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Describe the motion of free fall from different observators

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fluidistic
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Homework Statement


Describe the motion of a light bulb falling from the ceiling of a train with a constant acceleration. The train's motion is horizontal with respect to the ground.
Describe the motion of the light bulb if you are at rest on the rails. If you are at rest on the train. If you're experiencing a free fall motion with respect to the rails.


Homework Equations


None given.


The Attempt at a Solution


My intuition: On the rails, I'd see a parabolic motion. On the train I wouldn't see a vertical straight line, I'm not 100% sure about what I'd see. On a free fall motion with respect to the rails, I'd see a horizontal straight line.

I don't know how to set up the equations of motion in the different frames of reference.
I'm tempted to say that for the guy on the rails, [tex]\vec a (t)=a \hat i - g \hat j[/tex] but I wouldn't answer this. When the light bulb is cut from the ceiling, there's nothing that could apply a force and hence an acceleration on the bulb, except gravity. Hence I'd answer [tex]\vec a (t)=- g \hat j[/tex].
So [tex]\vec v(t)=v_0 \hat i -gt \hat j[/tex].
Thus [tex]\vec r(t)=v_0 t \hat i +\left (H-\frac{gt^2}{2} \right) \hat j[/tex] where [tex]v_0[/tex] is the velocity of the train when the light bulb is cut. Thus, it is the same motion (parabola) as if the train was moving with a constant velocity [tex]v_0[/tex] and no acceleration; according to this observer.
With the frame of reference falling with respect to the previous one, [tex]\vec a (t)=0 \hat j[/tex]. Thus I can directly write [tex]\vec r(t)=(v_0 t+d) \hat i[/tex], a horizontal straight line.

Now... on the train:
[tex]\vec a(t)=a \hat i - g \hat j[/tex].
Hence [tex]\vec v(t)=a t \hat i - gt \hat j[/tex].
Therefore [tex]\vec r(t)=\frac{at^2}{2} \hat i + \left (H- \frac{gt^2}{2} \right ) \hat j[/tex] which seems to me a straight line (not horizontal though), but I'm not 100% sure. I've taken vector calculus so I MUST be able to answer this with ease, but I should re read my course I guess. Unless you help me recover from this memory-loss.

Are my results good? I didn't use any Galilean transformation, I guess I should have? I only used an intuition, very un-rigorous. Is there a rigorous way to solve this problem? Like for example describing the 3 reference frames from only 1 and then use a Galilean transformation to describe everything they ask for...
 

Answers and Replies

  • #2
ideasrule
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My intuition: On the rails, I'd see a parabolic motion. On the train I wouldn't see a vertical straight line, I'm not 100% sure about what I'd see. On a free fall motion with respect to the rails, I'd see a horizontal straight line.
No, that's not right. On the rails, you're in an inertial frame, and the only force on the object is gravity. In the train's frame, a fictitious force ma seems to be pushing the object backwards as gravity pulls it down.
 
  • #3
fluidistic
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No, that's not right. On the rails, you're in an inertial frame, and the only force on the object is gravity.
I know about the only force acting on the bulb (I wrote
When the light bulb is cut from the ceiling, there's nothing that could apply a force and hence an acceleration on the bulb, except gravity. Hence I'd answer [tex]\vec a (t)=- g \hat j [/tex].
). I reached [tex]\vec r(t)=v_0 t \hat i +\left (H-\frac{gt^2}{2} \right) \hat j[/tex].
From this I now reach [tex]y(x)=H-\frac{gx^2}{v_0^2}[/tex].
This was by assuming [tex]\vec a (t)=- g \hat j[/tex].
ideasrule said:
In the train's frame, a fictitious force ma seems to be pushing the object backwards as gravity pulls it down.
Yes I know... that's why I wrote [tex]\vec a(t)=a \hat i - g \hat j[/tex]. Is this all wrong? If so, I don't see where in the equations.

By the way I just remember how to find the motion of [tex]\vec r(t)=\frac{at^2}{2} \hat i + \left (H- \frac{gt^2}{2} \right ) \hat j[/tex], which is indeed a straight line.
Now, have I used the good equations?
 

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