Centripetal force - Ferris wheel... Opposite results

  • #1
Iamconfused123
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Homework Statement
A person with a mass of 50kg is spinning on a Ferris wheel. The radius is 5m. If the wheel spins with a constant speed of 5m/s, what will be the force of the chair acting on the person (Normal force) at the highest and lowest point of the spin?
Relevant Equations
##F_{cp}=mv^2/r##, ##F=ma##, ##a=v^2/r## ##g=10m/s^2##
I don't understand why am I getting opposite answers. I get 250N for the lowest point and 750 for the highest point.

For the highest point: ##F_{net}=F_{cp}+F_{mg}-F_N## and then ##F_N=ma_{cp}+mg = 50*(25/5)+500=750N## because I've been told that ##F_{cp}## always acts towards the center of the circle and that that is always positive direction, which in this case is downwards which also makes ##F_g## positive.
But even if I reverse the signs. If I make upwards positive and downwards negative I'll get the same answer; ##F_{net}= -F_{cp}-F_{mg}+F_N## because when ##F_N## gets to ghe left it becomes negative and then all forces are negative. Multiply by -1 and we get 750N again as a result.

for the lowest point: ##F_{net}=F_{cp}-F_{mg}+F_N## and that is ##F_N=-ma_{cp}+mg = -50*(25/5)+500 = 250N## and if I reverse the signs and make upwards negative I will still get same result .

Can someone please tell me where I am wrong. Than you very much.
 
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  • #2
I always take from the center of rotation out to the mass as positive direction, then the acceleration is negative:

$$\sum F = -m\frac{v^2}{R} $$

Doing this works out for me in terms of the appropriate answer for each scenario.

Edit: just so you aren’t confused about taking the opposite convention, as @kuruman points out it won’t matter if you are consistent.
 
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  • #3
Draw a free body diagram of the sitting person.
Put in the 2 and only 2 forces acting on the person and label them.
Enclose the FBD with a dotted box and draw the acceleration outside the box.
Set the sum of the two arrows inside the box equal to mass times acceleration. Note that depending on your sign convention, the acceleration can be positive or negative.

If you do all this, you should be able to sort it out. If not, post your diagram and the equations you get from it for troubleshooting.
 
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  • #4
kuruman said:
Draw a free body diagram of the sitting person.
Put in the 2 and only 2 forces acting on the person and label them.
Enclose the FBD with a dotted box and draw the acceleration outside the box.
Set the sum of the two arrows inside the box equal to mass times acceleration. Note that depending on your sign convention, the acceleration can be positive or negative.

If you do all this, you should be able to sort it out. If not, post your diagram and the equations you get from it for troubleshooting.
I got the correct answers. Thank you very much for that. But, why my approach did not work? was it because the centripetal force is not a real force? just as with a car in a turn, the only force acting on the car is the friction force and no other force, so it would be wrong to write ##F_{net}=F_{cp} - F_{fr}## when ##F_{net}## is what we call in this case ##F_{cp}##

I used the acceleration of the point on the wheel as a positive direction, so for the top position downwards is positive, and for the bottom position upwards is positive, that way I got both ##F_N## as positive.

Regarding this direction thing, this might sound stupid, but have I done it correctly in this OR approach on he left side of the photo? Was I supposed to put - in front of ##ma_{cp}## on the left? I always mess up this direction stuff. I am not sure if I was supposed to put - in front of ##ma_{cp}## or if I should have just put negative acceleration without the minus in front of ##ma_{cp}##. But then I'd end up with ##ma_{cp}=F_N-F_g -> F_N=ma_{cp}+F_g ->F_N=-250-500 =-750N## so I guess I did it correctly.

Here is the photo. The results are correct, so the sketch should be fine as well, but just in case.
7.jpg
Once again thank you very much :bow:
 
  • #5
erobz said:
I always take from the center of rotation out to the mass as positive direction, then the acceleration is negative:

$$\sum F = -m\frac{v^2}{R} $$

Doing this works out for me in terms of the appropriate answer for each scenario.

Edit: just so you aren’t confused about taking the opposite convention, as @kuruman points out it won’t matter if you are consistent.
Thanks for helping me :oldsmile:
 
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  • #6
Iamconfused123 said:
centripetal force is not a real force
Centripetal force is very much a real force. You are thinking of centrifugal force. Your error is in assuming that it is a force of its own. The centripetal force in uniform circular motion is the net force acting on the object so this
Iamconfused123 said:
For the highest point: Fnet=Fcp+Fmg−FN and then
makes little sense. It seems you are using this with the assumption of Fnet = 0 so you would get Fmg - FN = - Fcp instead of Fmg - FN = Fnet = Fcp.
 
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  • #7
Iamconfused123 said:
But, why my approach did not work? was it because the centripetal force is not a real force?
The centripetal force is not a force in the same sense that the net force is not a force. In both cases they are the sum of some forces. The word "centripetal" is an adjective used to describe the direction of the net force. So ##mv^2/r## is called the centripetal force. Likewise, you could refer to ##mg## as the "downward" force.
 
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  • #8
I define the centripetal force as that component of the resultant of all the applied forces which is normal to the velocity.
 
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  • #9
Thanks everyone🙏, I believe I understand it now. 😅
 
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