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Moment of Inertia/Torque 
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#1
Dec609, 11:24 PM

#2
Dec709, 09:42 PM

P: 3

Can anyone at least confirm that my solution is right?



#3
Dec809, 08:14 AM

Mentor
P: 12,074

First, welcome to Physics Forums
2) Yes to both HOWEVER ... this I is the M.O.I. for the arm only. You need to add the contributions of the projectile+basket and counterweight, to get I for the entire catapult. These would be m·d^{2} for each mass, using the distance from the axis of rotation. 


#4
Dec809, 03:00 PM

P: 3

Moment of Inertia/Torque
Thank You Red Belly, that was extremely helpful. I have now come across another problem, I need to find the centre of mass of this very same arm. Is it always the centre? I placed it there by the way, seemed right. Does it have anything to do with the Masses*r aswell? I wonder if the counterweight and the basket affect it.



#5
Dec809, 03:33 PM

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P: 12,074

You can find the center of mass for yourself by seeing where the arm balances.
Is the arm thicker or heavier at one end? If not, and it is a uniform crosssection size along it's whole length, then the center of mass should be in the middle of the arm. If you can verify this by checking where the balance point is (with no other masses attached), all the better. Don't worry about the other masses when calculating the arm's center of mass. You are accounting separately for those masses' contribution to torque and moment of inertia. 


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