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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?
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just need the FBD thank you!!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442581163215:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 50.67 answer but I feel so discontent to my work. It is like Ineed to know something

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0my prof just gave a formula of mg=mv^2/r but I want to know where did he derive it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know mg is always down sooo mv^2r < kinda confuse

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1just imagine you are the car or in the car

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw: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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