Angular Frequency of a two-object system

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Homework Help Overview

The discussion revolves around the calculation of angular frequency in a two-object system, specifically involving a ladder and a woman. Participants are examining the role of mass in the formula and the moment of inertia related to the system's rotation.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning the inclusion of the ladder's weight in the mass used in the angular frequency formula. There are discussions about the moment of inertia and its calculation, as well as the distance 'd' in relation to the center of mass of the system.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the calculations and the physics involved. Some guidance has been provided regarding the moment of inertia and the significance of the center of mass, but no consensus has been reached on the correct approach.

Contextual Notes

There are indications of potential errors in the original calculations and assumptions, particularly regarding the mass and moment of inertia, which are being critically examined by participants.

hidemi
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Homework Statement
A 2.00 m-long 6.00 kg ladder pivoted at the top hangs down from a platform at the circus. A 42.0 kg trapeze artist climbs to a point where her center of mass is at the center of the ladder and swings at the system's natural frequency. The angular frequency (in s −1) of the system of ladder and woman is
A) 1.01
B) 3.07
C) 4.03
D) 8.05
E) 16.2

The answer is B.
Relevant Equations
w= √(mgd/I)
I calculate as follow and get a correct answer, but I wonder why the weight of the ladder 6 kg is not included in the mass (m) in the numerator.

w= √(mgd/I)
= √ { (42*10*1)/ [(1/12)(6)(2^2)+42*1] }

= √ (420/44)​

= 3.06
 
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hidemi said:
but I wonder why the weight of the ladder 6 kg is not included in the mass (m) in the numerator
It should be.
 
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Doc Al said:
It should be.
If the 6kg is included into the numerator, the answer wouldn't be correct.
How did you calculate to get the given answer? Is there something wrong in my original calculation?
Thanks!
 
hidemi said:
Is there something wrong in my original calculation?
Your original calculation -- as you point out yourself -- uses incorrect physics. (The mass of the ladder should be included.) So that cannot be the answer unless the problem is mistaken (which does happen).

But there's another error in your calculation, besides the missing mass in the numerator.
 
Last edited:
hidemi said:
Is there something wrong in my original calculation?
What is the moment of inertia of the ladder if it’s rotating about one end?
 
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TSny said:
What is the moment of inertia of the ladder if it’s rotating about one end?
I = √(mgd/I)= √[(42+6)*9.8*1 / (1/3*6*2^2 + 42*1^2)] = 3.067
Is this correct? Should the 'd' be half of the ladder length for both the ladder and woman?
 
hidemi said:
Is this correct?
Yes, looks good now.

hidemi said:
Should the 'd' be half of the ladder length for both the ladder and woman?
That 'd' is the distance from the pivot point to the center of mass of the entire system. (In this case, the center of mass of the system is at the midpoint of the ladder, but only because the woman happens to be at that point. What if she were at some other point?)
 
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