# Find Normal and Tangential components of accelaration

1. Jun 15, 2015

### RyanH42

1. The problem statement, all variables and given/known data
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 ?
2. Relevant equations
$\vec{a}=d^2s/dt^2\vec{T}+k\vec{N}d^2s/dt^2$
$k=dθ/ds$

3. The attempt at a solution
I couldnt find $ds/dt$ I tried but I stucked.
I have no idea to find k but I couldnt find s

Last edited: Jun 15, 2015
2. Jun 15, 2015

### 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 $k$?

3. Jun 15, 2015

### RyanH42

Yeah I remember equations false I checked again it was $d\vec{R}/ds$
I remembered that false too it will be $k=dθ/ds$

4. Jun 15, 2015

### 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

Last edited: Jun 15, 2015
5. Jun 15, 2015

### ChrisVer

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

Last edited: Jun 15, 2015
6. Jun 15, 2015

### RyanH42

you can look the graph

7. Jun 15, 2015

### 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 ?

8. Jun 15, 2015

### ChrisVer

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.

9. Jun 15, 2015

### ChrisVer

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

10. Jun 15, 2015

### ChrisVer

It is just because the normal vector is:

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

Last edited: Jun 15, 2015
11. Jun 15, 2015

### 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 $\vec{N}=\vec{j}$

Last edited: Jun 15, 2015
12. Jun 15, 2015

### RyanH42

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

Last edited: Jun 15, 2015
13. Jun 15, 2015

### ChrisVer

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

14. Jun 15, 2015

### ChrisVer

I am using the parameter of your curve.

15. Jun 15, 2015

### RyanH42

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

16. Jun 15, 2015

### 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.

17. Jun 15, 2015

### ChrisVer

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

18. Jun 15, 2015

### RyanH42

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

19. Jun 15, 2015

### ChrisVer

so?

20. Jun 15, 2015

### RyanH42

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