Question

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 $x$ between the pivot point and the rod's centre.

(a) If the rod's length is $L = 2.20 m$ and its mass is $m = 20.5 g$ , what is the minimum period?

(b) chosen to minimize the period and then $L$ is increased, does the period increase, decrease, or remain the same?

(c) If, instead, $m$ is increased without $L$ increasing, does the period increase, decrease, or remain the same?

Accepted Answer

Given,
length of the rod, $L = 2.20 m$ ,
lass of the rod, $m = 20.5 g$ .

It oscillates as a physical pendulum, whose time period, $T$ is given by,
$T = 2 π \sqrt{\frac{I}{m g h}}$ ,
where, $I$ is the moment of inertia, $m$ is the mass of the pendulum and $h$ is the distance between the pivot point and the center of mass.

The moment of inertia of a rod rotating about its center is,
$I_{cm} = \frac{1}{12} m L^{2}$ .
We may find the inertia at distance of $x$ from the center (where it is pivoted) using the parallel axis theorem as,
$I = I_{cm} + m x^{2} = m (\frac{L^{2}}{12} + x^{2})$ .

Hence, we get the time period as,
$T = 2 π \sqrt{\frac{m (\frac{L^{2}}{12} + x^{2})}{m g x}} = 2 π \sqrt{\frac{(L^{2} + 12 x^{2})}{12 g x}}$

(a) Now, for the time period to be minimum, we have,
$m = 20.5 g$
$\Rightarrow \frac{d}{d x} \sqrt{\frac{(L^{2} + 12 x^{2})}{12 g x}} = 0$
$\Rightarrow \frac{1}{2} \sqrt{\frac{12 g x}{(L^{2} + 12 x^{2})}} (\frac{288 g x^{2} - 12 g L^{2} - 144 g x^{2}}{144 g^{2} x^{2}}) = 0$
$\Rightarrow 288 g x^{2} - 12 g L^{2} - 144 g x^{2} = 0$
$\Rightarrow x = \frac{L}{\sqrt{12}}$ .

Substituting the values we get,
$x = \frac{2.20}{\sqrt{12}} = 0.635 m$ .

Now, for $x = 0.635 m$ , the minimum value of time period $T_{\min}$ we get is,
$T_{\min} = 2 π \sqrt{\frac{((2.20)^{2} + 12 (0.635)^{2})}{12 (9.8) (0.635)}} = 2.26 s$ .

(b) Since, the time period is directly proportional to the length of the rod, it will increase if the length is increased.

(c) Since, the time period does not depend on the mass of the rod, it will remain same if the mass is increased.

Important Questions on Oscillations

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