CaptainOfSmug said:
Well I'm at a loss of how to even begin my calculations, I'm having trouble with how to approach the problem, my book isn't doing very well on this topic for me and my professor is essentially a proctor for tests, I tried khan academy with little luck. Should I be setting up three reference frames then each in accordance of each disk? I have no idea how to find the inertia of each disk so I assumed I just leave them as m1,m2, and m3 but I'm assuming there mass increases proportionally to their size since they are the same material, and each disk increases diameter by 1.2cm. So for the relevant equation I gave I assume the denominator will be 6m, but I'm having trouble with the numerator.
EDIT:
So here's what I've come if so far but I'm not sure if it's right:
center of mass= m1x1+2m2x2+3m3x3 /6m
=2.8cm. This answer seems somewhat reasonable, but my intuition tells me it should be larger
I think I can see one problem right off the bat. The disks are supposed to have diameters which increase in the progression 1.2 m, 2.4 m, and 3.6 m, yet in your formula for the center of the mass, you are assuming that if the mass of the smallest disk is
m, then the mass of the next larger disk is
2m, and the mass of the largest disk is
3m. This is incorrect.
Since the OP states that all three disks are made of sheet metal and are the same material, it would then be reasonable to assume that all three disks have the same thickness. This implies that the mass of each disk varies as the volume of metal in the disk, which, since each disk has the same thickness, also implies that the mass of each disk is proportional to the area of the disk. Since the area of the disk is in turn proportional to the
square of the diameter, then the mass of each disk does not increase as the ratio of the diameters, but as the ratio of the
square of the diameters.
The OP seems to imply that each disk is laid down in a line, with the smaller disk touching, but not overlapping, the next larger disk located to its right.
A simple sketch should be made showing three disks laid out thus: oO∅. Using this sketch, one can then determine where the center of mass of each disk is in relation to the origin, which the OP states is located at the leftmost point of the leftmost disk. Once the proper mass relationship and the location of the center of each disk is determined, then you may apply the formula to determine the center of mass of the three disks.
For part B of the problem, rinse and repeat the procedure above, except instead of disks, use spheres instead. (Hint: there will be a different relationship between each mass in the case of spheres as opposed to disks made out of sheet metal.)