91Ó°ÊÓ

The length of the stick is L = 1 m

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:

$I=I_{com}+m{h}^2$ …(¾±)

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

$T=2\mathrm{Ï€}\sqrt{\frac{I}{mgh}}$ …(¾±¾±)

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

The two sticks have equal mass. The center of mass of the stick is shown horizontally at A. The center of mass of the other stick is half of its length that is 0.50 m below A.

Consider rotational inertia of the horizontal stick as l₁ . The axis of rotation passing through its center and perpendicular to its plane is

$I_1=\frac{1}{12}m{L}^2$

And the rotational inertia for the vertical stick is I₂ . According to the parallel axis theorem,

$$\begin{gathered} I=I_{com}+m{h}^2 \\ {I}_2=I_1+m{h}^2 \\ =\frac{1}{12}m{L}^2+m{\left(\frac{1}{2}L\right)}^2 \\ =\frac{1}{3}m{L}^2 \end{gathered}$$

The total inertia of the system as shown in the figure is,

$$\begin{gathered} I=I_1+I_2 \\ =\frac{1}{12}m{L}^2+\frac{1}{3}m{L}^2 \\ =\frac{5}{12}m{L}^2 \end{gathered}$$

The total mass of the system is m = 2M . From the figure, the center of mass of the system is h . Let O be the center of the vertical stick.

The distance between A and O is

$$\begin{gathered} h=\frac{1}{2}L-\frac{1}{4}L \\ =\frac{L}{4} \end{gathered}$$

The expression of the period for the physical pendulum is

$$\begin{gathered} T=2\mathrm{π}\sqrt{\frac{I}{mgh}} \\ =2\mathrm{π}\sqrt{\frac{\left(\frac{5}{12}\right)M{L}^2}{2Mg\left(\frac{1}{4}L\right)}} \\ =2\mathrm{π}\sqrt{\frac{5L}{6g}} \\ =2\mathrm{π}\sqrt{\frac{5×\left(1.0\text{m}\right)}{6×9.8{\text{m/s}}^{\text{2}}}} \\ =1.83\text{s} \end{gathered}$$

Therefore, the period of oscillation of the system is 1.83 s .