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!

Centripetal Acceleration of rocket car

  1. Feb 5, 2015 #1
    1. The problem statement, all variables and given/known data
    A rocket car is constrained to move on an elliptical track (semi-major axis [itex] a [/itex] and semi-minor axis [itex] b [/itex]). The car is moving at a constant speed [itex] v_0 [/itex]. Determine the acceleration of the car in [itex] \frac{m}{s^2} [/itex]. [itex] a = 4 \hspace{2 mm} km [/itex], [itex] b = 2 \hspace{2 mm} km [/itex], and [itex] v_0 = 360 \frac{km}{hr} [/itex].

    2. Relevant equations


    3. The attempt at a solution
    My thought for this problem was to try and find the value of 'R' using the definition of curvature: [itex] \kappa = \frac{1}{R} [/itex]. Using the definition of an ellipse: [itex] \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 [/itex], I solved for 'y' and ended up with: [itex] y = b \sqrt{1- \frac{x^2}{a^2}}[/itex]. Then, I used the definition of curvature [itex] \kappa = \frac{\frac{d^2y}{dx^2}}{(1+(\frac{dy}{dx})^2)^\frac{3}{2}} [/itex]. Then, I took the first and second derivative of my equation for an ellipse (I will not write them here because it gets very messy). Then, I substituted my second derivative into the equation for curvature. Finally, using the fact that [itex] \kappa = \frac{1}{R} [/itex], therefore [itex] R = \frac{1}{\kappa} [/itex]. So I substituted my equation for R (1 over kappa) into the equation for centripetal acceleration( [itex] a = \frac{v^2}{R} [/itex] and used the given [itex] v_0 = 360 \frac{km}{hr} [/itex]. However, as you can probably guess, this is extremely messy. Furthermore, I do not see how I am to get rid of the x and y in my equations (the answer is meant to be a numeric value). So I am certain I have done something wrong but I'm not sure what...
     
  2. jcsd
  3. Feb 5, 2015 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    What doe "R" represent? You have to say... is the idea to use the equation ##a=v^2/R##?
    Note: the acceleration of the car won't be a constant since it's rate of change in direction varies but it's speed doesn't.
     
  4. Feb 5, 2015 #3

    gneill

    User Avatar

    Staff: Mentor

    You might get neater expressions for the first and second derivatives if you use implicit differentiation.
     
  5. Feb 5, 2015 #4
    Sorry, I realize I wasn't very clear. The idea was to use the equation [itex] a = \frac{v^2}{R} [/itex].. I attempted to find R using the equation [itex] \kappa = \frac{1}{R} [/itex]. So I solved for R using the curvature equation and then plugged this value into the equation for centripetal acceleration but it is just so messy that I think I probably am not on the right track. I hope this clears things up...
     
  6. Feb 5, 2015 #5
    I hadn't thought about that but I'll certainly give it a shot, thanks :D
     
  7. Feb 6, 2015 #6
    So in that case I am basically at the solution, I just need to clean things up algebraically I think. I have this bad habit where if my solution looks really messy I just assume I've done something wrong..
     
  8. Feb 6, 2015 #7

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I can't tell, because you didn't answer the questions.
     
  9. Feb 6, 2015 #8
    Which questions are you referring to? R is meant to represent radius of curvature, is there something wrong with how I've approached the problem?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Centripetal Acceleration of rocket car
  1. Acceleration of Rocket (Replies: 1)

Loading...