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I'm trying to solve exercises about relative motion and having hard times in finding out the initial conditions in respect to the inertial and non-intertial frame of reference, while I'm able to do everything else. Here's a fine example (pretty easy apparently):

Imagine being inside a train moving on a straight line railway with a constant speed of $v_0 = 90 km/h$

At $t_0 = 0$ the train starts a constant deceleration $A=-2m/s^2$ when a particle point starts falling from a table inside the train (with no friction). Determin the trajectory of the point when observed by the two systems.

My considerations are:

$\cdot$ Inside a non intertial frame of reference, when the point starts falling, i.e. there's no more interaction of the point particle with the table, the forces on the body are the weight $W = -mg$ and a fictitious force $F_t = -mA$ so that I can tell for sure that vertical acceleration is $a_y = -g$ and horizontal one is $a_x = -A = 2m/s^2$ so that

$\ddot{\gamma}'(t) = \begin{cases} \ddot{x}'(t) =-A \\ \ddot{y}'(t) = -g\end{cases} \implies \gamma'(t) = \begin{cases} x'(t) =x_0 + v_0t-A\frac{t^2}{2} \\ y'(t) = h-g\frac{t^2}{2}\end{cases}$

By selecting the origin in $x_0$ it becomes

$\gamma'(t) = \begin{cases} x'(t) =v_0t-A\frac{t^2}{2} \\ y'(t) = h-g\frac{t^2}{2}\end{cases}$

which would be a line because it's composition of two uniformly accelerated straight line motions. Here I'm usually not sure how to select the initial condition for velocity.

Remembering that $\vec{OO'}(t) = v_ot+A\frac{t^2}{2}$ and $\gamma(t) = \gamma'(t) + \gamma_{OO'}(t)$ which would result in

$\gamma(t) = \begin{cases} x(t) =2v_0t \\ y'(t) = h-g\frac{t^2}{2}\end{cases}$

which would turn me in a laugh to me cause it's really unlikely that this resould coult even be correct (The trajectory would be parabolic as composition of uniform and uniformly accelerated straight line motions).

Is there any trick that would lead me to significantly reduce this kind of errors in establishing the initial conditions? (Expecially velocity)

Baffo rasta
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  • Why should parabolic motion be an incorrect result (it's the calculation for inertial frame, right?)? – Ruslan May 22 '18 at 19:29
  • @Ruslan I'm pretty confident that the trajectory is correct but not sure at all about the horizontal equation... Anyway yep, for the inertial frame. – Baffo rasta May 22 '18 at 19:31
  • Your question is quite hard to follow. Could you define what your symbols like $\gamma, \gamma', x, y, x', y'$ and especially $\vec{OO'}$ mean? Otherwise one has to guess their physical meaning from the equations which might contain mistakes. – Ruslan May 22 '18 at 19:36
  • @Ruslan I'm using $\gamma$ to refer to the decomposition of the motion on the plane in horizontal and vertical directions, $x', y'$ is referred to the non-inertial frame, $x,y$ to the inertial one. $\vec{OO'}$ is the position vector for the origin of the non inertial frame as seen in the inertial one. – Baffo rasta May 22 '18 at 19:40

1 Answers1

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You added an extraneous initial velocity in the non-inertial frame in your expression for $\gamma'$:

$$\gamma'(t) = \begin{cases} x'(t) =x_0 \color{red}{{}+ v_0t}-A\frac{t^2}{2} \\ y'(t) = h-g\frac{t^2}{2}\end{cases}.$$

Remember that the particle was initially at rest in the train's reference frame, before the train began decelerating. So your first constant of integration of $\ddot\gamma'$ shouldn't be equal to $v_0$ – it should be zero. Then, after we include our choice of origin $x_0=0$, we get

$$x'(t)=-At^2/2.$$

And from this you'll get a sensible result of $$x(t)=v_0t.$$

Is there any trick that would lead me to significantly reduce this kind of errors in establishing the initial conditions?

Your mistake is because you became confused by notation. Apparently, you named the constant of integration $v_0$ because it "makes sense", but forgot that this symbol was already used. So one of the tricks is to keep track of your symbols, and use some more or less unique names for constants of integration (e.g. start with something like $\kappa_1$ and $\kappa_2$ instead of $x_0$ and $v_0$). You can always rename them later if you appear to need them further.

Ruslan
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  • Apart from the notations I feel like having problems in "detecting" the right conditions. With this correction if I decided to start the description from the inertial frame I'd have to understand that the initial velocity was the same as the train's one when I don't have this kind of intuition (indeed I was sure that horizontal acceleration was null as application of 2nd Newton's law) – Baffo rasta May 22 '18 at 20:15
  • @Bafforasta to determine the right initial conditions you should just start from the reference frame for which you know positions, velocities etc. for sure, and then apply corresponding Galilean transformation to convert to the frame you're interested in. – Ruslan May 22 '18 at 20:21
  • I know you're completely right and this is usually what I would do but I have problems understanding why the horizontal speed of the point in the inertial frame of reference equals the one of the train (I get the correct result applying the transformation but can't interpretate why) – Baffo rasta May 22 '18 at 20:26
  • @Bafforasta Consider that the point is stuck to the table (for all you can see inside the train before deceleration, it's really stuck). Then you can consider it part of the train for a while. What velocity do parts of the train have, if the train itself has velocity $v_0$ (assuming no tricks like rotation etc.)? – Ruslan May 22 '18 at 20:29
  • Yeah that's correct, no rotation for sure. Thanks for your help. – Baffo rasta May 22 '18 at 20:30