Angular Frequency of a two-object system

In summary, the correct calculation for the weight of the ladder should include its mass in the numerator. Additionally, the moment of inertia for the ladder when rotating around one end should be calculated separately. The distance 'd' in the equation represents the distance from the pivot point to the center of mass of the entire system.
  • #1
hidemi
208
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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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  • #2
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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  • #3
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!
 
  • #4
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:
  • #5
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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  • #6
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?
 
  • #7
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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What is the definition of angular frequency?

Angular frequency is a measure of the rate at which an object rotates or oscillates around a fixed point. It is represented by the symbol ω (omega) and is measured in radians per second.

How is angular frequency related to linear frequency?

Angular frequency and linear frequency are inversely proportional. This means that as the angular frequency increases, the linear frequency decreases, and vice versa. The relationship between the two can be expressed as ω = 2πf, where ω is the angular frequency and f is the linear frequency.

What is the formula for calculating the angular frequency of a two-object system?

The formula for calculating the angular frequency of a two-object system is ω = √(k/m), where k is the spring constant and m is the mass of the system.

How does the mass affect the angular frequency of a two-object system?

The mass of the system has an inverse relationship with the angular frequency. This means that as the mass increases, the angular frequency decreases, and vice versa. This is because a heavier mass requires more force to rotate or oscillate at the same frequency as a lighter mass.

What factors can affect the angular frequency of a two-object system?

The angular frequency of a two-object system can be affected by the mass, spring constant, and any external forces acting on the system. Changes in any of these factors can alter the angular frequency of the system.

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