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

verskalksv

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?

Answer & Explanation

Leory2000

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=1.2s
Solution:
The moment of inertia a uniform bar roating about an axis through one end is found from table
I=13mL2
where L is the length of the bar.
The period of the pendulum is found from equation (1):
T=2πImgd
Since the bar uniform, the center of mass is located at the center and d=L/2:
T=2πImgL2
=2π2ImgL
Substitute for I from equation (2):
T=2π2(13mL2)mgL
=2π2L3g
Solve for L:
L=3gT28π2
Substitute numeral values:
L=3(9.80ms2)(1.2s)28π2
=0.54m
Result: L=0.54 m

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?