So, yes... That becomes obvious because they’re asking about more than one reference frame in the problem. Okay, so to solve this without considering the aspect of special relativity, I need to look at both the reference frame of A and B, because if B were traveling at the same speed as A, than...
I understand why the answer would be for B to travel at 4 m/s, but I don’t understand how I would have arrived at the answer myself.
If I take the reference frame of B, where A and the space station are moving in that reference frame, A is moving towards B at 8 m/s, and the space station is...
Well, then A approaches B at 8 m/s, and A would also approach the spaceship at 8 m/s. In the reference frame of A this means that spaceship B is moving at 8 m/s relative to A, and the space station also moves at 8 m/s towards A, relative to A.
Similarily, in the reference frame of B... A...
Oh, so the question gives me the speed of the spaceship A in the reference frame of the space station, and then I’m considering everything to be in the reference frame of B when solving the problem, because everything is moving relative to the spaceship B.
Okay, so the goal here would be to find a relative velocity of A according to B, such that the relative velocity of A is equal to the speed of spaceship B. So, I would set v = (0.8c - v)/(1 - (((0.8c)(v))/(c^2))
Okay, so I took the relative velocity of spaceship A (according to B) to be equal to 0.8c + v/((1+((0.8c)(v)/(c^2))), where v is the speed of spaceship B, and the 0.8c is my relative velocity of A according to the space station, but since I assume that the space station is at rest, then 0,8c...
Well then, the station would be moving at the same speed as B relative to B... is that right? So, in that case, I would be looking for a relative velocity of A that is the same speed as the spaceship B?
I tried setting u' in the relative velocity formula to 0.8c, the speed of ship A, and then trying to somehow find an equation that has the speed of the space station relative to B. But for some reason I take the space station to be at rest. Because we want the relative speed of ship A and the...
Okay, so maybe that's where I screwed up. The force of gravity is 5(9.8) = 49 N regardless of the angle of the rod. So, the torque would be 5(9.8)(sin30)(0.15) = 3.675 N*m. Maybe I do understand the parallel axis theorem, then. So, I already know that the rotational inertia about the centre of...
It wasn't a helpful figure. It just showed two spaceships in a line, A behind B, and the space station a short distance away from B. I could have drawn something similar myself. I don't think it was too helpful to the question.
Homework Statement
A uniform rod (mass = 5.0 kg, length = 0.3 m; the rotational inertia of the rod relative to the axis through its centre of mass is I = (1/12) ML^2 ) is free to rotate about a frictionless pivot at one end. The rod is released from rest in the horizontal position. What is the...
So, I'm looking for units of a and b for which all of the s's in the numerator cancel, and there is only one s left in the denominator, and one m in the numerator?