`v = r\omega = \left(0.1\,\text{m}\right)\left(35000\,\frac{\text{rev}}{\text{min}}\right)\left(\frac{1\,\text{min}}{60\,\text{sec}}\right)\left(2\pi\,\frac{\text{rad}}{\text{rev}}\right)\approx 370\,\frac{\text{m}}{\text{s}}`

First, you need to convert the angular velocity to radians per second (or just 'per second' since radians are really just a ratio, and thus unit-less). Then you can multiply by the radial distance from the point of rotation to get the regular velocity or speed at that point on the disk.

That's my main comment on problem 1 of the exam. The second one being that the time dilation factor is just the ratio between the time elapsed at the rim of the disc to that at the center (which is at rest). The ratio is just (gamma) - we don't need to use the full Lorentz transformation, because there is no significant spatial separation, and we are talking about different parts of the same object.

Thats pretty simple. Wish I'd have thought about that on the test!

ReplyDeleteI was hoping that writing v=r(omega) on the board would trigger some memories, but for most of you PH105 was quite a while back ...

ReplyDeleteyeah a few years ago. and I personally didn't even look at the board. mental note!

ReplyDeleteI didn't see that connection but I did the same thing for some reason. Totally different thought process, same result I guess.

ReplyDeletewell, why isn't the equation v=2pi*rw instead of v=rw?

ReplyDeleteNormally I just think about frequency and circumference.. but I wrote down v = rw and left out the 2pi.. :C

The 2pi is just to convert between revolutions and radians, that's all. If you first figured the circumference and multiplied that by frequency, that works too -- it amounts to the same thing.

ReplyDeleteSeveral people did it that way on the exam, and that's fine. Whichever is easier for you to remember, really. For me, its converting revolutions to radians, but your mileage may vary.