Calculate tangential and normal components of acceleration

[MarkSA]

Hello,

Q: Given a particule that moves with a position function: r(t) = <cos(2t),1,-sin(2t)>,
d) Calculate tangential and normal components of acceleration when t=pi/8
e) Is the particle speeding up or slowing down , and turning or going straight when t=pi/8. Why?

This is a long multipart problem, so I just posted two parts. These are the two parts that I don't have a clue how to tackle.
I can get the acceleration for this particle when pi/8 by taking the second derivative r''(t), but i'm not sure if that corresponds to either the tangential or normal, or how to distinguish between the two. I am also clueless about how to solve part e.

I'm thinking this has to do with the unit tangent vector, principle unit normal vector. I'm not sure how to relate those to this problem though.

 

[Deleted member 4993]

Hello,

Q: Given a particule that moves with a position function: r(t) = <cos(2t),1,-sin(2t)>,
d) Calculate tangential and normal components of acceleration when t=pi/8
e) Is the particle speeding up or slowing down , and turning or going straight when t=pi/8. Why?

Normal (radial) unit vector

\(\displaystyle e_r \, = \, \frac{1}{\sqrt{2}} \cdot [\cos 2t, \, 1, \, -\sin 2t]\)

If you take a cross product of this vector with acceleration vector - you'll get the tangential component of the acceleration.

ofcourse dot product will give you the normal component.


This is a long multipart problem, so I just posted two parts. These are the two parts that I don't have a clue how to tackle.
I can get the acceleration for this particle when pi/8 by taking the second derivative r''(t), but i'm not sure if that corresponds to either the tangential or normal, or how to distinguish between the two. I am also clueless about how to solve part e.

I'm thinking this has to do with the unit tangent vector, principle unit normal vector. I'm not sure how to relate those to this problem though.

 

[MarkSA]

\(\displaystyle e_r \, = \, \frac{1}{\sqrt{2}} \cdot [\cos 2t, \, 1, \, -\sin 2t]\)

Hi, how did you calculate this? The 1/sqrt(2)?
I'm looking at these on a review sheet that someone worked out, and the work is kind of scattered. But this is what they put:

V = |V(t)| = sqrt[4sin^2(2t) + 0 + 4cos^2(2t)] = 2
a[sub:17kf9kix]T[/sub:17kf9kix] = v' = 0
a[sub:17kf9kix]N[/sub:17kf9kix] = KV^2 = 1(2)^2 = 4

But I haven't been able to make sense of it. Not sure why they are finding the magnitude of the velocity. Also, K appears to be the curvature equation... I don't understand why they are multiplying this by V^2, to get the normal component of acceleration.

 

[pka]

MarkSA, this may be too late help with this problem, but file it away because it will make these easier.
\(\displaystyle a_T=\frac{{r' \cdot r''}}{{\left\|{r'}\right\|}},\,a_N=\frac{{\left\|{r'\times r''}\right\|}}{{\left\|{r'}\right\|}},\,\&\,\kappa(t)=\frac{{\left\|{r'\times{r'}'}\right\|}}{{\left\|{r'}\right\|^3}}\)

These make things easy because you can evaluate the derivatives at the point before beginning any of the calculations on the above.

 

[MarkSA]

Yes, that makes this much easier. Thanks. I wish they had put those formulas in my calculus book.