- #1

JulienB

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## Homework Statement

Hi everybody! I'm really struggling with those exercises from my special relativity class, and I'd like to know what you think about that one because I come up with a strange result:

The motion of a point mass in a plane is given by:

[tex]x = a cos(ωt), y = b sin(ωt)[/tex]

Calculate the speed as well as the normal and tangential components of the acceleration for any time t.

**2. The attempt at a solution**

Okay so first I derivated x and y to calculate |v|:

[tex]

\dot{x} = -a \cdot ω \cdot sin(ωt), \dot{y} = b \cdot ω \cdot cos(ωt) \\

\implies |\dot{\vec{r}}| = \sqrt{-a^2 ω^2 sin^2(ωt) + b^2 ω^2 cos^2(ωt)} \\

= ω \sqrt{b^2 cos^2(ωt) - a^2 sin^2(ωt)}

[/tex]

Then I calculate the tangential vector e

_{T}:

[tex]

\vec{e_T} = \frac{\dot{\vec{r}}}{|\dot{\vec{r}}|} = \binom{\frac{-a sin(ωt)}{\sqrt{b^2 cos^2(ωt) - a^2 sin^2(ωt)}}}{\frac{b cos(ωt)}{\sqrt{b^2 cos^2(ωt) - a^2 sin^2(ωt)}}}

[/tex]

Then I use that result to calculate the tangential component of the acceleration:

[tex]

\dot{v} \vec{e_T} = \binom{aω^2 sin(ωt) \sqrt{\frac{a^2 +b^2 tan^2(ωt)}{b^2-a^2 tan^2(ωt)}}}{-bω^2cos(ωt) \sqrt{\frac{a^2 + b^2tan^2(ωt)}{b^2 - a^2tan^2(ωt)}}}

[/tex]

That's pretty strange... And then I subtract this result from the total acceleration to get the normal acceleration:

[tex]

v \dot{\vec{e_T}} = \ddot{\vec{r}} - \dot{v} \vec{e_T} = \binom{-aω^2(cos(ωt) - sin(ωt)\sqrt{\frac{a^2 +b^2 tan^2(ωt)}{b^2-a^2 tan^2(ωt)}})}{-bω^2 (sin(ωt) + cos(ωt)

\sqrt{\frac{a^2 + b^2tan^2(ωt)}{b^2 - a^2tan^2(ωt)}})}

[/tex]

Ehem... Doesn't look like a classical answer to a problem :D did I do something wrong?

Thanks a lot in advance for your answers.

Julien.

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