Uniform Circular Motion and the Force of a Car on a Bumpy Road

[nahanksh]

Homework Statement

A car, which weighs 1000 N, travels over a bumpy road with a constant speed. Gravity acts. The road at point B is in the shape of a circle with a radius of 100 meters.
http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/oldexams/exam1/sp08/fig21.gif

Which one of these statements correctly relates the force of the car on the road at points A (a valley) and B (the top of a hill)?

(a) The magnitude of the force of the car on the road is larger at point A than it is at point B.
(b) The magnitude of the force of the car on the road is larger at point B than it is at point A.
(c) You need to know the road radius at point A to answer this question.

Homework Equations

a = v^2/r^2 = w^2*r

The Attempt at a Solution

I thought at the point A, it's considered as the lowest point of circular path, while
the point B is considered as the highest point.

So, at A Weight is somewhat canceled with the tension(thinking about uniform circular path) but at B weight is added to the tension..

That's why i thought B has larger force than at point A...

What's wrong in my attempt..?

Please help me out here...

Thanks.

 

[rock.freak667]

At point B, the centripetal force equation is

F_c= mg-N_B

where N is the normal reaction. What is the equation at A?

 

[nahanksh]

Isn't it F= (N)a - mg at A?

But I'm still confused it seems to me that normal force is the same for both points...

It's not the same, is it?



Thanks for your reply.

 

[rock.freak667]

Isn't it F= (N)a - mg at A?

But I'm still confused it seems to me that normal force is the same for both points...

It's not the same, is it?



Thanks for your reply.

So at A, N_A= mv²/r +mg

and at B, N_B=mg-mv²/r

which one will be bigger?

 

[nahanksh]

Thanks a lot.
I've got it !