1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Describe the motion of free fall from different observators

  1. Mar 17, 2010 #1

    fluidistic

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    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.


    2. Relevant equations
    None given.


    3. 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...
     
  2. jcsd
  3. Mar 17, 2010 #2

    ideasrule

    User Avatar
    Homework Helper

    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.
     
  4. Mar 17, 2010 #3

    fluidistic

    User Avatar
    Gold Member

    I know about the only force acting on the bulb (I wrote
    ). 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].
    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Describe the motion of free fall from different observators
  1. Free Fall/Motion (Replies: 6)

  2. Free Fall Motion (Replies: 22)

Loading...