Find Normal and Tangential components of accelaration

[RyanH42]

Homework Statement

The Particle moves in the plane in a such a way that its polar equation of motion is $\vec{R}=t\vec{i}+(t^2+1)\vec{j}$.
a)What are its normal and tangential components of accelaration any time t ?
b)Whats the curvature of path any time t ?

Homework Equations

$\vec{a}=d^2s/dt^2\vec{T}+k\vec{N}d^2s/dt^2$
$k=dθ/ds$

The Attempt at a Solution

I couldn't find $ds/dt$ I tried but I stucked.
I have no idea to find k but I couldn't find s

 

[ChrisVer]

One note to avoid confusion. The \frac{ds}{dt} you wrote is the magnitude of the velocity, whereas the \frac{d\vec{R}}{dt} is its vector.

Where did you get that equation for k?

 

[RyanH42]

One note to avoid confusion. The dsdt\frac{ds}{dt} you wrote is the magnitude of the velocity, whereas the dR⃗ dt\frac{d\vec{R}}{dt} is its vector.

Yeah I remember equations false I checked again it was $d\vec{R}/ds$

Where did you get that equation for kk?

I remembered that false too it will be $k=dθ/ds$

 

[RyanH42]

And Is this equation true every time ? $d\vec{T}/dθ=\vec{N}$ (N is not unit vector).I now its a stupid qustion but the derivation of it come to me not generally true

 

[ChrisVer]

Let me put it else... what does \theta stand for?

 

[RyanH42]

you can look the graph

 

[RyanH42]

I got the idea in 21:49 look the magnitude of $d\vec{T}/ds=dθ/ds=k$ Is that mean the magnitude of $\vec{N}$ equal $k$.I am confused one time he says $d\vec{T}/dθ=\vec{N}$(In 10:49) now he says $d\vec{T}/ds=\vec{N}$ Is it a special case which I mentioned ?

 

[ChrisVer]

Is that mean the magnitude of

N→\vec{N} equal k

The magnitude of \vec{N} is unity (it's a unit vector). k= \Big| \frac{d\vec{T}}{ds} \Big| is some number and so:
\vec{N} = \frac{d \vec{T}}{ds} \Big/ \Big| \frac{d\vec{T}}{ds} \Big| has magnitude equal to 1.

 

[ChrisVer]

First of all, did you find the tangent and normal vector to the acceleration?(question a)

 

[ChrisVer]

I am confused one time he says dT→/dθ=N→d\vec{T}/dθ=\vec{N}(In 10:49) now he says dT→/ds=N→d\vec{T}/ds=\vec{N} Is it a special case which I mentioned ?

It is just because the normal vector is:

\vec{N} \equiv \frac{d\vec{T}}{dt} \Big/ \Big|\frac{d \vec{T}}{dt} \Big|

 

[RyanH42]

Tangent vector is \vec{T} = \frac{d \vec{R}}{dt} \Big/ \Big| \frac{d\vec{R}}{dt} \Big| so Its $\vec{T}=1/√5\vec{i}+2/√5t\vec{j}$.Normal vector will be then $\vec{N}=\vec{j}$

 

[RyanH42]

It is just because the normal vector is:

\vec{N} \equiv \frac{d\vec{T}}{dt} \Big/ \Big|\frac{d \vec{T}}{dt} \Big|

beg me but I didnt understand first ds (in your #8 post) now we can use dt. Whatever we want we can use that equation I guess.

In previous post I did something wrong cause dot product of T and N must be zero and here its not

 

[ChrisVer]

Your T looks wrong. What's dR/dt ? and what's |dR/dt| ?

 

[ChrisVer]

beg me but I didnt understand first ds (in your #8 post) now we can use dt. Whatever we want we can use that equation I guess

I am using the parameter of your curve.

 

[RyanH42]

dR/dt=i+4j If I divide the length which its √17 I find dT/dt=i/√17+2/√17j I made a huge mistake I must be an idiot soort

 

[ChrisVer]

Either you use the parameter t/\lambda,\theta or whatever you call it, or the arc-length s it doesn't matter...
\frac{d \vec{T}}{ds} will point to the same direction as \frac{d \vec{T}}{dt}. That can be seen by rewritting:
\frac{d \vec{T}}{ds} = \frac{d \vec{T}}{dt} \Big/ \frac{ds}{dt} so you are only getting a multiplicative factor of (ds/dt)^{-1} between the vector \frac{d \vec{T}}{ds} and \frac{d \vec{T}}{dt}. You don't care about that factor as long as you normalize them.

 

[ChrisVer]

dR/dt=i+2j

this is wrong, what's the derivative \frac{d}{dt} (t^2+1)?

 

[RyanH42]

I made a huge mistake I must be an idiot so so soory

 

[ChrisVer]

I made a huge mistake I must be an idiot so so soory

so?

 

[RyanH42]

dR/dt=i+4j If I divide the length which its √17 I find dT/dt=i/√17+2/√17j

 

[ChrisVer]

Still you are taking the derivative wrong... \frac{d}{dt} (t^2+1) \ne 4

 

[RyanH42]

4t

 

[ChrisVer]

4t

close but no...

 

[RyanH42]

what ??

 

[RyanH42]

2t

 

[RyanH42]

how can I so dump ı have no idea.Actually I wite there 2j but it will be 2tj I missed the t and then I didnt read so I confused

 

[ChrisVer]

Try practicing some derivatives...

 

[RyanH42]

No.I know derivatives really believe me.Just I am looking computer like 3 hours maybe 4 and my mind is melt

 

[RyanH42]

I will going to delete these one word posts.I can't do that

 

[RyanH42]

You don't have to help me If you have other works . I will check this question again .Thank you so much.Its your desicion to help me now

 

[RyanH42]

I found the answer The question is not about $\vec{T}$ or $\vec{N}$.We are in wrong way the equation is $\vec{a}=d^2s/dt^2(\vec{T}+k\vec{N})$ here the components is $d^2s/dt^2$ for $\vec{T}$ and $d^2s/dt^2k$ for $\vec{N}$.

$ds/dt=√(1+(dy/dx)^2)$
$y=(t^2+1)\vec{j}$ → $dy =2t\vec{j}$,
$x=t\vec{i}$ →$dx=t\vec{j}$
$ds/dt=√(1+(2t/1)^2)$
$d^2s/dt^2=8t/2√(1+4t^2)=4t/√(1+4t^2)$
Which I checked I found the right answer.I will do the same thing for $\vec{N}$ but there's k. k=magnitude of $d\vec{T}/ds$. $d\vec{T}/ds=d\vec{T}/dt/(ds/dt)$ $d\vec{T}/dt=2j$ and $ds/dt=√(1+(2t/1)^2)$ so $d\vec{T}/ds=2j/√(1+(4t^2)$ and the magnitude is $2/√(1+(4t^2)$ but the answer is 2/(1+(4t^2)^3/2
What I am missing ?

 

[RyanH42]

I guess I mixed everything

 

[ChrisVer]

Well I found some mistakes...
First of all avoid writting x,y as vectors when they are not...
It would be correct to write x=t \Rightarrow dx = dt and y=t^2+1 \Rightarrow dy= 2 t dt

Also the reason you don't get the right answer is because in \frac{d \vec{T}}{ds} you didn't take the derivatives of the unit-tangent vector, just a tangent vector...you used \vec{T} = \hat{i} + 2t \hat{j} \Rightarrow |\vec{T}|= \sqrt{1+4t^2} \ne 1...
If you do the derivative with the unit-tangent vector correctly (I did it) you will get the result you ask for.

 

[RyanH42]

Well I found some mistakes...
First of all avoid writting x,yx,y as vectors when they are not...
It would be correct to write x=t⇒dx=dtx=t \Rightarrow dx = dt and y=t2+1⇒dy=2tdty=t^2+1 \Rightarrow dy= 2 t dt

Ok,thanks for this

"Also the reason you don't get the right answer is because in \frac{d \vec{T}}{ds} you didn't take the derivatives of the unit-tangent vector, just a tangent vector...you used \vec{T} = \hat{i} + 2t \hat{j} \Rightarrow |\vec{T}|= \sqrt{1+4t^2} \ne 1...
If you do the derivative with the unit-tangent vector correctly (I did it) you will get the result you ask for."
Ok.As you said I did it and I understand.I want to ask one question what's the realtionship between $\vec{T}$ and $k$and $\vec{N}$.I thought or see $d\vec{T}/dt=k\vec{N}$ ?

 

[ChrisVer]

\frac{d \vec{T}}{ds} = k \vec{N}

 

[RyanH42]

I understand it thanks the magnitude will give us k but in the vector form this.This vector calculus is really hard.Thanks again for help.