Cloverleaf highway interchange - determine car acceleration

AI Thread Summary
The discussion revolves around calculating the acceleration of a car traveling along the curved portion of a cloverleaf highway interchange defined by the equation R^2 = b^2 sin(2θ). Participants point out errors in the initial attempt, specifically the omission of terms involving the second derivative of θ (denoted as ##\ddot{\theta}##) and the square of the first derivative of θ (##\dot{\theta}^2##). Dimensional analysis is suggested to verify the correctness of the equations being used. The conversation emphasizes the need to properly account for derivatives when differentiating the expressions involving trigonometric functions. Accurate calculations are crucial for determining the car's acceleration at point A.
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Homework Statement


13.34 The curved portion of the cloverleaf highway interchange is defined by
##R^2=b^2sin2θ##, 0<=θ<=90deg. If a car travels along the curve at the constant speed v0,
determine its acceleration at A

Fig13_34.jpg


Homework Equations

The Attempt at a Solution



Pytels_Dynamics071.jpg

[/B]
Is this wrong?
 
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Check your expression for ##\ddot{R}##. There should be a ##\ddot{\theta}## in it.​
 
and also you're missing a ##\dot \theta^2##.
 
kuruman said:
Check your expression for ¨RR¨\ddot{R}. There should be a ¨θθ¨\ddot{\theta} in it.

Is the following correct:

Pytels_Dynamics072.jpg


Marc Rindermann said:
and also you're missing a ˙θ2θ˙2\dot \theta^2
I cannot see where I am missing the ##\dot θ^2##
 
first of all you can check whether your equations are correct if you do a dimensional analysis.
Look at ##\ddot R = 2b\dot\theta##

##[\ddot R] = ms^{-2}##
##[2b\dot\theta] = ms^{-1}##

So you see something is not quite right.

When you take the 2nd derivative of ##\dot R = b\cos(2\theta)(\sin(2\theta))^{-1/2}\dot\theta## you need to remember that the derivative of both ##\cos(2\theta)## and ##(\sin(2\theta))^{-1/2}## produces another ##\dot\theta##.
 
Pytels_Dynamics088.jpg

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Can you check this?
 

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