# Elliptic path, normal and tangential acceleration

JulienB

## 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:

$$x = a cos(ωt), y = b sin(ωt)$$

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|:

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

Then I calculate the tangential vector eT :

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

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

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

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

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

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

Julien.

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It might be simpler to calculate some general formulas for the tangential and normal components of acceleration. For example, you can express ##a_T## and ##a_N## in terms of ##\vec{a}##, ##\vec{v}## and ##v##

RedDelicious

When taking the magnitude, you didn't square the x component correctly.

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You mean ##v## is wrong. Yes. That is a problem!

JulienB
Oops. I didn't include the - in the square, that's what you mean right? I'm gonna redo the calculations and post again... Apart from that, is the rest alright?

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Oops. I didn't include the - in the square, that's what you mean right? I'm gonna redo the calculations and post again... Apart from that, is the rest alright?

Have you ever seen this:

##\vec{e}_T = \frac{\vec{v}}{v}##

##\vec{e}_T' = \frac{\vec{a}}{v} - (\frac{\vec{v}.\vec{a}}{v^2}) \frac{\vec{v}}{v} ##

##\vec{a} = (\frac{\vec{v}.\vec{a}}{v}) \frac{\vec{v}}{v} + v \vec{e}_T' = (\frac{\vec{v}.\vec{a}}{v})\vec{e}_T + v |\vec{e}_T'| \vec{e}_N = a_T \vec{e}_T + a_N \vec{e}_N##

Where ##a_T = \frac{\vec{v}.\vec{a}}{v}##

Note also that:

##|\vec{a} \times \vec{v}| = a_N v \ \ ## hence ## \ \ a_N = \frac{|\vec{a} \times \vec{v}|}{v}##

Also useful is:

##v' = \frac{\vec{v}.\vec{a}}{v} = a_T##

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JulienB
@PeroK Thanks for your answer. I think that's what I've used, isn't it? I corrected the sign error but it doesn't look much better:

$$v = \sqrt{a^2 ω^2 sin^2(ωt) + b^2ω^2 cos^2(ωt)} = ω \sqrt{a^2 sin^2 (ωt) + b^2 cos^2 (ωt)} \\ \vec{e_T} = \frac{\vec{v}}{|\vec{v}|} = \binom{\frac{-a sin(ωt)}{\sqrt{a^2 sin^2(ωt) + b^2 cos^2 (ωt)}}}{\frac{b cos(ωt)}{\sqrt{a^2 sin^2(ωt) + b^2 cos^2 (ωt)}}} \\ \ddot{\vec{r}} = \dot{v} \vec{e_T} + v \dot{\vec{e_T}} \\ \implies \mbox{The tangential component of the acceleration is:} \\ \dot{v} \vec{e_T} = \binom{aω^2 sin(ωt) \sqrt{\frac{a^2 + b^2 tan^2(ωt)}{b^2 + a^2 tan^2(ωt)}}}{b ω^2 cos(ωt) \sqrt{\frac{a^2 + b^2 tan^2(ωt)}{b^2 + a^2 tan^2(ωt)}}} \\ \mbox{And the normal component is } \\ v \dot{\vec{e_T}} = \ddot{\vec{r}} - \dot{v} \vec{e_T} \\ = - ω^2 \binom{acos(ωt) + a sin(ωt) \sqrt{\frac{a^2 + b^2 tan^2(ωt)}{b^2 + a^2 tan^2(ωt)}}}{b sin (ωt) + b cos(ωt) \sqrt{\frac{a^2 + b^2 tan^2(ωt)}{b^2 + a^2 tan^2(ωt)}}}$$

It still looks very strange, what do you say? :/

Julien.

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@PeroK Thanks for your answer. I think that's what I've used, isn't it? I corrected the sign error but it doesn't look much better:

$$v = \sqrt{a^2 ω^2 sin^2(ωt) + b^2ω^2 cos^2(ωt)} = ω \sqrt{a^2 sin^2 (ωt) + b^2 cos^2 (ωt)} \\ \vec{e_T} = \frac{\vec{v}}{|\vec{v}|} = \binom{\frac{-a sin(ωt)}{\sqrt{a^2 sin^2(ωt) + b^2 cos^2 (ωt)}}}{\frac{b cos(ωt)}{\sqrt{a^2 sin^2(ωt) + b^2 cos^2 (ωt)}}} \\ \ddot{\vec{r}} = \dot{v} \vec{e_T} + v \dot{\vec{e_T}} \\ \implies \mbox{The tangential component of the acceleration is:} \\ \dot{v} \vec{e_T} = \binom{aω^2 sin(ωt) \sqrt{\frac{a^2 + b^2 tan^2(ωt)}{b^2 + a^2 tan^2(ωt)}}}{b ω^2 cos(ωt) \sqrt{\frac{a^2 + b^2 tan^2(ωt)}{b^2 + a^2 tan^2(ωt)}}} \\ \mbox{And the normal component is } \\ v \dot{\vec{e_T}} = \ddot{\vec{r}} - \dot{v} \vec{e_T} \\ = - ω^2 \binom{acos(ωt) + a sin(ωt) \sqrt{\frac{a^2 + b^2 tan^2(ωt)}{b^2 + a^2 tan^2(ωt)}}}{b sin (ωt) + b cos(ωt) \sqrt{\frac{a^2 + b^2 tan^2(ωt)}{b^2 + a^2 tan^2(ωt)}}}$$

It still looks very strange, what do you say? :/

Julien.

I would have assumed that you only need to calculate the components of acceleration, not the unit vectors themselves. Also, if ##a=b## then you have circular motion hence ##a_T = 0## and ##a_N = aw^2##, which is something you should do to check your answers.

I agree with your answer for ##v##. After that, you've lost me.

The point of what I gave you is that you can use ##\vec{v}.\vec{a}## and ##\vec{v} \times \vec{a}## to get the components. ##\vec{v} \times \vec{a}## in particular simplifies.

PS Even if you do calculate the unit vectors, then I would keep the components and the unit vectors themselves separate. That's the point of the exercise: that you have known components in the direction of each unit vector.

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JulienB

$$|\vec{a} \times \vec{v}| = -abω^3 cos^2(ωt) - a^2ω^3sin^2(ωt) = -aω^3 (b cos^2 (ωt) - a sin^2 (ωt)) \\ \implies a_N = \frac{-aω^2 (b cos^2 (ωt) - a sin^2 (ωt))}{\sqrt{a^2 sin^2(ωt) + b^2 cos^2(ωt)}}$$

I can probably simplify further, but I wanted to ask you if that's correct first. I have never used this formula before, where does it come from?

Thanks a lot.

Julien.

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$$|\vec{a} \times \vec{v}| = -abω^3 cos^2(ωt) - a^2ω^3sin^2(ωt) = -aω^3 (b cos^2 (ωt) - a sin^2 (ωt)) \\ \implies a_N = \frac{-aω^2 (b cos^2 (ωt) - a sin^2 (ωt))}{\sqrt{a^2 sin^2(ωt) + b^2 cos^2(ωt)}}$$

I can probably simplify further, but I wanted to ask you if that's correct first. I have never used this formula before, where does it come from?

Thanks a lot.

Julien.

You've got two mistakes in there. Both terms in the cross product should be ##ab##. And then you factored out the negative incorrectly.

Re this formula, I derived it in post #6. In this case, it's easier to work with general vectors than with the complicated specific vectors you have.

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##|\vec{a} \times \vec{v}| = a_N v \ \ ## hence ## \ \ a_N = \frac{|\vec{a} \times \vec{v}|}{v}##

##\vec{a} \times \vec{v} ## is a vector perpendicular to both ##\vec v## and ##\vec a##. In this problem, it is the normal of the x,y plane, the plane of the ellipse.
You might have meant ##\vec v \times [\vec a\times \vec v]= \vec {a_N} v^2##.

JulienB
I gave it another go, for the tangential acceleration this time. I detailed it more so that you can spot easier where my mistakes are...

$$\dot{\vec{v}} = |\dot{\vec{v}}| \vec{e_T} + |\vec{v}| \dot{\vec{e_T}} \\ \mbox{The tangential acceleration is the first element of the sum:} \\ \implies a_T = \frac{\vec{v} \cdot \dot{\vec{v}}}{|\vec{v}|} \\ = \frac{1}{ω \sqrt{a^2 sin^2 (ωt) + b^2 cos^2(ωt)}} \binom{-aω sin(ωt)}{bω cos(ωt)} \binom{-aω^2 cos(ωt)}{-b ω^2 sin(ωt)} \\ = \frac{ω^2 cos(ωt) sin(ωt)}{\sqrt{a^2 sin^2(ωt) + b^2 cos^2 (ωt)}} \binom{a^2}{-b^2}$$

Am I still doing something wrong?? :( Shouldn't it be the magnitude of the acceleration in my equation for aT, like it says on Wikipedia?
https://en.m.wikipedia.org/wiki/Acceleration#Tangential_and_centripetal_acceleration

Julien.

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##\vec{a} \times \vec{v} ## is a vector perpendicular to both ##\vec v## and ##\vec a##. In this problem, it is the normal of the x,y plane, the plane of the ellipse.
You might have meant ##\vec v \times [\vec a\times \vec v]= \vec {a_N} v^2##.

##\vec{a} = a_T \vec{e_T} + a_N \vec{e_N}##

And take the cross product with ##\vec{v}##, then as ##\vec{e_T}## is parallel to ##\vec{v}## and ##\vec{e_N}## is orthogonal to ##\vec{v}##, this gives:

##\frac{|\vec{v} \times \vec{a}|}{v} = a_N## (as ##a_N## is always positive).

There's no need to take the cross product twice as you suggest. In any case, I derived all this in post #6.

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I gave it another go, for the tangential acceleration this time. I detailed it more so that you can spot easier where my mistakes are...

$$\dot{\vec{v}} = |\dot{\vec{v}}| \vec{e_T} + |\vec{v}| \dot{\vec{e_T}} \\ \mbox{The tangential acceleration is the first element of the sum:} \\ \implies a_T = \frac{\vec{v} \cdot \dot{\vec{v}}}{|\vec{v}|} \\ = \frac{1}{ω \sqrt{a^2 sin^2 (ωt) + b^2 cos^2(ωt)}} \binom{-aω sin(ωt)}{bω cos(ωt)} \binom{-aω^2 cos(ωt)}{-b ω^2 sin(ωt)} \\ = \frac{ω^2 cos(ωt) sin(ωt)}{\sqrt{a^2 sin^2(ωt) + b^2 cos^2 (ωt)}} \binom{a^2}{-b^2}$$

Am I still doing something wrong?? :( Shouldn't it be the magnitude of the acceleration in my equation for aT, like it says on Wikipedia?
https://en.m.wikipedia.org/wiki/Acceleration#Tangential_and_centripetal_acceleration

Julien.

That's correct, except you failed to take the dot product! It should be the scalar ##a^2 - b^2## instead of a vector that comes out of the dot product.

Also, now if you try ##a=b## you get ##a_T = 0## as it should be.

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##\vec{a} = a_T \vec{e_T} + a_N \vec{e_N}##

And take the cross product with ##\vec{v}##, then as ##\vec{e_T}## is parallel to ##\vec{v}## and ##\vec{e_N}## is orthogonal to ##\vec{v}##, this gives:
a vector perpendicular to the plane of motion.
##\frac{|\vec{v} \times \vec{a}|}{v} = a_N## (as ##a_N## is always positive).

Somehow we consider the centripetal acceleration negative.
In case of motion in plane, you can get the magnitude of the normal acceleration with this method. In 3D motion, ##\vec a\times \vec v ## gives the direction of the binormal vector, and you get the normal acceleration from the triple vector product, or by subtracting the tangential acceleration from the acceleration vector. ## \vec {a_N}=\frac{1}{v^2}[\vec v \times [\vec a\times \vec v]]=\vec a - \vec v \frac {\vec a \cdot \vec v}{v^2} ## .

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a vector perpendicular to the plane of motion.

You mean the centripetal acceleration is positive?
In case of motion in plane, you can get the magnitude of the normal component of acceleration with this method. In 3D motion, ##\vec a\times \vec v ## gives the direction of the binormal vector, and you get the normal acceleration from the triple vector product, or by subtracting the tangential acceleration from the acceleration vector. ## \vec {a_N}=\frac{1}{v^2}[\vec v \times [\vec a\times \vec v]]=\vec a - \vec v \frac {\vec a \cdot \vec v}{v^2} ## .

Nevertheless,

##a_N = \frac{|\vec{v} \times \vec{a}|}{v}## (for 2D and 3D motion)

We're just trying to get the magnitude of the components to begin with. That seems like a good first step for the OP.

JulienB
@PeroK @ehild Thanks for your answers. I guess I got confused with the vector operations :( Here is where I stand now:

$$a_T = \frac{ω^2 \cos{(ωt)} \sin{(ωt)}}{\sqrt{a^2 \sin^2{(ωt)} + b^2 \cos^2{(ωt)}}} (a^2 - b^2) \\ \implies |\vec{a_N}| = |\vec{a} - \vec{a_T}| \\ = |\dot{\vec{v}} - \vec{v} \frac{\vec{v} \cdot \vec{a}}{|\vec{v}|^2}| \\ = |\binom{-aω^2 \cos{(ωt)}}{-bω^2 \sin{(ωt)}} - \frac{ω^2 \cos{(ωt)} \sin{(ωt)}}{ω \sqrt{a^2 \sin^2{(ωt)} + b^2 \cos^2{(ωt)}}} (a^2 - b^2) \binom{-aω \sin{(ωt)}}{bω \cos{(ωt)}}| \\ = \frac{ω^2 \cos{(ωt)} \sin{(ωt)}}{a^2 \sin^2{(ωt)} + b^2 \cos^2{(ωt)}} (a^2 - b^2) |\binom{a(\sin{(ωt)} - \cos{(ωt)}}{-b(\cos{(ωt)} + \sin{(ωt)})}| \\ = \frac{ω^2 cos{(ωt)} sin{(ωt)}}{a^2 \sin{(ωt)} + b^2 \cos^2{(ωt)}} (a^2 - b^2) \sqrt{a^2(\sin{(ωt)} - \cos{(ωt)})^2 + b^2 (cos{(ωt)} + sin{(ωt)})^2}$$

That's crazy, there must be a mistake again somewhere... :( Using the other method with the cross product:

$$|\vec{v} \times \vec{a}| = |\binom{-aω \sin{(ωt)}}{bω \cos{(ωt)}} \times \binom{-aω^2 \cos{(ωt)}}{-bω^2 \sin{(ωt)}}| \\ = abω^3 (\cos^2{(ωt)} + \sin^2{(ωt)}) \\ = abω^3 \\ \implies a_N = \frac{|\vec{v} \times \vec{a}|}{|\vec{v}|} \\ = \frac{abω^2}{\sqrt{a^2 sin^2{(ωt)} + b^2 cos^2{(ωt)}}}$$

That looks nice, but is it correct? Unless there is a mistake or simplification I don't see using the first method, one is right and the other is wrong!

Julien.

JulienB
Nevermind I factored wrongly the first method, I'm doing it again...

JulienB
Here is what here is what I get using the 1st method now. Still not looking good, I've attached it because it's really hard to type it with LaTex...

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@PeroK @ehild Thanks for your answers. I guess I got confused with the vector operations :( Here is where I stand now:

$$a_T = \frac{ω^2 \cos{(ωt)} \sin{(ωt)}}{\sqrt{a^2 \sin^2{(ωt)} + b^2 \cos^2{(ωt)}}} (a^2 - b^2) \\ \implies |\vec{a_N}| = |\vec{a} - \vec{a_T}| \\ = |\dot{\vec{v}} - \vec{v} \frac{\vec{v} \cdot \vec{a}}{|\vec{v}|^2}| \\ = |\binom{-aω^2 \cos{(ωt)}}{-bω^2 \sin{(ωt)}} - \frac{ω^2 \cos{(ωt)} \sin{(ωt)}}{ω \sqrt{a^2 \sin^2{(ωt)} + b^2 \cos^2{(ωt)}}} (a^2 - b^2) \binom{-aω \sin{(ωt)}}{bω \cos{(ωt)}}| \\ = \frac{ω^2 \cos{(ωt)} \sin{(ωt)}}{a^2 \sin^2{(ωt)} + b^2 \cos^2{(ωt)}} (a^2 - b^2) |\binom{a(\sin{(ωt)} - \cos{(ωt)}}{-b(\cos{(ωt)} + \sin{(ωt)})}| \\ = \frac{ω^2 cos{(ωt)} sin{(ωt)}}{a^2 \sin{(ωt)} + b^2 \cos^2{(ωt)}} (a^2 - b^2) \sqrt{a^2(\sin{(ωt)} - \cos{(ωt)})^2 + b^2 (cos{(ωt)} + sin{(ωt)})^2}$$

That's crazy, there must be a mistake again somewhere... :( Using the other method with the cross product:

$$|\vec{v} \times \vec{a}| = |\binom{-aω \sin{(ωt)}}{bω \cos{(ωt)}} \times \binom{-aω^2 \cos{(ωt)}}{-bω^2 \sin{(ωt)}}| \\ = abω^3 (\cos^2{(ωt)} + \sin^2{(ωt)}) \\ = abω^3 \\ \implies a_N = \frac{|\vec{v} \times \vec{a}|}{|\vec{v}|} \\ = \frac{abω^2}{\sqrt{a^2 sin^2{(ωt)} + b^2 cos^2{(ωt)}}}$$

That looks nice, but is it correct? Unless there is a mistake or simplification I don't see using the first method, one is right and the other is wrong!

Julien.

That second answer is correct: ##|\vec{v} \times \vec{a}| = abω^3##

If you want the full vector form: ##a_N \vec{e_N} = \frac{\vec{v} \times (\vec{a} \times \vec{v})}{v^2}##

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No need to do that terrible calculation if you need only the magnitude of aN. Just use
$$=\vec a{_N}= \vec a - \vec{v} \frac{\vec{v} \cdot \vec{a}}{|\vec{v}|^2}$$

You need the magnitude. Take the product with itself and take the square root. You get
$$a_N^2= \left( \vec a - \vec{v} \frac{\vec{v} \cdot \vec{a}}{|\vec{v}|^2}\right)^2 =\frac{( \vec a )^2 (\vec v)^2 - (\vec a \cdot \vec v)^2}{v^2}$$

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JulienB
JulienB

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Yes, I forgot the square. I corrected it. Do you understand it now?

JulienB
JulienB
The calculation is still horrible I find :/ I'm not done, but I'd be surprised to get the same as with the other method with the cross product. May I ask again where that cross product comes from @PeroK ? Thanks a lot both of you.

JulienB
I mean, that looks really bad (see attached picture).

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I mean, that looks really bad (see attached picture).
You don't need to use the vectors in the final formula. You know that ##(\vec a )^2=|\vec a|^2=a^2##. Determine the squares a2 and v2. You know the product ##(\vec a \cdot \vec v)## already. Just substitute into $$a_N^2= \frac{ a^2 v^2 - (\vec a \cdot \vec v)^2}{v^2}$$
Lot of things cancel.
By the way, you left out the square in the second formula, second term in the numerator. It should be ##(\vec a \cdot \vec v)^2##

JulienB
JulienB
@ehild Sorry but I don't see what cancels, maybe I have the wrong terms for a and v? I attached my equation.

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@ehild Sorry but I don't see what cancels, maybe I have the wrong terms for a and v? I attached my equation.
It is ω6 also in the second term. There is ω2 in a, ω in v, and av is squared. And you simplified by ω2 twice in the first term. Factor out the ω-s and simplify after.

JulienB
JulienB
@ehild I corrected that but the equation still gets bigger and bigger without anything major canceling out really :/

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@ehild I corrected that but the equation still gets bigger and bigger without anything major canceling out really :/
Why did you divide the first term with the denominator? Just do the multiplication, and the square in the second term.

JulienB
JulienB
@ehild Still no canceling out... I really don't see it. :/ (I wrote only the top line in the attached file because it's so long)

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@ehild Still no canceling out... I really don't see it. :/ (I wrote only the top line in the attached file because it's so long)
You forgot to square sin and cos in v^2

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JulienB
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@JulienB
##(\vec v)^2=w^2(a^2\sin^2(wt)+b^2\cos^2(wt))##
##(\vec a)^2=w^4(a^2\cos^2(wt)+b^2\sin^2(wt))##
What is their product?

JulienB
JulienB
@ehild aaaah it worked out now! I get the same result as with the other method! Thanks a lot, that was crazy! ;)

Julien.

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@ehild aaaah it worked out now! I get the same result as with the other method! Thanks a lot, that was crazy! ;)

Julien.
Well done. I must sleep now but tomorrow I show you how the formula you used is connected to vector products.