A car exits a freeway on a circularly-shaped ramp that is banked at an angle of q = 23.5°. The ramp has a radius of curvature R = 48.6 meters. The ramp is icy, so you can neglect the effects of friction. If the total mass of the car and driver is m = 1397 kilograms, what constant speed v must the driver maintain in order for the center of mass of the car to stay in the center of the ramp?
Note: Use gravity g = 9.81 m/s2.
Note: Use gravity g = 9.81 m/s2.
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You know that the centripetal acceleration must be equal to the component of the vehicle's normal force that is directed along the radius of curvature, because the only forces that act on the vehicle are the weight and normal force, and the weight has no component that lies along a horizontal radius. So the normal force must provide all of the centripetal acceleration.
That is to say: mv^2/r = mg * sin(q) * cos(q)
This is derived by finding the normal force as equal to the component of weight perpendicular to the ramp, N=mg*cos(q) and then the component of N that is parallel with the horizontal must be Nsin(q), or mg*sin(q)cos(q).
First of all, notice that the mass will cancel on both sides of the equation.
So now v^2 / r = g * sin(q) cos(q)
Plug in all of your values and you are left with
v^2 / (48.6m) = (9.81m/s^2) * sin(23.5) * cos(23.5)
Isolate v and simplify to get
v = sqrt( (48.6m) * (9.81m/s^2) * sin(23.5) * cos(23.5) )
v=13.2m/s
That is to say: mv^2/r = mg * sin(q) * cos(q)
This is derived by finding the normal force as equal to the component of weight perpendicular to the ramp, N=mg*cos(q) and then the component of N that is parallel with the horizontal must be Nsin(q), or mg*sin(q)cos(q).
First of all, notice that the mass will cancel on both sides of the equation.
So now v^2 / r = g * sin(q) cos(q)
Plug in all of your values and you are left with
v^2 / (48.6m) = (9.81m/s^2) * sin(23.5) * cos(23.5)
Isolate v and simplify to get
v = sqrt( (48.6m) * (9.81m/s^2) * sin(23.5) * cos(23.5) )
v=13.2m/s