What Is the Normal Force on a Rollercoaster Car at the Top of a Hill?

AI Thread Summary
The discussion revolves around calculating the normal force on a rollercoaster car at the top of a hill, given its mass, radius of the hill, and speed. For a mass of 1150 kg and speeds of 8.5 m/s and 20 m/s, the equations of motion are applied to find the normal force. The correct normal force at 20 m/s was calculated as -17468.5 N, while the calculation for 8.5 m/s resulted in an incorrect value of -6088.53 N. Participants suggest checking the sign of the force and considering the upward force needed to maintain the circular path. The discussion highlights the importance of accurately applying physics equations and understanding the forces involved in circular motion.
arrax
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Homework Statement


A roller-coaster car has a mass of 1150 kg when fully loaded with passengers. As the car passes over the top of a circular hill of radius 16 m, its speed is not changing. (a) At the top of the hill, what is the normal force (using the negative sign for the downward direction) FN on the car from the track if the car's speed is v = 8.5 m/s? (b) What is FN if v = 20 m/s?

m=1150kg
R=16m
v= 8.5 or 20


Homework Equations



a=v^2/R
F=m*[(v^2)/R]
-FN-mg=m[(-v^2)/R]


The Attempt at a Solution



Using the above equation I got b) correct with -17468.5 N. However, using the same equation with a) I get -6088.53 and it's incorrect. I don't know what I'm doing wrong.

Also, this is from Wiley Plus.
 
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arrax said:

The Attempt at a Solution



Using the above equation I got b) correct with -17468.5 N. However, using the same equation with a) I get -6088.53 and it's incorrect. I don't know what I'm doing wrong.

Also, this is from Wiley Plus.

Check the sign of the force.

ehild
 


aside from a possible wrong sign, did you get a different number?
 


Think: can the track pull the roller coaster? What happens if upward force is needed to keep it on circular path?

ehild
 
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