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Moment of Inertia/Torque

by JJX
Tags: inertia or torque, moment
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JJX
#1
Dec6-09, 11:24 PM
P: 3
Hi all, I'm working on a catapult for a project and I'm having some problems with some of the calculations, I'm trying to get the angular velocity of my catapult's arm by first obtaining the torque and moment of inertia.

1. The problem statement, all variables and given/known data

A diagram I made with all the information and the sum of torques solved.


2. Relevant equations
Torque = Force * Distance to axis of rotation
I=(1/12)ML^2 + MD^2
F = mg

3.Attempt at solution
I've solved Torque, but I have some doubts for Moment of Inertia:
1) Is M the mass of the arm by itself or the addition of all the masses involved (arm+counterweight+basket+projectile)
2) Is L the distance of the whole Arm and D the distance of the axis of rotation to the center of mass?

I=(1/12)ML^2 + MD^2
I=(1/12)(0.9)(.3650)^2 + (0.9)(.1)^2
I=.0189 kg*m^2
Is this right?
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JJX
#2
Dec7-09, 09:42 PM
P: 3
Can anyone at least confirm that my solution is right?
Redbelly98
#3
Dec8-09, 08:14 AM
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Redbelly98's Avatar
P: 12,074
First, welcome to Physics Forums
Quote Quote by JJX View Post
A diagram I made with all the information and the sum of torques solved.
Since the axis of rotation is not at the arm's center of mass, you need to add a torque term for the arm.

2. Relevant equations
Torque = Force * Distance to axis of rotation
I=(1/12)ML^2 + MD^2
F = mg

3.Attempt at solution
I've solved Torque, but I have some doubts for Moment of Inertia:
1) Is M the mass of the arm by itself or the addition of all the masses involved (arm+counterweight+basket+projectile)
2) Is L the distance of the whole Arm and D the distance of the axis of rotation to the center of mass?

I=(1/12)ML^2 + MD^2
I=(1/12)(0.9)(.3650)^2 + (0.9)(.1)^2
I=.0189 kg*m^2
Is this right?
1) M is just the arm mass
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 mĚd2 for each mass, using the distance from the axis of rotation.

JJX
#4
Dec8-09, 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.
Redbelly98
#5
Dec8-09, 03:33 PM
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Redbelly98's Avatar
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 cross-section 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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