The main objective of this question is to find the length 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:
\[L \space = \space 5.96 \space m\]