Find Normal and Tangential components of accelaration

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Homework Help Overview

The problem involves analyzing the motion of a particle described by a polar equation in the plane, specifically focusing on finding the normal and tangential components of acceleration and the curvature of the path at any given time t.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the derivation of the normal and tangential components of acceleration, questioning the equations used and the definitions of variables like k and θ. There are attempts to clarify the relationships between the tangent vector, normal vector, and their derivatives.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the equations and clarifying misunderstandings about derivatives and vector relationships. Some guidance has been offered regarding the correct forms of the tangent and normal vectors, but no consensus has been reached on the final approach to the problem.

Contextual Notes

Participants express confusion over the correct application of derivatives and the definitions of certain variables, indicating a need for clearer understanding of the underlying concepts. There are also references to mistakes made in previous calculations, highlighting the iterative nature of the problem-solving process.

RyanH42
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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
 
Last edited:
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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.
 
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First of all, did you find the tangent and normal vector to the acceleration?(question a)
 
  • #10
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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  • #11
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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  • #12
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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  • #13
Your T looks wrong. What's dR/dt ? and what's |dR/dt| ?
 
  • #14
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.
 
  • #15
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
 
  • #16
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.
 
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  • #17
RyanH42 said:
dR/dt=i+2j
this is wrong, what's the derivative \frac{d}{dt} (t^2+1)?
 
  • #18
I made a huge mistake I must be an idiot so so soory
 
  • #19
RyanH42 said:
I made a huge mistake I must be an idiot so so soory

so?
 
  • #20
dR/dt=i+4j If I divide the length which its √17 I find dT/dt=i/√17+2/√17j
 
  • #21
Still you are taking the derivative wrong... \frac{d}{dt} (t^2+1) \ne 4
 
  • #22
4t
 
  • #23
RyanH42 said:
4t
close but no...
 
  • #24
what ??
 
  • #25
2t
 
  • #26
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
 
  • #27
Try practicing some derivatives...
 
  • #28
No.I know derivatives really believe me.Just I am looking computer like 3 hours maybe 4 and my mind is melt
 
  • #29
I will going to delete these one word posts.I can't do that
 
  • #30
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
 

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