Understanding a book example about centripetal force

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SUMMARY

The discussion centers on the concept of centripetal force and its relationship with gravitational force and normal force in circular motion. Participants clarify that centripetal force is not an applied force but rather the radial component of the resultant force, which points towards the center of the circular path. The equation ΣFr=mv²/r is referenced to establish that the book treats this force as centripetal. The conversation emphasizes the importance of distinguishing between centripetal and centrifugal forces when analyzing forces in a non-inertial frame.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with circular motion concepts
  • Knowledge of force vectors and their components
  • Ability to interpret equations related to centripetal acceleration
NEXT STEPS
  • Study the derivation of the centripetal force equation ΣFr=mv²/r
  • Explore the differences between centripetal and centrifugal forces
  • Learn about non-inertial reference frames and their implications
  • Investigate real-world applications of centripetal force in engineering
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of circular motion and the forces involved in such systems.

Bunny-chan
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Homework Statement


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If the centripetal force is at the left side of the equation; that means if we move it over to the other side, it'll have a negative sign, which means it is opposite in sense in relation to N and mg. But how is that possible, considering that the centripetal force always points to the center of the circumference?

The Attempt at a Solution

 
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The centripetal force is the force that provides the centripetal acceleration. In this case, it is the vector sum of ##\vec{n}_{bot ~or ~top} +m\vec{g}##. Note that it points in the same direction as the centripetal acceleration in both diagrams which is all that is required of it.
 
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Bunny-chan said:
opposite in sense in relation to N and mg. But how is that possible, considering that the centripetal force always points to the center of the circumference?
Centripetal force is not an applied force, like N and mg are. It is the radial component of the resultant force.
If you shift it to the other side and combine it with real applied forces as though it is another such then in effect you are switching to a non-inertial view, replacing the centripetal force with a centrifugal one. That acts away from the centre.
 
Since gravitational force is down and it has a positive sign, down is positive. The acceleration on the left side of the equation is downward when at the top of the loop, so it needs to be positive. The rest follows.
 
FactChecker said:
Are you sure this is centripetal force and not centrifugal force?
The first quoted equation has ΣFr=mv2/r, so I would say the book treats it as centripetal at that point. But see post #3.
 
haruspex said:
The first quoted equation has ΣFr=mv2/r, so I would say the book treats it as centripetal at that point. But see post #3.
Yes. Sorry. I was correcting my post as you responded to it..
 
OK. I understand it now. Thanks for the inputs.
 

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