## 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?

1. anonymous

2. anonymous

is FBD necessary?

3. anonymous

I just need the FBD thank you!!

4. IrishBoy123

|dw:1442581163215:dw|

5. anonymous

I got 50.67 answer but I feel so discontent to my work. It is like Ineed to know something

6. anonymous

hmm okay

7. anonymous

my prof just gave a formula of mg=mv^2/r but I want to know where did he derive it

8. anonymous

I know mg is always down sooo mv^2r <- kinda confuse

9. IrishBoy123

just imagine you are the car or in the car

10. IrishBoy123

for derivation:

11. IrishBoy123

|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