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

verskalksv 2021-11-10 Answered
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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Expert Answer

Leory2000
Answered 2021-11-11 Author has 795 answers
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
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