Radial acceleration, car sircular path

[eventob]

Homework Statement

A car is traveling with constant speed over a hill and down a hill. The radius of the curve is the same. At the top of the hill, the driver experience no normal force from the ground. The mass of the driver is 70.0kg

a) calculate the value of the normalforce experienced by the driver at the bottom of the ground.

Homework Equations

A_c = (v^2/r)

The Attempt at a Solution

Drew a free-body-diagram of the forces acting on the car (modeled as a particle under uniform circular motion. Not sure how to approach this problem.

Thanks in advance. :)

 

[rock.freak667]

The Attempt at a Solution

Drew a free-body-diagram of the forces acting on the car (modeled as a particle under uniform circular motion. Not sure how to approach this problem.

Well at the bottom of the hill, what are the forces acting? What directions are they acting (towards or away from the center of the circle)?

 

[eventob]

At the bottom there is the force of gravity (70kg*9.80*(m/s^2), the centripetal acceleration towards the center of the circle, and the normal force also towards the enter of the circle. I think.

At the top the normal force is going in the opposite direction, away from the center of the circle? I've tried to put the data into Newtons second law, but I get to many unknown variables I think.

Thanks. :)

 

[rock.freak667]

The weight acts downwards and the normal reaction is opposite, so towards the center of the circle.

What is the resultant force then ? In terms of N and mg.

 

[eventob]

sigma F_y= n-mg = m*(v^2/r)

Which means that the normal force acting on the driver at the bottom of the hill is:

n=m*(v^2/r)+mg
=m[(v^2/r)+g]

But, in order to calculate the speed, can I make a similar equation for the top of the hill and substitute for v?
:)

 

[rock.freak667]

sigma F_y= n-mg = m*(v^2/r)

Which means that the normal force acting on the driver at the bottom of the hill is:

n=m*(v^2/r)+mg
=m[(v^2/r)+g]

But, in order to calculate the speed, can I make a similar equation for the top of the hill and substitute for v?
:)

Yes but at the top remember, if the normal reaction is not present, then centripetal force mv²/r is?

 

[eventob]

Sorry, but I still don't get it. At the top, the only force that acts on the object is the force of gravity? But if there is no normal force, what is preventing the car from falling through the ground?

Thanks for your time.

 

[rock.freak667]

Sorry, but I still don't get it. At the top, the only force that acts on the object is the force of gravity? But if there is no normal force, what is preventing the car from falling through the ground?

Thanks for your time.

Can't say. There should be a normal force, but I am just going with what the question is saying.