The main objective of this question is to **find** the l**ength of the steel bar**. This question uses the **concept of the pendulum**. A **pendulum** is simply the **weight suspended** from a **pivot or shaft** so that it will **move freely**. The **period **of the** pendulum** is **mathematically** equal to:

\[T\space = \space 2 \pi \space \sqrt \frac{I}{mgd}\]

## Expert Answer

The** following information** is given:

The **period** of the **pendulum** is equal to $1.2s$.

We have to find the **length** of the bar.

We **know** that:

\[I \space = \space \frac{1}{3}mL^2\]

**Where** the **length bar** is $L$.

The **time period** of the **pendulum** is:

\[T\space = \space 2 \pi \space \sqrt \frac{I}{mgd}\]

As the **bar is uniform**, so:

\[T\space = \space 2 \pi \space \sqrt \frac{I}{mg \frac{L}{2}}\]

\[= \space 2\pi \sqrt \frac{2I}{mgL}\]

By** substituting** the values, we get:

\[T\space = 2\pi \sqrt \frac{2/3ml^2}{mgL}\]

\[= \space 2\pi \sqrt \frac{2L}{3g}\]

**Solving** it for L results in:

\[L \space = \space \frac{3gt^2}{8\pi^2}\]

By **putting** the **values**, we get:

\[L \space = \space \frac{3(9.80)(1.2)^2}{8 \pi^2}\]

\[= \space 0.54m\]

**Hence** the length is:

\[L \space = \space 0.54m\]

## Numerical Answer

The **length** of the **steel bar** is $0.54$ m, whose **period** is $1.2 s$.

## Example

Find the length of a uniform steel bar whose one side is fixed to the pivot with time periods set at $2 s$ and $4 s$.

The following **information** is given:

The **time period** of the **pendulum** is equal to $2s$ and $4s$.

We have to find the **length of the bar**.

We **know** that:

\[I \space = \space \frac{1}{3}mL^2\]

**Where** the **length of the bar** is L.

First, we will solve it for some time of $2 s$.

The time period of the **pendulum** is:

\[T\space = \space 2 \pi \space \sqrt \frac{I}{mgd}\]

As the bar is **uniform**, so:

\[T\space = \space 2 \pi \space \sqrt \frac{I}{mg \frac{L}{2}}\]

\[= \space 2\pi \sqrt \frac{2I}{mgL}\]

By **substituting** the **values**, we get:

\[T\space = 2\pi \sqrt \frac{2/3ml^2}{mgL}\]

\[= \space 2\pi \sqrt \frac{2L}{3g}\]

**Solving** it for $L$ results in:

\[L \space = \space \frac{3gt^2}{8\pi^2}\]

By **putting** the values, we get:

\[L \space = \space \frac{3(9.80)(2)^2}{8 \pi^2}\]

\[= \space 1.49 \space m\]

**Hence** the length is:

\[L \space = \space 1.49 \space m\]

Now **calculate the length** for a time period of $4 s$.

The following **information** is given:

The time period of the pendulum is equal to $4 s$.

We have to find the **length of the bar**.

We **know** that:

\[I \space = \space \frac{1}{3}mL^2\]

Where the length bar is L.

First, we will solve it for a **time period** of $2 s$.

The time period of the** pendulum** is:

\[T\space = \space 2 \pi \space \sqrt \frac{I}{mgd}\]

As the bar is **uniform**, so:

\[T\space = \space 2 \pi \space \sqrt \frac{I}{mg \frac{L}{2}}\]

\[= \space 2\pi \sqrt \frac{2I}{mgL}\]

By **substituting** the values, we get:

\[T\space = 2\pi \sqrt \frac{2/3ml^2}{mgL}\]

\[= \space 2\pi \sqrt \frac{2L}{3g}\]

\[L \space = \space \frac{3gt^2}{8\pi^2}\]

\[L \space = \space \frac{3(9.80)(4)^2}{8 \pi^2}\]

\[= \space 5.96 \space m\]

Hence, the **length** is:

\[L \space = \space 5.96 \space m\]