Rotational Kinetic energy in a loop

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Homework Help Overview

The problem involves a marble rolling down a track and around a loop-the-loop, with the goal of determining the minimum height required for the marble to complete the loop without falling off. The discussion centers around concepts of mechanical energy, forces at the top of the loop, and the relationship between the marble's mass, radius, and the loop's radius.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conservation of mechanical energy and the forces acting on the marble at the top of the loop. There are attempts to derive the minimum height using energy equations and force balance, with some questioning the correctness of substitutions made in the equations.

Discussion Status

Some participants have pointed out potential errors in the original poster's reasoning, specifically regarding the use of variables and equations. There is acknowledgment of these mistakes, and some participants express confidence in the overall approach while focusing on clarifying specific details. However, there is no explicit consensus on the final solution.

Contextual Notes

Participants reference a specific textbook, HC Verma, suggesting that the problem may be sourced from there. This indicates a potential context for the problem's complexity and expectations.

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[tex]\omega[/tex], so mgh = mgR + 1/2mv^2 + 1/2I[tex]\omega[/tex]. I then substituted 2/5mr^2 for I (general I of a sphere), and v/r for [tex]\omega[/tex]. 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[tex]\omega[/tex],

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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