Calc Moment of Inertia of Spinning Disc at Angle

[marooned]

to calculate the moment of inertia of a spinning disc (a spinning top for example) i know the formula is I=(mR^2)/2, but in order to find the moment of inertia of a spinning disc at and angle to the vertical, how would this be done? does the moment of inertia change if the disc is not spinning parallel to the ground? The disc in question is spinning about its centre, and precessing a point at which it stays a constant angle to the vertical from.

 

[marooned]

On thinking about this problem overnight, i decided that the moment of inertia for the disc will be the same no matter what plane it is spinning in, as long as the axis around which it spins is constant. So that means the moment of inertia for the disc will be the same on an angle as it is parallel to the ground. Is this right?

 

[DaTario]

no, it seems not. You have to consider the cossines which appear when the angle is different from zero. Moment of inertia has to do with the distances from the point masses to the axis.

Best regards

DaTario

 

[marooned]

But if the disc remains spinning on the same axis, the distance from the point masses to the axis remains the same. If the axis is then tilted the distance from the point masses to the axis is still the same, as the disc tilts with the axis. I realize this is probably wrong, but I can´t seem to get past that idea.

 

[Yegor]

But if the disc remains spinning on the same axis, the distance from the point masses to the axis remains the same. If the axis is then tilted the distance from the point masses to the axis is still the same, as the disc tilts with the axis. I realize this is probably wrong, but I can´t seem to get past that idea.

I think you aren't right here. Distance to the axis doesn't remains the same for each point (it doesn't equals the distance to the centre of disc, but shortest distance to the axis!).
If you know I according to 3 orthogonal axis, then you can calculate I for each axis (through the same point) using $$I=I_1 (\cos{\alpha})^2+I_2 (\cos{\beta})^2+I_3 (\cos{\gamma})^2$$. Here $$I_k$$ is I according to k-axis and $$\alpha$$ is angle between your new axis and k-axis.

 

[marooned]

That sounds correct, the distance to the x-axis is the distance we are after, not the distance to the centre of the disc. However, if the moment of inertia of the disc was to be found with respect to its shaft, no matter what angle it is on, the distance from any point to the shaft is the same, so the moment of inertia would be I=(MR^2)/2, right?

 

[Crosson]

Moment of Inertia is more complex then that. We have the equation:

$$\tau = I \alpha$$

We want this equation to be valid all the time, but the torque and angular acceleration vectors are not always in the same direction, so MoI cannot always be represented by a scalar. In general, MoI can be represented by a 3x3 matrix (infacta second rank tensor).

http://scienceworld.wolfram.com/physics/MomentofInertia.html

 

[marooned]

sorry i meant distance to the axis of rotation, not the x-axis which is completely irrelevant if it is not the axis around which the object is spinning.. sorry. But yes i agree, the torque and angular acceleration vectors change so a scalar will not work. Thanks for your help,
marooned