A uniform steel bar swings from a pivot

In summary: When I do the math out of \frac{I}{Mgh} and substitute \frac {ML^2}{3}for I, I get 2\pi\sqrt{\frac {L}{3g}}instead of 2\pi\sqrt{\frac{2L}{3g}} Where did the 2 come from in the numerator in the correct equation?The time period of a physical pendulum is T=2\pi \sqrt{\frac{I}{mgh}}where I is the moment of inertia of the swinging body with respect to the pivot, m is the mass, h is the distance of the centre of mass from
  • #1
chicagobears34
37
0

Homework Statement


A uniform steel bar swings from a pivot at one end with a period of 1.3 sec.
Solve for Length of bar

Homework Equations


T=2pi*sqrt(L/g)

The Attempt at a Solution


since the period is 1.3 seconds, I just plug in 1.3 for T and 9.8m/s^2 for g and solve for L
i get L= .42m, which is wrong.
What am I doing wrong?
 
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  • #2
So are you solving for the length of the bar? It is unclear in the information above what you are solving for.
 
  • #3
Yosty22 said:
So are you solving for the length of the bar? It is unclear in the information above what you are solving for.

yes I am supposed to solve for the length of the bar, sorry about that
 
  • #4
chicagobears34 said:

Homework Statement


A uniform steel bar swings from a pivot at one end with a period of 1.3 sec.
Solve for Length of bar

Homework Equations


T=2pi*sqrt(L/g)

Your formula is valid for a simple (mathematical) pendulum. It is a "physical pendulum" now.

ehild
 
  • #5
I used T=2pi * sqrt(2L/3G) and got the correct answer.
is this the formula for period of a pendulum with a bar or something?
 
  • #7
ehild said:
Yes, it is correct for a rod pivoted at one end. See http://cnx.org/content/m15585/latest/


I read your source and can't figure out how they came to that equation. When I do the math out of [tex] \frac{I}{Mgh} [/tex] and substitute [tex]\frac {ML^2}{3}[/tex]for I, I get [tex]2\pi\sqrt{\frac {L}{3g}}[/tex]instead of [tex]2\pi\sqrt{\frac{2L}{3g}}[/tex] Where did the 2 come from in the numerator in the correct equation?

EDIT: Was missing the equation higher up in your source article where it stated that [tex]h=\frac{L}{2}[/tex] where I was assuming h was the same as L.
 
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  • #8
The time period of a physical pendulum is

[tex]T=2\pi \sqrt{\frac{I}{mgh}}[/tex]

where I is the moment of inertia of the swinging body with respect to the pivot,
m is the mass,
h is the distance of the centre of mass from the pivot.

In case of a homogeneous thin bar of length L, I=mL2/3, and the CM is at the middle, so h=L/2.ehild
 
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FAQ: A uniform steel bar swings from a pivot

1. How does the length of the bar affect its swing?

The length of the bar affects its swing by changing the period of the pendulum. A longer bar will have a longer period, while a shorter bar will have a shorter period. This is because the length of the bar determines the distance the pendulum travels in one swing, which affects the time it takes to complete one swing.

2. What factors affect the swing of the bar in addition to its length?

In addition to the length, the mass of the bar and the angle at which it is released also affect its swing. A heavier bar will have a slower swing, while a lighter bar will have a faster swing. The angle at which the bar is released also affects the amplitude and direction of the swing.

3. Can the pivot point affect the swing of the bar?

Yes, the pivot point can affect the swing of the bar. The pivot point should be frictionless and allow the bar to swing freely, without any external forces acting on it. If the pivot point is not well lubricated or is uneven, it can affect the swing of the bar and cause it to deviate from its expected motion.

4. How is the period of the bar's swing calculated?

The period of the bar's swing can be calculated using the formula T = 2π√(L/g), where T is the period, L is the length of the bar, and g is the acceleration due to gravity. This formula assumes that the angle of release is small (less than 10 degrees) and the amplitude of the swing is also small.

5. Does the type of steel used in the bar affect its swing?

Yes, the type of steel used can affect the swing of the bar. Different types of steel have different densities and strengths, which can affect the mass and stiffness of the bar. This, in turn, can affect the period and amplitude of the swing of the bar.

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