Past Cambridge, conservation of energy problem.

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Homework Help Overview

The problem involves a conservation of energy scenario related to a ball moving in a circular path, specifically addressing the conditions at the top of the circle and the implications of height and radius in the energy equations.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the conservation of energy equation and its components, questioning the inclusion of the ball's radius and the critical speed at the top of the circle. There are attempts to clarify the relationship between height, radius, and energy terms.

Discussion Status

Participants are actively engaging with each other's reasoning, with some providing guidance on the need to clarify calculations and assumptions. There is a focus on ensuring that all relevant factors, such as the radius and critical speed, are correctly accounted for in the energy equations.

Contextual Notes

There are indications of missing steps in the original poster's calculations, and assumptions about the relationship between height and radius are being questioned. The problem context suggests a need for careful consideration of the rolling condition and its impact on energy terms.

xdrgnh
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Homework Statement


http://www-teach.phy.cam.ac.uk/dms/dms_getFile.php?node=5735 problem B8, the one with the with finding h>11(R-a)/4

Homework Equations


(1/2)mv^2=mgh
ac=(v^2/R)
w=V/R
(1/2)mv^2+(1/2)Iw^2

The Attempt at a Solution



At the top of the circle, the velocity squared is GR. And because the a is above ground, conservation of energy is written as. MG(h+a)=(3/4)MGR+MG(2R-2a). When I solve it out I get h=(11/4)R-a

Help would be greatly appreciated
 
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You are leaving out some steps so I can't see all your workings, but possibly your error is in forgetting to include the ball's radius when determining critical speed and potential energy. Please write out the conservation of energy equation and show your numbers term by term.
 
PhanthomJay said:
You are leaving out some steps so I can't see all your workings, but possibly your error is in forgetting to include the ball's radius when determining critical speed and potential energy. Please write out the conservation of energy equation and show your numbers term by term.

well I got the 3/4 because of the rolling condition.

(1/2)(mv^2)+(1/2)(ma^2*.5*(V^2/a^2) so when that is added you get (3/4)mv^2.

the critical speed squared at the top of the circle is just GR, because I just need to know the min speed, which is just GR. So v^2 at the top of the circle can just be replaced by GR. The M and G's cancel out of the mgh equations. The problem first states, that a is above h initial. So MG(h+a)= (3/4)(mGR)+MG(2R-2a). Does the a, affect what the radius would be. Maybe the radius is (R-a)?
 
Yes, the radius is R-a, and the critical speed is not root RG, but rather, ___?
 
PhanthomJay said:
Yes, the radius is R-a, and the critical speed is not root RG, but rather, ___?

root (R-a)G
 
Last edited:

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