Question: In Figure, the pendulum consists of a uniform disk with radius r = 10.cmand mass 500 gm attached to a uniform rod with length L =500mm and mass 270gm.

1. Calculate the rotational inertia of the pendulum about the pivot point.
2. What is the distance between the pivot point and the center of mass of the pendulum?
3. Calculate the period of oscillation.
4. 
   

1. The radius of uniform disk is $r=10.0\text{cm} \text{or}10.0\times {10}^{-2}\text{m}$
2. The mass of the disk is $M=500\text{g or}500\times {10}^{-3}\text{kg}$
3. The length of the uniform rod is $L=500\text{mm or}500\times {10}^{-3}\text{m}$
4. The mass of the uniform rod is $m=270\text{g or}270\times {10}^{-3}\text{kg}$

The moment of inertia about an axis of rotation is equal to the sum of moment of inertial about a parallel axis passing through the center of mass and 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 moment of inertia about any axis, I_com is 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, 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 rotational inertia of the pendulum about the pivot point:

The rod is pivoted at one end hence, the axis of rotation passing through 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 the parallel axis theorem,

$$\begin{gathered} I_1=I_{com}+m{h}^2 \\ =\frac{1}{12}m{L}^2+m{\left(\frac{1}{2}L\right)}^2 \\ {I}_1=\frac{{\displaystyle 1}}{{\displaystyle 3}}m{L}^2 \end{gathered}$$

…(iii)

The uniform disk is pivoted at the center and the axis of rotation passing through its center and perpendicular to its plane is

${\text{I}}_{com}=\frac{1}{2}{\text{Mr}}^2$

The distance between the center of disk to the pivoted point of the rod at the other end is L +r ,

Thus

h = L+r

According to the parallel axis theorem,

$I_2=I_{com}+M{h}^2$

…(iv)

The total rotational inertia of the system is

$$\begin{gathered} I=I_1+I_2 \\ =\frac{1}{3}m{L}^2+\frac{1}{2}M{r}^2+M{\left(L+r\right)}^2 \\ =\frac{1}{3}\times \left(270\times {10}^{-3}\text{kg}\right)\times {\left(500\times {10}^{-3}\text{m}\right)}^2+\frac{1}{2}\times \left(500\times {10}^{-3}\text{kg}\right)\times {\left(10.0\times {10}^{-2}\text{m}\right)}^2 \\ +\left(500\times {10}^{-3}\text{kg}\right)\times {\left[\left(500\times {10}^{-3}\text{m}\right)+\left(10.0\times {10}^{-2}\text{m}\right)\right]}^2 \\ =0.205\text{kg}·{\text{m}}^2 \end{gathered}$$

The distance between the pivot point and the center of mass of the pendulum:

The distance between the pivot point and the center of the rod is

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

The distance between the pivot point and the center of the disk is

$l_d=L+r$

The distance from the pivot point to the center of mass of the disk-rod system is

$$\begin{gathered} d=\frac{m{l}_r+M{l}_d}{m+M} \\ =\frac{m\left(\frac{L}{2}\right)+M\left(L+r\right)}{m+M} \\ =\frac{270\times {10}^{-3}\text{kg}\times \left(\frac{500\times {10}^{-3}\text{m}}{2}\right)+\left(500\times {10}^{-3}\text{kg}\right)\times \left(500\times {10}^{-3}\text{m}\right)+\left(10.0\times {10}^{-2}\text{m}\right)}{\left(270\times {10}^{-3}\text{kg}\right)+\left(500\times {10}^{-3}\text{kg}\right)} \\ =0.477\text{m} \end{gathered}$$

The period of oscillation of the system:

The expression of the period for the physical pendulum is

$T=2\pi \sqrt{\frac{I}{mgh}}$

Here the system is rod and disk, hence mass of the system is M + m . The distance from the pivot point to the center of mass of the system is This system can act as a physical pendulum hence,

$$\begin{gathered} T=2\pi \sqrt{\frac{I}{\left(m+M\right)gd}} \\ =2\times 3.14\sqrt{\frac{0.205\text{kg}·{\text{m}}^2}{\left(270\times {10}^{-3}\text{kg}\right)+\left(500\times {10}^{-3}\text{kg}\right)\times 9.8\text{m}/{\text{s}}^2\times 0.477\text{m}}} \\ =1.50\text{s} \\ \Rightarrow \text{T}=2\times 3.14\sqrt{\frac{0.205\text{kg}.{\text{m}}^2}{\left(270\times {10}^{-3}\text{kg}\right)+\left(500\times {10}^{-3}\text{kg}\right)\times 9.8\text{m}/{\text{s}}^2\times 0.477\text{m}}} \\ \Rightarrow \text{T}=1.50\text{s} \end{gathered}$$