What Is the Normal Component of Acceleration for Curvilinear Motion?

[aolse9]

Homework Statement

Given, y² = 9x³ + 6x, where x and y are in metres and y is positive.
What is the normal component of the acceleration when x= 3m, \dot{x} = 7ms^-1 and \ddot{x} = 8 ms^-2?

Homework Equations

V = V_xi + V_yj
V_x = \dot{x}
V_y = \dot{y}

a = a_xi + a_yj
a_x = \ddot{x}
a_y = \ddot{y}

The Attempt at a Solution

I was fairly confident that my following attempt would yield the correct solution, however it did not. I still fail to see any problems with my attempt, so I thought I'd post it. Any feedback on my work would be greatly appreciated.

http://hdimage.org/images/irbxrbb7r7fgf42658at.jpg

 

[aolse9]

Anyone have any ideas?

 

[gabbagabbahey]

Hmmm... it's difficult to spot the error in your calculations, but you might have an easier time if you don't mess about with too many angles...

Everything up to and including your calculation of \theta looks correct to me. From that point, you should be able to basically look at diagram and read off the vector expression for the unit tangent and unit normal; \textbf{T}=\cos\theta\textbf{i}+\sin\theta\textbf{j} and \textbf{N}=\cos(\frac{\pi}{2}-\theta)\textbf{i}-\sin(\frac{\pi}{2}-\theta)\textbf{j}=\sin\theta\textbf{i}-\cos\theta\textbf{j}...The advantage of expressing the unit normal like this is that once you calculate \textbf{a}=\ddot{x}\textbf{i}+\ddot{y}\textbf{j}, all you need to do to find its normal component is take the dot product a_N=\textbf{a}\cdot\textbf{N}.

 

[aolse9]

Thanks for the reply. I have found a mistake in my calculation of a, more specifically the second order time derivative of y. I agree, when you start to work with these angles it can get a tad confusing, so I'll try your dot product suggestion. I should be on the right track now, I'll post back if I have any troubles, thanks for your help.