Magnitude and direction of the force of the seat on you?

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Homework Help Overview

The problem involves analyzing the forces acting on a rider at the bottom of a vertical loop on a roller coaster, specifically focusing on the magnitude and direction of the normal force exerted by the seat. The context is centered around concepts of circular motion and forces, including centripetal force and gravitational force.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between centripetal force and gravitational force, with some attempting to derive the normal force using equations from their textbooks. Others express confusion about the application of these equations and the underlying concepts.

Discussion Status

There is an ongoing exploration of the forces involved, with some participants providing equations and reasoning to clarify the situation. While attempts to derive the normal force are being made, there is no explicit consensus on the best approach or understanding of the concepts involved.

Contextual Notes

Some participants emphasize the importance of understanding the derivation of formulas rather than applying them blindly, indicating a focus on deeper comprehension of the physics principles at play.

charan1
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Homework Statement


One fine spring day you decide to skip class and go to the amusement park to ride the loopthe-
loop roller coaster. At the bottom of the loop, you’re moving with a speed of 20 m/s in a vertical circle 20 m
in diameter. What’s the magnitude and direction of the force of the seat on you? Assume your mass is 50 kg.

Homework Equations



Vt=omega x radius
Fr=mass x omega2 x radius

The Attempt at a Solution



Vt=omega x radius
20m/s=omega x 10m
omega=10 rad/s

Fr=mass x omega2 x radius
Fr=50kg x 10rad/s x 10m
=5,000 N

I'm not sure what I'm doing at all on this problem please help me out.

I drew the FBD and the only forces are the Normal which I assume is the seats force on the rider which I further moved towards thinking that would be the Force towards the center thus the equation
Fr=mass x omega2 x radius, and the other force is (of course) g.

Please help me out I am in a bit of confusion.

Thank you for reading!

-Charan
 
Last edited:
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F(centripetal)=V^2/R because you are moving in a circle there is a centripetal force, always. At the bottom it is opposite of weight and the normal you feel is w-Fc at the top it is creating a normal if I remember correctly. so It is Fc+W=N
 


Oh I see thank you I also just found it in my book

at the bottom N=mg+mv2/r
and at the top N=mv2/r - mg

so in this case it is asking for the Normal force at the bottom and that is the first equation.

so:

N=(50kg)(g) + (50kg)(20m/s)2/10m
N=2,490N
 


Charan,

Just so you know how to get that equation and not memorize it, here are some steps you might take to help you understand why that is the case.

First draw a free-body and analyze the forces acting on you due to the seat. After drawing the diagram you will see that you have a Fn going up and a Fw going down. So if you use Newtons second law you get.

Efy=ma(c)

Fn-Fw=m(V^2/r) -----> Fn=Fw+m(V^2/r)

It will probably stick better if you understand why those equations are true
 


charan1 said:
Oh I see thank you I also just found it in my book

at the bottom N=mg+mv2/r
and at the top N=mv2/r - mg

so in this case it is asking for the Normal force at the bottom and that is the first equation.

so:

N=(50kg)(g) + (50kg)(20m/s)2/10m
N=2,490N
That's all very well, but please don't blindly find formulas in the book and try to apply them without understanding them.
 


When you are at the bottom, then Fseat is upward and your weight Mg is downward.
Therefore Fr = Fseat - Mg
Also Fr = Mv^2/r
Therefore Mv^2/r = Fseat - Mg
Fseat = Mv^2/r + Mg
Solve the above to calculate Fseat.

For centripetal force, you have used M*omega^2*r. That is also correct, but in this problem, v is given. So it is easier to use Mv^2/r
 

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