Normal forces for small car performing a vertical loop

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Homework Help Overview

The problem involves a small car with a mass of 0.800 kg traveling at a constant speed of 12 m/s on a vertical circular track with a radius of 5.0 m. The discussion centers around determining the normal force exerted on the car at the bottom of the track, given the normal force at the top is 6.00 N.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between normal force, weight, and centripetal force, with some attempting to derive the acceleration and normal force at the bottom of the track. Questions arise about the correctness of the given constant speed and its implications for the calculations.

Discussion Status

Some participants have provided calculations related to the centripetal acceleration and have expressed confusion regarding the values presented. There is an ongoing exploration of how to calculate the normal force at the bottom of the track, with guidance offered on reconsidering the assumptions made about the speed of the car.

Contextual Notes

There is a noted discrepancy regarding the constant speed of the car, with suggestions that it may not be accurately stated. Participants are encouraged to clarify the implications of this on their calculations.

Anatalbo
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Homework Statement



A small car with mass .800 kg travels at a constant speed of 12m/s on the inside of a track that is a vertical circle with radius 5.0m. If the normal force exerted by the track on the car when it is at the top of the track is 6.00N, what is the normal force at the bottom of the track?

Reference https://www.physicsforums.com/threa...mics-car-traveling-in-vertical-circle.656316/

Homework Equations

The Attempt at a Solution


I got that the acceleration towards the center of the circle is 17.31 m/s^2
 
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Greg Bernhardt said:
Please show your work.
N+w=FC (normal force, weight, centripetal force)

6N+(9.81)(0.8)=0.8a
a=17.31 m/s^2
 
Anatalbo said:
N+w=FC (normal force, weight, centripetal force)

6N+(9.81)(0.8)=0.8a
a=17.31 m/s^2
Ignore the given constant speed of 12m/s, as it is incorrectly stated as such, which you will discover by reading the thread you referenced. So you now have the correct centripetal acceleration. So now show how you would calculate the normal force at the bottom, assuming the car's speed is constant.
 
PhanthomJay said:
Ignore the given constant speed of 12m/s, as it is incorrectly stated as such, which you will discover by reading the thread you referenced. So you now have the correct centripetal acceleration. So now show how you would calculate the normal force at the bottom, assuming the car's speed is constant.
Thanks! I was confused about why the values didn't add up
 

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