Curvilinear Motion n/t coordinates

by aolse9
Tags: curvilinear motion, mechanics, n/t coordinates
aolse9 is offline
Aug26-09, 02:44 AM
P: 23
1. The problem statement, all variables and given/known data
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, [tex]\dot{x}[/tex] = 7ms-1 and [tex]\ddot{x}[/tex] = 8 ms-2?

2. Relevant equations
V = Vxi + Vyj
Vx = [tex]\dot{x}[/tex]
Vy = [tex]\dot{y}[/tex]

a = axi + ayj
ax = [tex]\ddot{x}[/tex]
ay = [tex]\ddot{y}[/tex]

3. 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.

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aolse9 is offline
Aug26-09, 06:47 PM
P: 23
Anyone have any ideas?
gabbagabbahey is offline
Aug26-09, 08:11 PM
HW Helper
gabbagabbahey's Avatar
P: 5,004
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 [itex]\theta[/itex] 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; [itex]\textbf{T}=\cos\theta\textbf{i}+\sin\theta\textbf{j}[/itex] and [itex]\textbf{N}=\cos(\frac{\pi}{2}-\theta)\textbf{i}-\sin(\frac{\pi}{2}-\theta)\textbf{j}=\sin\theta\textbf{i}-\cos\theta\textbf{j}[/itex]...The advantage of expressing the unit normal like this is that once you calculate [itex]\textbf{a}=\ddot{x}\textbf{i}+\ddot{y}\textbf{j}[/itex], all you need to do to find its normal component is take the dot product [itex]a_N=\textbf{a}\cdot\textbf{N}[/itex].

aolse9 is offline
Aug27-09, 12:52 AM
P: 23

Curvilinear Motion n/t coordinates

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.

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