# Find Normal and Tangential components of accelaration

• RyanH42
In summary, the Particle moves in a plane in a such a way that its polar equation of motion is ##\vec{R}=t\vec{i}+(t^2+1)\vec{j}##. a)Its normal and tangential components of accelaration are unknown, but it has a curvature of path. b)The magnitude of the velocity is equal to the magnitude of the normal vector, and the equation holds every time.
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

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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 $k$?

ChrisVer said:
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##
ChrisVer said:
Where did you get that equation for kk?
I remembered that false too it will be ##k=dθ/ds##

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

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

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you can look the graph

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 ?

RyanH42 said:
Is that mean the magnitude of
RyanH42 said:
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.

RyanH42
First of all, did you find the tangent and normal vector to the acceleration?(question a)

RyanH42 said:
I am confused one time he says dT→/=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:

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

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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}##

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ChrisVer said:
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

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

RyanH42 said:
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.

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

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.

RyanH42
RyanH42 said:
dR/dt=i+2j
this is wrong, what's the derivative $\frac{d}{dt} (t^2+1)$?

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

RyanH42 said:
I made a huge mistake I must be an idiot so so soory

so?

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

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

4t

RyanH42 said:
4t
close but no...

what ??

2t

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

Try practicing some derivatives...

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

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

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

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 ?

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I guess I mixed everything

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
ChrisVer said:
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}## ?

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

RyanH42
<h2>1. What is the difference between normal and tangential acceleration?</h2><p>The normal acceleration is the component of acceleration that is perpendicular to the velocity vector, while the tangential acceleration is the component that is parallel to the velocity vector. In other words, the normal acceleration changes the direction of the velocity, while the tangential acceleration changes the magnitude of the velocity.</p><h2>2. How do you find the normal and tangential components of acceleration?</h2><p>To find the normal and tangential components of acceleration, you need to first determine the velocity vector and the acceleration vector. Then, use vector decomposition techniques to find the components of acceleration that are perpendicular and parallel to the velocity vector.</p><h2>3. What is the formula for calculating the normal and tangential components of acceleration?</h2><p>The formula for calculating the normal acceleration is <strong>a<sub>n</sub> = |a| * cos(θ)</strong>, where |a| is the magnitude of the acceleration vector and θ is the angle between the acceleration and velocity vectors. The formula for calculating the tangential acceleration is <strong>a<sub>t</sub> = |a| * sin(θ)</strong>.</p><h2>4. Why is it important to find the normal and tangential components of acceleration?</h2><p>Knowing the normal and tangential components of acceleration allows us to understand how an object's velocity is changing. This information is crucial in many fields of science, such as physics and engineering, as it helps us predict and analyze the motion of objects.</p><h2>5. Can the normal and tangential components of acceleration be negative?</h2><p>Yes, the normal and tangential components of acceleration can be negative. A negative value for the normal acceleration means that the object is accelerating towards the center of the circular path, while a negative value for the tangential acceleration means that the object is slowing down in the direction of motion.</p>

## 1. What is the difference between normal and tangential acceleration?

The normal acceleration is the component of acceleration that is perpendicular to the velocity vector, while the tangential acceleration is the component that is parallel to the velocity vector. In other words, the normal acceleration changes the direction of the velocity, while the tangential acceleration changes the magnitude of the velocity.

## 2. How do you find the normal and tangential components of acceleration?

To find the normal and tangential components of acceleration, you need to first determine the velocity vector and the acceleration vector. Then, use vector decomposition techniques to find the components of acceleration that are perpendicular and parallel to the velocity vector.

## 3. What is the formula for calculating the normal and tangential components of acceleration?

The formula for calculating the normal acceleration is an = |a| * cos(θ), where |a| is the magnitude of the acceleration vector and θ is the angle between the acceleration and velocity vectors. The formula for calculating the tangential acceleration is at = |a| * sin(θ).

## 4. Why is it important to find the normal and tangential components of acceleration?

Knowing the normal and tangential components of acceleration allows us to understand how an object's velocity is changing. This information is crucial in many fields of science, such as physics and engineering, as it helps us predict and analyze the motion of objects.

## 5. Can the normal and tangential components of acceleration be negative?

Yes, the normal and tangential components of acceleration can be negative. A negative value for the normal acceleration means that the object is accelerating towards the center of the circular path, while a negative value for the tangential acceleration means that the object is slowing down in the direction of motion.

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