Question

Question: In Figure, a stick of lengthL = 1.85oscillates as a physical pendulum.

1. What value of distance x between the stick’s center of mass and its pivot pointOgives the least period?
2. What is that least period?

Short answer

Answer

1. The value of distance x between the stick’s center of mass and its pivot point gives the least period is x = 0.53
2. The least period is T = 2.1 s

Step-by-step solution

Identification of given data

The length of the stick is L = 1.85 m

Understanding the concept

The moment of inertia about an axis of rotation is equal to the sum of the moment of inertial about a parallel axis passing through the center of mass and the product of mass and a square of perpendicular distance between two axes. The time period of the physical pendulum can be defined in terms of its moment of inertia, mass, gravitational acceleration, and height.

Use the concept of parallel axis theorem and expression of the period for the physical pendulum.

Formulae:

$I=I_{com}+m{h}^2$

…(i)

Here, is a moment of inertia about any axis, is a moment of inertia about a parallel axis passing through the center of mass, is mass, and is the perpendicular distance between the two axes.

$T=2\pi \sqrt{\frac{I}{mgh}}$ …(ii)

Here, T is the time period, g is the gravitational acceleration

(a) Determining the value of distance x  between the stick’s center of mass and its pivot point 0  gives the least period

The axis of rotation passing through its center of mass and perpendicular to its plane is,

$I_{com}=\frac{1}{12}m{L}^2$

The distance from the center of mass and pivot point is,

h = x

The thin uniform rod swings about an axis passing through one end and perpendicular to its plane, then the rotational inertia by using parallel axis theorem,

$$\begin{gathered} I=I_{com}+m{h}^2 \\ I=\frac{1}{12}m{L}^2+m{x}^2 \\ I=m\left(\frac{L^2}{12}+x^2\right) \end{gathered}$$

The expression of the period for the physical pendulum is

$$\begin{gathered} T=2\pi \sqrt{\frac{I}{mgh}} \\ T=2\pi \sqrt{\frac{m\left(\frac{L^2}{12}+x^2\right)}{mgx}} \\ T=2\pi \sqrt{\frac{\left(\frac{L^2}{12}+x^2\right)}{gx}} \end{gathered}$$

For least period of oscillation, the first derivative of period is zero. Hence

$$\begin{gathered} \frac{dT}{dx}=\frac{2\pi }{\sqrt{g}}\sqrt{-\frac{L^2}{12x^2}+1} \\ 0=\frac{2\pi }{\sqrt{g}}\sqrt{-\frac{L^2}{12x^2}+1} \\ 0=\sqrt{-\frac{L^2}{12x^2}+1} \\ -\frac{L^2}{12x^2}+1=0 \\ {x}^2=\frac{L^2}{12} \\ x=\frac{L}{\sqrt{12}} \\ x=\frac{1.85\text{m}}{\sqrt{12}} \\ x=0.53\text{m} \end{gathered}$$

(b) Determining the least period

The least period:

The least period of the rod is

$$\begin{gathered} T=\frac{2\pi }{\sqrt{g}}\sqrt{\frac{L^2}{12x}+x} \\ =\frac{2\times 3.14}{\sqrt{9.8{\text{m/s}}^{\text{2}}}}\sqrt{\frac{{\left(1.85\text{m}\right)}^2}{12\times 0.53\text{m}}+0.53\text{m}} \\ =2.1\text{s} \end{gathered}$$