Calculating the A value of a physical pendulum

AI Thread Summary
To calculate the A value for a person's arms in a physical pendulum, the moment of inertia is expressed as I = AmL^2, where L is the distance from the center of mass. The arm length is given as 31.30 cm, but the center of mass for a uniform rod is located at its midpoint, meaning L should be half of the arm length. Therefore, L should be 0.1565 m (half of 0.313 m). The equation A = (g/L)(T/2π)^2 is used to compute A, and the correct value for L must be applied to avoid errors in the calculation. The discussion emphasizes the importance of accurately identifying L as the distance from the center of mass to the pivot point for correct results.
ttk3
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Homework Statement



The moment of inertia for an arm or leg can be expressed as I = AmL^2, where A is a unitless number that depends on the axis of rotation and the geometry of the limb and L is the distance from the center of mass. Say that a person has arms that are 31.30 cm in length and legs that are 40.69 cm in length and that both sets of limbs swing with a period of 1.20 s. Assume that the mass is distributed uniformly in both the arms and legs.

Calculate the value of A for the person's arms.


L arm = .313 m
T = 1.20

Homework Equations



A = (g/L) (T/2pi)^2

The Attempt at a Solution




(9.8/.313) (1.20/2pi)^2 = 1.142

I'm not sure where I'm going wrong with this problem. After the derivation of the equation it's plug and chug. I looked up the equation I derived and it's correct. Can anyone lend me a hand?
 
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Hi ttk3,

What does L represent in the equation? You plugged in the length of the arm but I don't think that's correct.
 
It says that L is the distance from the center of mass. Wouldn't the center of mass be located at the top of the arm (the point from which the pendulum swings)?

The equation is derived from the attached equation, and L would be the arm length in that one I thought.
 

Attachments

ttk3 said:
It says that L is the distance from the center of mass. Wouldn't the center of mass be located at the top of the arm (the point from which the pendulum swings)?

The equation is derived from the attached equation, and L would be the arm length in that one I thought.

I can't view the attachment yet, but the center of mass would not be at the top of the arm. The problem indicates that the mass is uniformly distributed in the arm.

Pretend the arm is a uniform rod. Where is the rod's center of mass? It's not at either end of the rod.

Also, when you say that L is the distance from the center of mass, is that all it said? That does not sound like it is complete. Shouldn't it be something like, L is the distance from the center of mass to the shoulder (arm's pivot point)? Because distances are between two points.
 
The question is a direct copy and paste from the program. So if the center of mass is the center of the arm, would I the length of the arm divided by two for my L value?
 
If L is the distance from the center of mass to the shoulder, then that sounds right to me. What do you get?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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