1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Curvilinear Motion n/t coordinates

  1. Aug 26, 2009 #1
    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.

    http://hdimage.org/images/irbxrbb7r7fgf42658at.jpg [Broken]
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 26, 2009 #2
    Anyone have any ideas?
  4. Aug 26, 2009 #3


    User Avatar
    Homework Helper
    Gold Member

    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].
  5. Aug 27, 2009 #4
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook