# normal component of accelleration

• Sep 22nd 2010, 11:57 PM
heatly
normal component of accelleration
Iam having issues with this one iff anybody can help?

Given y^2=3x^3+8x
where x and y are in meters and y is positive,what is the normal component of the accelleration when x=2m,x=3m/s and x=7m/s^2.

I have so far subed in 2 and got the y coordinate but dont know what to do from here?
• Sep 23rd 2010, 05:37 AM
Ackbeet
Is this two-dimensional motion? If so, are y and x coordinates of the position? If so, what is the time parametrization of the curve? Your equation has no t's in it. Without knowing how the coordinates depend on t, you can't solve this problem.
• Sep 23rd 2010, 06:58 AM
heatly
Thanks for the reply,
Yes 2d
x,and y are coordinates I presume
This is exactly what the question says as I read it.
I had a browse around and found you have to implicitly differentiate the whole equa (d/dt).
dx/dt is given =3 and x= 2 so you can solve for dy/dt.
you now differentiate again and solve for d^2y/dx^2 because we are given d^2/dx^2 = 7 and x=2.

My aclculations say when x = 3m/s ,y = 10.4m/s
when x= 7m/s^2
y=27.14m/s^2
I then found accel tangent = 28.0327m/s^2

DONT KNOW HOW TO FIND THE ACCELL NORMAL BUT??
ANY IDEAS?
• Sep 23rd 2010, 08:07 AM
Ackbeet
Your notation is a bit off, I think. I think you mean to ask the question:

Given that $y^{2}=3x^{3}+8x,$ find the normal component of acceleration when $x(t_{0})=2\,\text{m},\;\dot{x}(t_{0})=3\,\text{m/s},$ and $\ddot{x}(t_{0})=7\,\text{(m/s)/s}.$

According to this web page, the normal component of acceleration is given by

$a_{N}=\dfrac{\|\mathbf{v}\times\mathbf{a}\|}{\|\ma thbf{v}\|}=\dfrac{\|\mathbf{v}\|\,\|\mathbf{a}\|\, \sin(\theta)}{\|\mathbf{v}\|}.$

Here the $\|\mathbf{v}\|$ notation means, for example, the length of vector $\mathbf{v}.$ Also, the angle $\theta$ is the smallest angle between $\mathbf{v}$ and $\mathbf{a}.$

Can you see your way forward from here? You can compute the magnitude of a cross product, even if you're only working in two dimensions, by the way. Just follow the formula on the right.
• Sep 23rd 2010, 11:55 PM
heatly
thanks,

I tried that but still no luck.
To get the resultant 'a' vector you need the At and An,and I dont have An.
• Sep 24th 2010, 01:13 AM
heatly
Well after 10 hrs on this Q I finally solved it,ans 2.323m/s^2
• Sep 24th 2010, 03:52 AM
Ackbeet
I would generally agree with your answer. I got 2.31968 m/s^2. You probably just have a little round-off error if you introduced numerical computations earlier than I (I used Mathematica and got symbolic exact answers until the very end.)

Good work!
• Sep 25th 2010, 04:35 AM
heatly
Thanks Ackbeet
• Sep 27th 2010, 03:50 AM
Ackbeet
You're welcome. Have a good one!
• Oct 3rd 2010, 01:56 PM
Techro
I have a very similar question. What is your second derivative? What is the equation you are deriving?
• Oct 4th 2010, 05:22 AM
heatly
Its lost in a pile of paperwork somewhere,I cant remember now.