Cloverleaf highway interchange - determine car acceleration

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Alexanddros81
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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

Diagram of a looping curved road with a car labeled A at the upper right moving with speed v0 along the curve; radius marked R = b, angle 45 degrees at the lower left, and an x axis along the bottom. Fig. P13.34.


Homework Equations

The Attempt at a Solution



Handwritten solution to a polar-coordinates problem: R squared = b squared sin 2θ, differentiated with respect to time, leading to R-dot = -2b and acceleration components a_R = -2b - b·θ-dot squared and a_θ = 0.

[/B]
Is this wrong?
 
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:

Handwritten polar-coordinate derivation for problem 13.34 with solution 3u0^2/b at 45 degrees, deriving radial and transverse velocity and acceleration terms for a particle with r = b sqrt(sin 2θ).


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##.
 
Handwritten derivation of d/dt of R = sqrt(b^2 sin(2θ)) using the chain rule, power rule and differentiation of a sine function, ending with dR/dt = |b| cos(2θ) / sqrt(sin(2θ)) · θ-dot.

Handwritten derivation: v_R = Ṙ = 0, v_θ = Rθ̇, so v_θ = v_0 and bθ̇ = v_0, giving θ̇ = v_0/b; then v_0² = v_θ² + v_R² = Ṙ² + R²θ̇², differentiated with respect to time to give 0 = 2ṘR̈ + 2Rṙθ̇² + 2R²θ̇θ̈, leading to θ̈ = 0.

Handwritten derivation of radial acceleration in polar coordinates, with R double-dot minus R theta-dot squared, reduced step by step to a_R = -3 v0^2/b.


Can you check this?
 
Attachments
  • Handwritten derivation of d/dt of R = sqrt(b^2 sin(2θ)) using the chain rule, power rule and differentiation of a sine function, ending with dR/dt = |b| cos(2θ) / sqrt(sin(2θ)) · θ-dot.
    Pytels_Dynamics088.jpg
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  • Handwritten derivation of radial acceleration in polar coordinates, with R double-dot minus R theta-dot squared, reduced step by step to a_R = -3 v0^2/b.
    Pytels_Dynamics089.jpg
    31.9 KB · Views: 621
  • Handwritten derivation: v_R = Ṙ = 0, v_θ = Rθ̇, so v_θ = v_0 and bθ̇ = v_0, giving θ̇ = v_0/b; then v_0² = v_θ² + v_R² = Ṙ² + R²θ̇², differentiated with respect to time to give 0 = 2ṘR̈ + 2Rṙθ̇² + 2R²θ̇θ̈, leading to θ̈ = 0.
    Pytels_Dynamics093.jpg
    18.7 KB · Views: 678