Past Cambridge, conservation of energy problem.

In summary: Now, let's summarize the conversation. In summary, the problem involves finding the value of h for a ball rolling on a track with a given radius. The equation for conservation of energy is used, with a correction factor of 3/4 due to the rolling condition. The critical speed at the top of the circle is determined to be root (R-a)G. The error in the attempt at a solution may be due to forgetting to include the ball's radius in determining critical speed and potential energy.
  • #1
xdrgnh
417
0

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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  • #2
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.
 
  • #3
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)?
 
  • #4
Yes, the radius is R-a, and the critical speed is not root RG, but rather, ___?
 
  • #5
PhanthomJay said:
Yes, the radius is R-a, and the critical speed is not root RG, but rather, ___?

root (R-a)G
 
Last edited:
  • #6
yes..
 

Related to Past Cambridge, conservation of energy problem.

1. What is the "Past Cambridge, conservation of energy problem"?

The "Past Cambridge, conservation of energy problem" refers to a physics problem first proposed by William Thomson, the 1st Baron Kelvin, at Cambridge University in the 1850s. It involves the concept of energy conservation in a closed system, where the initial and final energies are known, but the process between them is unknown.

2. Why is the "Past Cambridge, conservation of energy problem" important?

This problem is important because it demonstrates the fundamental principle of energy conservation, which is a cornerstone of modern physics. It also challenges scientists to think critically and creatively to solve a complex problem with limited information.

3. How is the "Past Cambridge, conservation of energy problem" solved?

The problem can be solved using mathematical equations and principles of energy conservation, such as the conservation of mechanical energy, work-energy theorem, and the first and second laws of thermodynamics. It may also involve making assumptions and simplifications to the system.

4. What are some real-world applications of the "Past Cambridge, conservation of energy problem"?

The principles involved in solving this problem are applicable to many real-world scenarios, such as designing energy-efficient systems, understanding the behavior of physical systems, and predicting the outcomes of chemical reactions. It also has applications in fields like engineering, astronomy, and environmental science.

5. Are there any limitations to the "Past Cambridge, conservation of energy problem"?

While the problem is a useful tool for understanding energy conservation, it does have some limitations. For example, it assumes that the system is isolated and there are no external forces or sources of energy. In reality, most systems are not perfectly isolated, and there may be other factors at play that can affect the energy balance. Therefore, the solutions to this problem should be interpreted with caution and may not always accurately reflect real-world situations.

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