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anonymous

  • one year ago

In the figure, a stuntman drives a car (without negative lift) over the top of a hill, the cross section of which can be approximated by a circle of radius R = 262 m. What is the greatest speed at which he can drive without the car leaving the road at the top of the hill?

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    is FBD necessary?

  3. anonymous
    • one year ago
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    I just need the FBD thank you!!

  4. IrishBoy123
    • one year ago
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    |dw:1442581163215:dw|

  5. anonymous
    • one year ago
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    I got 50.67 answer but I feel so discontent to my work. It is like Ineed to know something

  6. anonymous
    • one year ago
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    hmm okay

  7. anonymous
    • one year ago
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    my prof just gave a formula of mg=mv^2/r but I want to know where did he derive it

  8. anonymous
    • one year ago
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    I know mg is always down sooo mv^2r <- kinda confuse

  9. IrishBoy123
    • one year ago
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    just imagine you are the car or in the car

  10. IrishBoy123
    • one year ago
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    for derivation:

  11. IrishBoy123
    • one year ago
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    |dw:1442656237102:dw| for a particle accelerating in a circle, radius r \[\vec s = r < \cos \theta, \sin \theta>\] \[\vec { \dot s } = \vec v = r<-\sin \theta, \cos \theta> \dot \theta\] \[\vec a = \vec{ \dot v} = r<-cos \theta, -\sin \theta>\dot \theta ^2 + r<-\sin \theta, \cos \theta> \ddot \theta\] \[= r<-\cos \theta \, \dot \theta ^2 - \sin \theta \, \ddot \theta, -\sin \theta \, \dot \theta ^2+ \cos \theta \, \ddot \theta>\] projecting that along the radial vector \(\vec r = <\cos \theta, \sin \theta>\) \[a_r = r<-\cos \theta \, \dot \theta ^2 - \sin \theta \, \ddot \theta, -\sin \theta \, \dot \theta ^2+ \cos \theta \, \ddot \theta> \bullet <\cos \theta, \sin \theta>\] \[= -r \cos^2 \theta \, \dot \theta ^2 - r \sin \theta \cos \theta \ddot \theta - r \sin^2 \theta \, \dot \theta^2 + r \sin \theta \cos \theta \, \ddot \theta\] \[= - r \dot \theta ^2 = -\omega^2 r \] the particle is accelerating inwardly with \(a = \omega^2 r = \frac{v^2}{r}\) the car going over the hump must do the same or else it will continue in a straight line along the tangent to the hump. gravity must provide the force to accelerate the car in a circle

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