Question

Question: In the Figure below, a meter stick initially swings about a pivot point at one end, at distance h from the stick’s center of mass.

1. If this physical pendulum is inverted and suspended at point P , what is its new period of oscillation?
2. Is the period now greater than, less than, or equal to its previous value?

Short answer

1. The period of oscillation of the physical pendulum is 1.64s
2. The period of oscillation is equal to its previous value.

Step-by-step solution

Identification of given data

The length of the stick is

Understanding the concept of the parallel axis theorem

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, I is a moment of inertia about any axis, I_c0m is a moment of inertia about a parallel axis passing through the center of mass, and is the perpendicular distance between the two axes.

$T=2\pi \sqrt{\frac{I}{mgh}}$ …(ii)

Here, is the time period and is the gravitational acceleration

(a) Determining the new period of oscillation if this physical pendulum is inverted and suspended at the point  

The distance between P and C is,

$$\begin{gathered} h=\frac{2}{3}L-\frac{1}{2}L \\ =\frac{1}{6}L \end{gathered}$$

According to the parallel axis theorem

$$\begin{gathered} I=I_{com}+m{h}^2 \\ =\frac{1}{12}m{L}^2+m{\left(\frac{1}{6}L\right)}^2 \\ =\left(\frac{1}{12}+\frac{1}{36}\right)m{L}^2 \\ =\frac{1}{9}m{L}^2 \end{gathered}$$

The expression of the period for the physical pendulum is,

$$\begin{gathered} T=2\pi \sqrt{\frac{I}{mgh}} \\ =2\pi \sqrt{\frac{\left(\frac{1}{9}\right)m{L}^2}{mg\left(\frac{1}{6}L\right)}} \\ =2\pi \sqrt{\frac{\left(\frac{1}{9}\right)L}{g\left(\frac{1}{6}\right)}} \\ =2\pi \sqrt{\frac{2L}{3g}} \\ =2\pi \sqrt{\frac{2\times \left(1.0\text{m}\right)}{3\times 9.8{\text{m/s}}^{\text{2}}}} \\ =1.64\text{s} \end{gathered}$$

Therefore, the time period is 1.64 s.

Find out if the period now greater than, less than, or equal to its previous value

The period of oscillation of the physical pendulum remains the same when it is suspended and inverted.