Calculating Moment of Inertia of a Sphere

AI Thread Summary
The discussion focuses on calculating the moment of inertia for a sphere, specifically using the formula 2/5 Ma^2 for rotation about a diameter. The user expresses confusion over the significance of the calculated values, particularly when adding a point mass, which resulted in different inertia values depending on its position. There is a question about whether the larger inertia value should correspond to greater resistance to rotation. The conversation also touches on the concept that moment of inertia is calculated with respect to an axis and that dimensions affect inertia values, while also questioning how to find inertia along non-axis-aligned vectors. Understanding these principles is crucial for applying inertia in practical scenarios.
Bucky
Messages
79
Reaction score
0
Hi, I'm in the middle of programming an inertia system and am only really just starting to appreciate what the heck inertia is :) I have been taught inertia, but I haven't actually applied it in a real situation (all my exams and tutorials have resulted in formulae giving answers like 2/7Ma^2).

So I've written a function to find inertia of a sphere and I've plugged some numbers in. I'm a bit confused over what formulae to use for starters.

I can sort of see that inertia of the spheres centre might be useless? given that they're totally symetrical? Am I off with this?

For inertia of a sphere about the diamater I'm using the equation 2/5 Ma^2. Is this appropriate?

Also I've plugged in numbers (like I said). I've never got a numerical answer for a system before so I'd appreciate some guidance as to wether or not I'm correct.

sphere at 0,0,0
radius 2
mass 1

rotation about point on diameter

2/5 Ma^2
0.4 * 1 * 4
MI = 1.6 ?
 
Physics news on Phys.org
This should be useful: http://hyperphysics.phy-astr.gsu.edu/Hbase/isph.html" .
 
Last edited by a moderator:
sorry to bump an old thread, but I've done some other bits and pieces and have some results that I'd like to ask about.

I took my mass and added a point mass 1kg to it. If m is at the 'point' of inertia the resulting answer is: 1.6kgm sq

if m is at the opposite side of the sphere from the point of inertia the resulting answer is: 17.6kgm sq

firstly should it be this way round? i can't help but think that the bigger number should be for the higher resistance

also, what is the meaning of what i have found? at the end of the day, what do these numbers actually mean?
 
Ok after reading through another book I've found equations defining inertia of cylinders and cubeoids. These are given as inertia in x, y and z dimensions. Basically...this wasn't what I was expecting. Is a bodies inertia usually provided as a value in X, Y, Z dimensions?

Does inertia have nothing to do with where on the body a force to move it would act?
 
It has nothing to do with any dimensions. Moment of inertia is calculated with respect to some axis, typically x, y and z.
 
thanks for the responce!...but i would asume that the width/depth/height are the values that affect its inertia in each axis? for example the inertia of a cube along the x-axis is
<br /> \frac{1}{12} m ( b^2 + c^2)<br />where m = mass, and b and c are the length of the cube along y and z axis.

also, if inertia is calculated through XYZ axis, how is the inertia along a non axis-aligned vector found? interpollation? do you even need to find this? To be honest I've never used inertia in anything (i've just "found" the inertia) so I'm finding it hard to learn.
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top