Question

Help Please!!!! Will give medal!! At the outer edge of a rotating space habitat, 130 m from its center, the rotational acceleration is g. What is the rotational acceleration at a distance of 65 m from the center of the habitat?

Answer 1

@hartnn

Answer 2

@MrNood

Answer 3

@radar

Answer 4

@abb0t

Answer 5

@esshotwired

Answer 6

@ashore0482

Answer 7

@ganeshie8 please help

Answer 8

In a rotating system, the angular velocity is constant among the disc. What we know is the angular acceleration at a distance b from the center and we want to know the angular acceleration at a distance a from the center.\[\alpha = \omega ^{2}r\]Solving for the angular velocity, we get: \[\omega_{outer} = \sqrt{\alpha_{outer}/b}\]This is also true for the inner radius.\[\omega_{inner} = \sqrt{\alpha_{inner}/a}\]However,\[\omega_{inner} = \omega_{outer}\]Therefore,\[\sqrt{\alpha_{inner}/a} = \sqrt{\alpha_{outer}/b}\]Rearranging the equation, we find\[\alpha_{inner}=\alpha_{outer}*(a/b)\]From here it is simply plug-and-chug with: \[\alpha_{outer}=g, a=65m, b=130m\]I will leave the rest to you.

Answer 9

I understand that the answer is 130m, right? But I really don't understand how you go to it. Is there a simpler formula that could be used?

Answer 10

From the problem, you already know that the outer distance is 130m and you are solving for rotational acceleration (which is in units of 1/s^2). The problem is utilizing your knowledge that rotational velocity is constant on a spinning disc and that the formula for rotational acceleration (which I used a poorly chosen symbol) is:\[a_{r} = \omega^{2}r\]In my previous response, I used these two concepts to derive a formula for the rotational acceleration at one part of the disc knowing it at another part and the distance from the center at the two points in question.

Answer 11

*I meant the units are m/s^2, this is rotational, not radial.