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A uniform steel bar swings from a pivot at one end with a period of 1.2 s. How long is the bar?

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:

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

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