# A uniform steel bar swings from a pivot at one end with a period of 1.2 s. How l

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

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Leory2000
Given:
We have a uniform steel bar swining from a pivot at one end.
The period of the physical pendulum is $$\displaystyle{T}={1.2}{s}$$
Solution:
The moment of inertia a uniform bar roating about an axis through one end is found from table
$$\displaystyle{I}={\frac{{{1}}}{{{3}}}}{m}{L}^{{2}}$$
where L is the length of the bar.
The period of the pendulum is found from equation (1):
$$\displaystyle{T}={2}\pi\sqrt{{{\frac{{{I}}}{{{m}{g}{d}}}}}}$$
Since the bar uniform, the center of mass is located at the center and d=L/2:
$$\displaystyle{T}={2}\pi\sqrt{{{\frac{{{I}}}{{{m}{g}\frac{{L}}{{2}}}}}}}$$
$$\displaystyle={2}\pi\sqrt{{{\frac{{{2}{I}}}{{{m}{g}{L}}}}}}$$
Substitute for I from equation (2):
$$\displaystyle{T}={2}\pi\sqrt{{{\frac{{{2}{\left({\frac{{{1}}}{{{3}}}}{m}{L}^{{2}}\right)}}}{{{m}{g}{L}}}}}}$$
$$\displaystyle={2}\pi\sqrt{{{\frac{{{2}{L}}}{{{3}{g}}}}}}$$
Solve for L:
$$\displaystyle{L}={\frac{{{3}{g}{T}^{{2}}}}{{{8}\pi^{{2}}}}}$$
Substitute numeral values:
$$\displaystyle{L}={\frac{{{3}{\left({9.80}\frac{{m}}{{s}^{{2}}}\right)}{\left({1.2}{s}\right)}^{{2}}}}{{{8}\pi^{{2}}}}}$$
$$\displaystyle={0.54}{m}$$
Result: L=0.54 m