Rotational Motion - Centripetal force

[Ch3m_]

Homework Statement

A hump backed bridge is in the form of a circular arc of radius 35m.
What is the greatest speed with which a car can cross the bridge without leaving the ground at its highest point?

Homework Equations

F = m v²/r = mrω²

The Attempt at a Solution

I've tried using the equation above but no speed, angular velocity or mass is given.
Am I missing something?

 

[Doc Al]

I've tried using the equation above but no speed, angular velocity or mass is given.

You're supposed to find the speed!

Am I missing something?

Do a force analysis of the car as it tops the bridge. Hint: What force tells you when the car is just about to leave contact with the ground?

 

[Ch3m_]

Do a force analysis of the car as it tops the bridge. Hint: What force tells you when the car is just about to leave contact with the ground?

The reaction force..? I still don't get it :(

 

[Doc Al]

The reaction force..?

Yes, the normal force that the bridge exerts on the car. What will its value be when the car is barely making contact with the road? What will ΣF be in that case?

 

[Ch3m_]

Yes, the normal force that the bridge exerts on the car. What will its value be when the car is barely making contact with the road? What will ΣF be in that case?

It must be greater than the weight of the car = mg? I don't know how to get an exact value though

 

[haruspex]

It must be greater than the weight of the car = mg? I don't know how to get an exact value though

You are making a false assumption. Start from scratch:
What are the forces acting on the car at the highest point?
If the car is moving at speed v, what is the acceleration of the car?
What equation does that give you?
If the car went a little bit too fast it would leave the ground (you are told). What would the reaction force be then?

 

[Ch3m_]

What are the forces acting on the car at the highest point?

You have the normal force acting upwards, the cars weight acting downwards and the centripetal force acting downwards?

If the car is moving at speed v, what is the acceleration of the car?

a = v²/r ?

If the car went a little bit too fast it would leave the ground (you are told). What would the reaction force be then?

Weight + centripetal force so mg + (mv²)/r ?

 

[haruspex]

You have the normal force acting upwards, the cars weight acting downwards and the centripetal force acting downwards?

No, centripetal force is not an applied force. It is that component of the resultant force which produces the centripetal acceleration.

a = v²/r ?
Weight - centripetal force so mg - (mv²)/r ?

Yes, that is the normal force whilst in contact with the ground. But if it just on the point of losing contact, what extra does that tell you about the normal force?

 

[Ch3m_]

Wait so is it v²=gr?

 

[haruspex]

Wait so is it v²=gr?

Yes.

 

[Ch3m_]

Thanks, that must mean the normal force = 0 when the car is on the point of losing contact yeah?

 

[haruspex]

Thanks, that must mean the normal force = 0 when the car is on the point of losing contact yeah?

Well, it is because the normal force is zero at that point.