Max Velocity for Banked Road Problem

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To find the maximum velocity of a car on a banked, frictionless circular track at an angle of 31 degrees, the centripetal acceleration must be considered. The equation (mv^2)/r = n*sin(theta) is used to relate the normal force to the centripetal force required for circular motion. The normal force, n, cannot simply be equated to mg*cos(theta) because the car experiences centripetal acceleration, which affects the forces acting on it. The net force equation perpendicular to the road shows that n - mg*cos(theta) equals the mass times the centripetal acceleration. Understanding this relationship is crucial for solving the problem correctly.
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Okay, so I am doing a problem involving a car driving on a banked, frictionless, circular track (theta=31degrees) and i am supposed to find the maximum velocity that the car can drive. I know that to find the velocity, i have to find the centripetal acceleration by saying that (mv^2)/r = nsin(theta). Then, I have to solve for n by saying that ncos(theta)=mg. However, I am confused... why can't n=mgcos(theta)? My understanding is that two forces are equal in magnitude if the object doesn't move in either direction. The car doesn't move into the road or out of the road... or does it?? please help! I have a test on monday.
 
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It is true that the car doesn't move into the road or out of the road. However, the component of acceleration into the road is not 0.

In other words... suppose you take the \Sigma{F} = ma equation perpendicular to the road. N - mgcos(theta) = ma... here a is not 0! this occurs because of the centripetal acceleration.
 

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