A car goes around a vertical circle (Uniform Circular Motion)

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SUMMARY

The discussion focuses on calculating the normal force exerted on a remote-control car with a mass of 1.51 kg moving at a constant speed of 12.0 m/s in a vertical circle with a radius of 5.00 m. At point A (the bottom of the circle), the normal force is calculated using the equation N - mg = ma, resulting in a normal force of 43.488 N. At point B (the top of the circle), the forces acting on the car differ, requiring a different approach to determine the normal force, which involves considering centripetal acceleration.

PREREQUISITES
  • Understanding of Newton's Second Law (Fnet = ma)
  • Knowledge of centripetal acceleration (a_y = v^2/r)
  • Ability to draw and interpret free-body diagrams
  • Familiarity with gravitational force calculations (w = mg)
NEXT STEPS
  • Study the derivation of centripetal force in circular motion
  • Learn how to analyze forces in vertical circular motion scenarios
  • Explore examples of normal force calculations in different contexts
  • Investigate the effects of varying speed on normal force in circular motion
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of objects in circular motion, particularly in analyzing forces acting on objects in vertical circles.

Chandasouk
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Homework Statement




A small remote-control car with a mass of 1.51 kg moves at a constant speed of v = 12.0 m/s in a vertical circle inside a hollow metal cylinder that has a radius of 5.00 m.

yf_Figure_05_76.jpg



What is the magnitude of the normal force exerted on the car by the walls of the cylinder at point A (at the bottom of the vertical circle)?

What is the magnitude of the normal force exerted on the car by the walls of the cylinder at point B (at the top of the vertical circle)?


Do I utilize Fnet = ma which would give me 43.488N as Fnet, but I do not see how that would help? I don't know what to do.
 
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Try drawing a force diagram when the car in at the bottom of the cylinder and when its
at the top of the cylinder.
 
For Part A, it would just be Fn pointing up and w pointing down.

w=mg so (1.51kg)(-9.8) = -14.798N so Fn must be 14.798N ?


Would part B basically be the same?
 
F_y = m*a_y
a_y = v^2/r

Does that help?
 
For part A it should be N - mg = 0, right ? Since mg is pointing down thus N has to be
pointing up.

For part B its not quite the same.
 

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