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