituting the equation for moment of inertia in this equation, we get

$\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{{\displaystyle \frac{{\mathrm{mL}}^2}{12}}+{\mathrm{mx}}^2}{\mathrm{mgx}}}$

Now, to get the minimum period, we differentiate this equation with respect to x

So,

$\frac{d\mathrm{T}}{d\mathrm{x}}=0$

By solving this equation, we get

$\mathrm{x}=\frac{\mathrm{L}}{\sqrt{12}}$

Now, substituting this equation, we can get T_min as,

$\begin{array}{rcl}{\mathrm{T}}_{\min }& =& 2\mathrm{\pi }\sqrt{\frac{{\displaystyle \frac{{\mathrm{mL}}^2}{12}+\mathrm{m}{\left(\frac{\mathrm{L}}{\sqrt{12}}\right)}^2}}{\mathrm{mg}\left({\displaystyle \frac{\mathrm{L}}{\sqrt{12}}}\right)}}\\ & =& 2\mathrm{\pi }\sqrt{\frac{{\displaystyle 2\mathrm{L}}}{\sqrt{12}\mathrm{g}}}\\ & =& 2\mathrm{\pi }\sqrt{\frac{{\displaystyle 2\times 2.20\mathrm{m}}}{\sqrt{12}\times 9.8\mathrm{m}/{\mathrm{s}}^2}}\\ & =& 2.26\mathrm{s}\end{array}$

We have derived the equation for minimum period as

${\mathrm{T}}_{\min }=2\mathrm{\pi }\sqrt{\frac{{\displaystyle 2\mathrm{L}}}{\sqrt{12}\mathrm{g}}}$

As, the term L is in the numerator, the period will increase with increase in length of pendulum.

The equation for minimum period is

${\mathrm{T}}_{\min }=2\mathrm{\pi }\sqrt{\frac{{\displaystyle 2\mathrm{L}}}{\sqrt{12}\mathrm{g}}}$

From this equation, we can conclude that there is no effect of mass on the period as there is no mass term in this equation so that the period will remain same