91Ó°ÊÓ

1. The mass of thin uniform rod is m = 0.50 kg
2. The period of oscillation of the rod is T = 1.5 s
3. The angular amplitude is $\mathrm{θ}={10}^{0}0r0.174$

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, I is a moment of inertia about any axis, I_com 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.

$T=2\mathrm{Ï€}\sqrt{\frac{I}{mgh}}$

…(¾±¾±)

Here, T is the time period, I is the moment of inertia, m is mass, g is the gravitational acceleration and h is the perpendicular distance between the center of mass and the pivot point.

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

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

The distance from the center of mass is

$h=\frac{L}{2}$

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,

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

The expression of the period for the physical pendulum is

$$\begin{gathered} T=2\mathrm{Ï€}\sqrt{\frac{I}{mgh}} \\ =2\mathrm{Ï€}\sqrt{\frac{\frac{1}{3}m{L}^2}{mg\left(\frac{1}{4}L\right)}} \\ =2\mathrm{Ï€}\sqrt{\frac{2L}{3g}} \end{gathered}$$

Squaring of both sides

$$\begin{gathered} T^2=4{\mathrm{π}}^2\left(\frac{2L}{3g}\right) \\ L=\frac{3g{T}^2}{8{\mathrm{π}}^2} \\ =\frac{3×9.8{\text{m/s}}^{\text{2}}×{\left(1.5\text{s}\right)}^2}{8×{\left(3.14\right)}^2} \\ =0.84\text{m} \end{gathered}$$

The maximum kinetic energy of the rod as it swings

By the energy conservation law,

Kinetic energy at the bottom of the swing = potential energy at the end of swing

The expression for potential energy is

$$\begin{gathered} P.E.=mgd \\ P.E.=mgl\left(1-\cos \mathrm{θ}\right) \end{gathered}$$

…(¾±¾±¾±)

is the distance from the axis of the rotation to the center of mass, hence

For small angle approximation,

$1-\text{cos}\mathrm{θ}=\frac{1}{2}{\mathrm{θ}}_m^2$

The equation (iii) becomes as,

$$\begin{gathered} P.E.=mg\left(\frac{L}{2}\right)\frac{1}{2}{\mathrm{θ}}_m^2 \\ =\frac{0.50\text{kg}×9.8{\text{m/s}}^2×0.84\text{m}×{\left(0.174 \text{rad}\right)}^2}{4} \\ =0.031\text{J} \end{gathered}$$

The maximum kinetic energy is

$$\begin{gathered} {\left(K.E.\right)}_m=P.E. \\ =0.031\text{J} \end{gathered}$$

Therefore, the maximum kinetic energy is 0.031J .