Do the trajectories found in primed and unprimed are the same?

Question

I am trying to show the Newtons law is valid in any frame of reference. Hence testing with a projectile motion. This is what I got in K and K' frame if I solve the differential equation and apply the initial condition for the figure above.

in K frame: $$x_{2} = \frac{-g}{2v_{0}^2\cos^2(\theta )}x_{1}^2 + \tan(\theta)x_{1}$$

in K' frame: $$x_{1}^{'} = \tan(\theta)x_{2}^{'} + v_{0}\left ( \frac{2x_{2}^{'}}{-g\cos(\theta )} \right )^{1/2}$$

If I plot these two equation for theta = 30, v = 30 and g= 10, I get the following plots, which I dont see the same laws of physics.

In K frame plot:

in K' frame plot:

Why I dont see the same parabolic type graph in both cases? What I am missing to understand?. Thanks for any input.

Answer 1

Newton's Laws imply that the path of a projectile in a non-rotating frame with a uniform gravitational field and no other forces on the projectile during flight will be a parabola whose axis is parallel to the direction of the gravitational field.

In your $K'$ frame the direction of the field is at an oblique angle, not parallel to the axes. Therefore you should expect a parabola whose axis is at an oblique angle.

The use of the square root eliminates part of the solution space so you see only part of the parabola. The other part is given by

$$x'_1 = x'_2 \tan(\theta) - v_0\left(\frac{2x'_2}{-g\cos(\theta)}\right)^{1/2}$$

(that is, by choosing the other square root of $2x'_2/(-g\cos(\theta))$).

You can also rearrange the equation so the term with the square root is alone on the right side,

$$x'_1 - x'_2 \tan(\theta) = v_0\left(\frac{2x'_2}{-g\cos(\theta)}\right)^{1/2},$$

and then square both sides:

$$\left(x'_1 - x'_2 \tan(\theta)\right)^2 = v_0^2\left(\frac{2x'_2}{-g\cos(\theta)}\right).$$

This will give the complete parabola.

Answer 2

Let $\underline{r}=[x,y]^\mathrm{T}$ be the position vector in the K frame and $\underline{r'}=[x',y']^\mathrm{T}$ the position vector in the K' frame. We can define a rotation matrix $$\mathbf{A} =\begin{bmatrix} \cos (-\theta) & -\sin (-\theta) \\ \sin (-\theta) & \cos (-\theta) \end{bmatrix} =\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$$ And state that $$\underline{r'}=\mathbf{A}\underline{r}$$ I can elaborate on the reasoning behind this if you want; it's not that hard.

We can solve an IVP to find $\underline{r}(t)$: $$\ddot{\underline{r}}=[0,-g]^\mathrm{T} ~~|~~\underline{r}(0)=[0,0]^\mathrm{T}~;~\dot{\underline{r}}(0)=\underline{v}_0$$ In polar coordinates, $\underline{v}_0=(v_0,\theta)^\mathrm{T}$ therefore in Cartesian coordinates $\underline{v}_0=[v_0\cos\theta,v_0\sin\theta]^\mathrm{T}$. We can solve this IVP as $$\underline{r}(t)=\big[v_0\cos(\theta)t,v_0\sin(\theta)t-\frac{1}{2}gt^2\big]^\mathrm{T}$$ Therefore $$\underline{r'}(t)=\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}\begin{bmatrix} v_{0}\cos( \theta ) t\\ v_{0}\sin( \theta ) t-\frac{1}{2} gt^{2} \end{bmatrix}$$ $$=\begin{bmatrix} v_{0}\cos^{2}( \theta ) t+v_{0}\sin^{2}( \theta ) t-\frac{g\sin( \theta )}{2} t^{2}\\ -v_{0}\cos( \theta )\sin( \theta ) t+v_{0}\sin( \theta )\cos( \theta ) t-\frac{g\cos( \theta )}{2} t^{2} \end{bmatrix}$$ $$=\begin{bmatrix} v_{0} t-\frac{g\sin( \theta )}{2} t^{2}\\ -\frac{g\cos( \theta )}{2} t^{2} \end{bmatrix}$$ Using $v_0=30$, $\theta=30^{\circ}=\frac{\pi}{2}$ and plotting in $0\leq t\leq 3$, you get the following plots:

The black line is the unprimed and the orange line is the primed frame. Notice that the orange curve is just the black curve rotated clockwise $30^{\circ}$, as expected.

Answer 3

After $t$ elimination in

$$ \cases{ x-x_0 = v_0\cos\theta t\\ y-y_0 = v_0\sin\theta t-\frac g2 t^2 } $$

we have

$$ y-y_0 -\left((x-x_0)\tan\theta - \frac{g(x-x_0)^2}{2\cos^2\theta v_0^2}\right) = 0 $$

which defines a conic called parabola.

Making a coordinates change (rotation over $(x_0,y_0)$) with angle $\phi$ as

$$ \cases{ x-x_0 = (X-x_0) \cos \phi-(Y-y_0)\sin\phi\\ y-y_0 = (X-x_0) \sin \phi+(Y-y_0) \cos\phi } $$

we have

$$ (X-x_0,Y-y_0)^{\dagger}\cdot M_{\phi}\cdot (X-x_0,Y-y_0)+B_{\phi}\cdot (X-x_0,Y-y_0) = 0 $$

with

$$ M_{\phi} = \left( \begin{array}{cc} -\frac{g \cos ^2\phi \sec ^2\theta}{2 v_0^2} & \frac{g \sin \phi \cos\phi \sec ^2\theta}{2 v_0^2} \\ \frac{g \sin \phi\cos\phi \sec ^2\theta}{2 v_0^2} & -\frac{g \sin ^2\phi \sec ^2\theta}{2 v_0^2} \\ \end{array} \right) $$

and

$$ B_{\phi}=\left( \begin{array}{c} \cos \phi \tan \theta -\sin \phi \\ -\sin\phi \tan\theta-\cos\phi \\ \end{array} \right) $$

Note that for $\phi = 0$ the matrix $M_0$ reads

$$ M_0 = \left( \begin{array}{cc} -\frac{g \sec ^2\theta}{2 v_0^2} & 0 \\ 0 & 0 \\ \end{array} \right) $$

with eigenvalues $\left\{-\frac{g \sec ^2\theta}{2 v_0^2},0\right\}$ and that $M_{\phi}$ eigenvalues are also $\left\{-\frac{g \sec ^2\theta}{2 v_0^2},0\right\}$ which define the conic as a parabola. Concluding, the referential rotations as defined, instantiate diverse rotated captions from the same parabola.