Rotational Kinetic energy in a loop

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SUMMARY

The discussion focuses on determining the minimum height (h) required for a marble of mass (m) and radius (r) to successfully navigate a loop-the-loop of radius (R) without falling off. The total mechanical energy at the starting height is expressed as mgh, while at the top of the loop, it is mgR plus the kinetic energy components. The user initially calculated h as R + 0.7r, which was incorrect due to miscalculations involving the moment of inertia (I) and the centripetal force equations. Corrections pointed out that the angular velocity should be squared and that the centripetal acceleration should reference R instead of r.

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  • Understanding of mechanical energy conservation principles
  • Familiarity with rotational dynamics and moment of inertia (I)
  • Knowledge of centripetal force and acceleration concepts
  • Basic calculus for solving equations involving motion
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  • Study the derivation of the conservation of mechanical energy in rotational systems
  • Learn about the moment of inertia for various shapes, focusing on spheres
  • Explore centripetal force calculations in circular motion
  • Review problems from "Concepts of Physics" by HC Verma for practical applications
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Students studying physics, particularly those focusing on mechanics and rotational dynamics, as well as educators seeking to clarify concepts related to energy conservation and motion in loops.

Metalsonic75
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The marble rolls down a track and around a loop-the-loop of radius R. The marble has mass m and radius r (see picture). What minimum height h must the track have for the marble to make it around the loop-the-loop without falling off?


This is driving me nuts. I know that the total mechanical energy at the top of the track (starting point) is mgh, and at the top of the loop, it is mgR + 1/2mv^2 + 1/2I\omega, so mgh = mgR + 1/2mv^2 + 1/2I\omega. I then substituted 2/5mr^2 for I (general I of a sphere), and v/r for \omega. Leaving that for a moment, I found that the sum of forces at the top of the loop are F = ma_c = m*(v^2/r) = mg - n. Since I'm trying to find the point where n=0, I substituted 0 for n, and solved for g = v^2/r. Then I substituted v^2/r for g in my earlier equation that I obtained for total mechanical energy. Simplifying everything gave me h = R + 0.7r, which is incorrect. I would really like to know where my error lies. The answer must include the variables r and R. Any help would be appreciated. Thank you.
 

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Metalsonic75 said:
This is driving me nuts. I know that the total mechanical energy at the top of the track (starting point) is mgh, and at the top of the loop, it is mgR + 1/2mv^2 + 1/2I\omega,

Should be omega^2.

Leaving that for a moment, I found that the sum of forces at the top of the loop are F = ma_c = m*(v^2/r) = mg - n. Since I'm trying to find the point where n=0, I substituted 0 for n, and solved for g = v^2/r.

Should be R here, not r.
 
Everything is done properly except those mistakes which Shooting mentioned.

And by the way,is this a sum from HC Verma?
 
FedEx said:
Everything is done properly except those mistakes which Shooting mentioned.

And by the way,is this a sum from HC Verma?

Thanks for the help, Shooting star. I was able to solve the problem. And, sorry FedEx, I have no idea what you're talking about.
 
Metalsonic75 said:
Thanks for the help, Shooting star. I was able to solve the problem. And, sorry FedEx, I have no idea what you're talking about.

I was just asking whether the problem was from a book named Concepts Of Physics by HC Verma.
 

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