- #1
JulienB
- 408
- 12
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 eT :
[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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