Angular speed at bottom of loop

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SUMMARY

The discussion centers on calculating the angular speed required for a circular ring with a mass of 11 kg and a radius of 0.66 meters to successfully navigate a vertical loop of radius 10 meters without slipping. The solution involves applying the moment of inertia formula, I = M R², and utilizing conservation of energy principles. The final calculated angular speed at the bottom of the loop is definitively 25.99 rad/s.

PREREQUISITES
  • Understanding of moment of inertia, specifically I = M R²
  • Knowledge of conservation of energy principles in physics
  • Familiarity with angular velocity and its relationship to linear speed
  • Basic concepts of rolling motion and frictionless surfaces
NEXT STEPS
  • Study the conservation of energy in rotational dynamics
  • Learn about the relationship between linear speed and angular velocity
  • Explore the dynamics of rolling motion without slipping
  • Investigate the effects of mass distribution on moment of inertia
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in the dynamics of rolling objects and rotational motion.

Paulie71199
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Homework Statement


A circular ring with a mass of 11 kg and radius of 0.66 meters is to roll without slipping about a vertical loop of radius 10 meters which is frictionless. If it is to just barely make it around the top of the loop, what must its angular speed be at the bottom of the loop in rad/s?


Homework Equations



The moment of inertia for the ring about it's center is I = M R2.

The Attempt at a Solution



The answer comes to be 25.99 rads/sec. Although I don't know how to get to that answer
 
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Welcome to PF!

Hi Paulie! Welcome to PF! :smile:

(have a theta: θ and try using the X2 icon just above the Reply box :wink:)

Use conservation of energy (together with the rolling equation relating speed of centre of mass to angular velocity, and putting hte normal reaction force equal to zero at the top). :smile:
 
Thank you for the welcome and the help!

Thanks for the tip lol.
 

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