Expert answers to "A uniform steel bar swings from a pivot..."

verskalksv

Answered question

2021-11-10

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

Leory2000

Beginner2021-11-11Added 10 answers

Given:
We have a uniform steel bar swining from a pivot at one end.
The period of the physical pendulum is

T = 1.2 s

Solution:
The moment of inertia a uniform bar roating about an axis through one end is found from table

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):

T = 2 π \sqrt{\frac{I}{m g d}}

Since the bar uniform, the center of mass is located at the center and d=L/2:

T = 2 π \sqrt{\frac{I}{m g \frac{L}{2}}}

= 2 π \sqrt{\frac{2 I}{m g L}}

Substitute for I from equation (2):

T = 2 π \sqrt{\frac{2 (\frac{1}{3} m L^{2})}{m g L}}

= 2 π \sqrt{\frac{2 L}{3 g}}

Solve for L:

L = \frac{3 g T^{2}}{8 π^{2}}

Substitute numeral values:

L = \frac{3 (9.80 \frac{m}{s^{2}}) {(1.2 s)}^{2}}{8 π^{2}}

= 0.54 m

Result: L=0.54 m

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