Circular motion - Please help, I think the book's wrong?

AI Thread Summary
The discussion revolves around calculating the forces acting on a car of mass 400kg traveling over a bump with a radius of 10m at a speed of 6 m/s. The initial calculation for the net force exerted by the track was incorrect, leading to confusion about the correct answer. Clarification was provided that the net force should account for both the centripetal force and the normal force acting in opposite directions. After further calculations, it was determined that the correct force exerted by the track is approximately 2500N, considering the gravitational acceleration of 9.8 m/s². The issue was resolved, and the participant expressed gratitude for the assistance received.
TheKovac
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Circular motion - Please help, I think the book's wrong!?

Homework Statement


A car of mass 400kg travels over a bump of radius 10m on a track in a fun park.

a) What force is exerted by the track on the car when it travels over the top of the hump at a speed of 6 m/s.

b) What is the minimum speed needed for the car to leave the track at the top of the hump?

Homework Equations


F = mv^2/t
a=V^2/r
P=2\prodr
A=\prodr^2

The Attempt at a Solution


a) v= 6m/s , Fnet=?
F=mv^2/r
F= (400)(6^2)/ 10
Fnet = 1440 N - WRONG!
RIGHT - 2600N

How did they get that answer? - Could someone please help with this issue.

Kindest Regards,
TheKovac
 

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There should be a centripetal force on the car pointing into the hump. At the top of the hump, the answer you got is the centripetal force, but at the same time there is the normal force of the car pointing in the opposite direction. The net force exerted on the car by the hump should be that of the centripetal force subtracted from the normal force.
 
Gear300 said:
There should be a centripetal force on the car pointing into the hump. At the top of the hump, the answer you got is the centripetal force, but at the same time there is the normal force of the car pointing in the opposite direction. The net force exerted on the car by the hump should be that of the centripetal force subtracted from the normal force.

Thank you very much for your answer, you really have got me thinking now.

Does that mean:

F=ma
F= 400*9.8
Fnet = 3920N
Centripital Force = 1440

=>Fnet - Fc
=> 3920 - 1440
F= 2500N

Am I correct?
 
Seems right to me. The answer would be between 2500N and 2600N, so I'm supposing your book may have rounded or used 10 instead of 9.8 for the acceleration due to gravity.
 
THANK YOU TO ALL WHO ASSISTED WITH MY PROBLEM!

The matter is solved, and I am very happy with all the kind hearted assistance I received.

Have a great afternoon.

Kindest Regards,
TheKovac
 
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