Centripetal Acceleration of rocket car

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Homework Statement


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].

Homework Equations

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...
 
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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.
 
You might get neater expressions for the first and second derivatives if you use implicit differentiation.
 
Simon Bridge said:
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.
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...
 
gneill said:
You might get neater expressions for the first and second derivatives if you use implicit differentiation.
I hadn't thought about that but I'll certainly give it a shot, thanks :D
 
Simon Bridge said:
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.
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..
 
Simon Bridge said:
I can't tell, because you didn't answer the questions.
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?