Recent content by OrlandoLewis

Oh, that's an interesting concept :smile:

Oh yeah, missed that. So theoretically, whatever angle I fold, the rod's moment of inertia will just be the same?

Yes, I meant decrease by the former value.

Then it should remain the same right?

Homework Statement Consider one rod of length L which is spun at its center perpendicular to its length; it will have a certain moment of inertia. Now, if the same rod is folded in the middle creating a certain angle and still spun at its center, what happens to its moment of inertia? Homework...

Yeah, I was pretty confused too. Just a follow-up question; what if the arrangement is not in a straight line. For example, the central disk has 7 adjacent disks to its left and 7 adjacent disks on top of it, making the arrangement form a 90 degree angle? Will the arrangement's rotational...

Isn't that what I've done? Using the parallel axis theorem, I calculate each flat disk's rotational inertia. Thereafter, I've added each of their inertia to find the real moment of inertia of the arrangement. Thus, the inertia of the whole arrangement equals to the rotational inertia of the...

So I should still consider the radius when calculating the moment of the rod? Why so?

Oh, sorry, the exponent are raised to a negative power :smile:

Found the inertia using that equation to be 8.352*10^-6 kg-m^2 When it is considered as a thin rod of length 1 m, the inertia becomes 8.333*10^-6 kg-m^2 The percent error becomes: 0.2217% (pretty low) BTW, I didn't get the image from a manual =( I used a math editor 'cause I don't know how to...

Sorry, I was confused at first... nevermind the former equation: 0.5mdiskrdisk2 + 0.5mdiskrdisk2(22 + 42 + 62 + 82 + 102 + 122 + 142) Is the other equation, the one in the image, the correct one I should use? Is that how parallel axis theorem applied in this problem?

This is what I'm referring to... The first term is for the center disk while the remaining is for the next ones, times two because there are rods from the left and right of the axis. This is letting m be the mass of each disk and r be the radius of each disk too.

Yeah, what I'm referring to d is the distance of one disk to the axis of rotation O. What I mean about moment of a single disk is its moment when spun at its center. So for example, the outermost disk has a distance of 14r from its center, then can I consider its moment to be 1/2 mr^2 +...

Yeah, I do know about it. So I should just find the moment of a single disk then add it to its md2 for each disk? Is that correct? Making the solution for a as: 0.5mdiskrdisk2 + 0.5mdiskrdisk2(22 + 42 + 62 + 82 + 102 + 122 + 142)

I tried to separate the rotational inertia of the center disk and the remaining fourteen disks. Is that how it is done? I'm still having trouble finding the rotational inertia of the arrangement.