If an object 40' wide and 80' long is spinning sideways at 9 trillion revolutions per second, what speed is that in miles per hour? It would be great if you could either do it for me or at least show me how to do it ya know? Anyway, thanks in advance.
Well, you need a radius. Is it fixed at it's center? Do you need the speed on the edge or towards the center?
Ben mentions symmetry. Take the distance of the point in question from the rotation axis, multiply it by 2*pi, and you end up with the path length it carves out per revolution. Multiply that by the number of revolutions per second, and you end up with the speed of the point. ex: If the point is 1 metre from the axis of revolution, its path length is 2*pi*1 metres. If the object spins once per second, then the point travels at 2*pi*1*1 metres per second. So if your sphere has a radius of 80 metres, and it rotates at 9000000000000 times per second, then the speed of a point on the equator is 80*2*pi*9000000000000 = 4523893421169302.26 metres per second. That's about 15 million times faster than the speed of light, so this is not a possible scenario. Either way, you can probably gather that a point on a rotating sphere's equator carves out a much longer path than the path carved by a point near the top or bottom of the sphere. I think Einstein referred to this as "clocks on the equator running slower than that at the poles", because the higher speed implies greater time dilation (Special Relativity).
The speed of the farthest point from the axis is anywhere between 689.4 trillion m/s (1542.2 trillion mph) and 1378,9 trillion m/s (3084,5 trillion mph), depending on the placement of the axis.
Forget the sidways part. Thanks for filling me in on the math! It's a sci fi story thing, the speed of light isn't really an issue... Please Register or Log in to view the hidden image! Anyway, thanks again!
That would be foot/s Please Register or Log in to view the hidden image! Ahem, as Fraggle pointed out, that should be feet/s.
I guess we all assumed he meant spinning in the plane of the largest cross section, i.e., rotating about its shortest axis. Since he did not give the thickness of the object it's conventional to assume that it is at least two orders of magnitude less than the other two dimensions and therefore unimportant to the problem. The plural of foot is feet, even as a unit of measurement. Please Register or Log in to view the hidden image!
Would it be safe to say that the thickness of a space ship (think space shuttle) is ordinarily two orders of magnitude less than the other dimensions? (jeez i haven't been on this site for ages, out of practice with the learning of physics...)
