Angular Frequency of a two-object system

AI Thread Summary
The discussion revolves around the calculation of angular frequency in a two-object system, specifically addressing the inclusion of the ladder's weight in the mass (m) of the formula. Initially, the user calculated the angular frequency without including the ladder's 6 kg mass, leading to confusion about the correctness of their answer. It was clarified that the mass of the ladder should indeed be included in the numerator for accurate results. Additionally, the moment of inertia calculations were discussed, confirming that 'd' represents the distance from the pivot to the center of mass of the entire system. The final consensus is that proper inclusion of all masses and correct interpretation of distances is essential for accurate calculations.
hidemi
Messages
206
Reaction score
36
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
 
Physics news on Phys.org
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.
 
  • Like
Likes TSny
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?
 
  • Like
Likes Doc Al
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?)
 
  • Like
Likes TSny
Back
Top