Question

A rectangular block, with face lengths $\mathbf{a}\mathbf{=}\mathbf{35}\mathbf{}\mathbf{cm}$ and$\mathbf{b}\mathbf{=}\mathbf{45}\mathbf{cm}$ , is to be suspended on a thin horizontal rod running through a narrow hole in the block. The block is then to be set swinging about the rod like a pendulum, through small angles so that it is in SHM. Figure shows one possible position of the hole, at distance$\mathbf{r}$ from the block’s center, along a line connecting the center with a corner.

1. Plot the period of the pendulum versus distance$\mathbf{r}$ along that line such that the minimum in the curve is apparent.
2. For what value of $\mathbf{r}$does that minimum occur? There is actually a line of points around the block’s center for which the period of swinging has the same minimum value.
3. What shape does that line make?

Short answer

1. The period verses distance $\mathrm{r}$along that line such that the minimum in the curve is apparent is plotted.
2. The value of$\mathrm{r}$for minimum value of $\mathrm{T}$ is,$\text{r}=0.16\text{\hspace{0.17em}m}$
3. The shape made by the line is a circle.

Step-by-step solution

Given

The face lengths of the rectangular block is $\text{a}=35\text{cm}=35\times {10}^{-2}\text{m}$ and $\text{b}=45\text{cm}=45\times {10}^{-2}\text{m}$

Understanding the concept

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

Formula:

$\begin{array}{c}\mathbf{I}\mathbf{=}{\mathbf{I}}_{\mathbf{com}}\mathbf{+}{\mathbf{mh}}^{\mathbf{2}}\\ \mathbf{T}\mathbf{=}\mathbf{2}\mathbf{\pi }\sqrt{\frac{\mathbf{I}}{\mathbf{mgh}}}\end{array}$

Step 3: Plot the period of the pendulum versus distance r  along that line such that the minimum in the curve is apparent 

(a)

According to the parallel axis theorem,

$\text{I}={\text{I}}_{\mathrm{com}}+{\text{mh}}^2\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\left(1\right)$

Here, the distance of the hole from the center of the rectangular block is $h=r$

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

${\text{I}}_{\mathrm{com}}=\frac{\text{m}({\text{a}}^2+{\text{b}}^2)}{12}$

The equation (1) becomes as,

$\text{I}=\frac{\text{m}({\text{a}}^2+{\text{b}}^2)}{12}+{\text{mr}}^2$

The period of oscillation of the physical pendulum is

$\begin{array}{c}\text{T}=2\pi \sqrt{\frac{\text{I}}{\text{mgh}}}\\ \Rightarrow \text{T}=2\pi \sqrt{\frac{\frac{\text{m}({\text{a}}^2+{\text{b}}^2)}{12}+{\text{mr}}^2}{\text{mgr}}}\\ \Rightarrow \text{T}=2\pi \sqrt{\frac{\frac{({\text{a}}^2+{\text{b}}^2)}{12}+{\text{r}}^2}{\text{gr}}}\\ \Rightarrow \text{T}=\frac{2\pi }{\sqrt{\text{g}}}\sqrt{\frac{({\text{a}}^2+{\text{b}}^2)}{12\text{r}}+\text{r}}\end{array}$

We plot the graph $T$ verses$\mathrm{r}$ for the given $a$ and $b$ as

Calculate the value of r at which the minimum occurs

(b)

For the minimum value of $\mathrm{T}$, its first derivative is zero.

$\begin{array}{c}\frac{\text{dT}}{\text{dr}}=\frac{2\mathrm{\pi }}{\sqrt{\text{g}}}\sqrt{\frac{({\text{a}}^2+{\text{b}}^2)}{12}-{\text{r}}^2}\\ \Rightarrow \text{0}=\frac{2\mathrm{\pi }}{\sqrt{\text{g}}}\sqrt{\frac{({\text{a}}^2+{\text{b}}^2)}{12}-{\text{r}}^2}\\ \Rightarrow 0=\sqrt{\frac{({\mathrm{a}}^2+{\mathrm{b}}^2)}{12}}-\mathrm{r}\\ \Rightarrow \text{r}=\sqrt{\frac{[{(35\times {10}^{-2}\mathrm{m})}^2+{(45\times {10}^{-2}\mathrm{m})}^2]}{12}}\end{array}$

$\Rightarrow \text{r}=0.16\text{m}$

Calculate the shape made by the line

(c)

The direction from the center is not important. Hence, the locus of the point is a circle around the center of radius
$\sqrt{\frac{({\text{a}}^2+{\text{b}}^2)}{12}}$