Length Contraction Problem for a Rod in a Spaceship

AI Thread Summary
The discussion centers on the length contraction of a rod moving in a spaceship, with participants debating whether the rod is at rest relative to the spaceship. One participant argues that the problem statement indicates the rod is moving at 0.4c relative to Earth, suggesting it cannot be assumed to be at rest in the spaceship. Another points out that while the observer is at rest relative to the spaceship, the spaceship's velocity relative to the rod is unspecified. This ambiguity leads to different interpretations of the scenario, particularly regarding the implications of the rod's motion. The conversation highlights the complexities of relativistic physics and the importance of clearly defined reference frames.
Skyxplorer
Messages
2
Reaction score
0
Homework Statement
A rod has length of 1m. It is moving in a spaceship with velocity 0.4c relative to the earth. The length of the rod measured by an observer in the spaceship will be

(A) 1m

(B) 0.9m

(C) 0.5m

(D) 0.25m
Relevant Equations
-
Shoudnt answer be A since for observer in spaceship rod is at rest.
 
Physics news on Phys.org
Yes
 
Thank you for confirming
 
One might take is as a good assumption, but nowhere do I see it mentioned that the rod is at rest relative to the spaceship.

(In fact, it kind of says "it (the rod) is moving in a spceship" so I wouldn't be too quick with that assumption.)
 
DaveC426913 said:
One might take is as a good assumption, but nowhere do I see it mentioned that the rod is at rest relative to the spaceship.
In fact it clear from the statement that it (the rod) is moving
Skyxplorer said:
##\dots## with velocity 0.4c relative to the earth.
It is safe to assume that the observer is at rest relative to the spaceship, but we are not told the velocity of the spaceship (and observer) relative to the rod.

Answer (B) is consistent with the spaceship being parked on the Earth and the rod passing through two diametrically opposing openings at 0.4c but that's stretching it. :oldsmile:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top