Centripetal Acceleration Quick Question

AI Thread Summary
The maximum speed for an automobile on a bridge with a radius of curvature of 41 m is calculated to be 20.06 m/s, derived from the centripetal acceleration formula. The discussion highlights confusion regarding the normal force being zero at the point of leaving the bridge, clarifying that this occurs when the car is no longer in contact with the bridge. It is noted that the limiting speed applies specifically at the apex of the bridge, and speeds must be lower at other points along the arc. To accurately determine the speed throughout the entire bridge, additional information about the bridge's length is necessary. Understanding these dynamics is crucial for ensuring safety while navigating curved roadways.
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Homework Statement



A bridge over a small river has a roadway which is in the shape of an arch having radius of curvature of 41 m. What is the maximum speed at which an automobile can travel across the bridge without leaving the bridge?

Homework Equations



Fc = (mv^2)/r

Fg= mg

The Attempt at a Solution



Fnet = ma

Fc = Fg - Fn

(mv^2) / r = mg - fn

(v^2) / r = g

root ( 41 * 9.81) = 20.06 m/s

I know that this is the right answer, but I don't understand why the normal force is zero. Isn't the bridge exerting a force equal to that of gravity on the car so that it doesn't go through the bridge?
 
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The point at which the automobile leaves the bridge is when it's no longer in contact with the bridge i.e. there is no normal force. There's no force preventing it going through the bridge because it isn't in contact with it.
 
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Speedking96 said:

Homework Statement



A bridge over a small river has a roadway which is in the shape of an arch having radius of curvature of 41 m. What is the maximum speed at which an automobile can travel across the bridge without leaving the bridge?

Homework Equations



Fc = (mv^2)/r

Fg= mg

The Attempt at a Solution



Fnet = ma

Fc = Fg - Fn

(mv^2) / r = mg - fn

(v^2) / r = g

root ( 41 * 9.81) = 20.06 m/s

I know that this is the right answer, but I don't understand why the normal force is zero. Isn't the bridge exerting a force equal to that of gravity on the car so that it doesn't go through the bridge?
That's only if the bridge is flat. Did you ever feel like, when you were going over a bump in the road, you were almost flying out of your seat. Part of the gravitational force is used up just trying to hold you down in the curved trajectory.
 
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Speedking96 said:
(v^2) / r = g

I know that this is the right answer,
Actually, it isn't. That's the limiting speed for it to stay in contact at the apex of the bridge, but if it's a constant arc then the speed at start and end of the arc will need to be lower. To find out what that is you'd need to know the length of the bridge.
 
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