Why Can't Conservation of Energy Be Used to Solve the Loop Problem in Physics?

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SUMMARY

The discussion centers on the physics problem of determining the height of a ramp required for a cart to successfully traverse a circular loop. The initial approach using conservation of energy led to an incorrect conclusion of h = 5r/2, while the correct solution derived from kinematics yields h = r/2. The key takeaway is that equating the speed at the bottom of the ramp to that required at the top of the loop is a fundamental error in applying energy conservation principles in this scenario.

PREREQUISITES
  • Understanding of basic physics concepts, including potential and kinetic energy.
  • Familiarity with kinematic equations, specifically vf² = vi² + 2ax.
  • Knowledge of circular motion dynamics, particularly the forces acting on an object in a loop.
  • Ability to manipulate algebraic equations to solve for unknown variables.
NEXT STEPS
  • Study the principles of circular motion and the forces acting on objects in loops.
  • Learn more about the application of kinematic equations in solving physics problems.
  • Explore the limitations of conservation of energy in dynamic systems.
  • Review examples of similar physics problems involving energy conservation and kinematics.
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Students studying physics, educators teaching mechanics, and anyone interested in understanding the application of energy conservation and kinematics in real-world scenarios.

squirrelschaser
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Homework Statement



Bob starts at rest from the top of a frictionless ramp. At the bottom of the ramp, he enters a frictionless circular loop. The total mass of the child and the cart he sits in his m. What must the height of the ramp be in order for the cart to successfully traverse the loop.

r = radius of loop
h = height of ramp
theta = angle of the ramp (irrelevant though)

Homework Equations

The Attempt at a Solution


[/B]
I solved for the minimum speed at the top of the loop.

Fy = F + mg = mv^2/r

v= sqrt(rg)

I then used conservation of energy.

Initial : mgh
Final : mg2r + (m(sqrt(rg))^2)/2

mgh = mg2r + mrg/2

mgh = 5mgr/2

Cancel stuff out h = 5r/2 (WRONG)

Instead the solution calls for using kinematics not energy conservation.

v= sqrt(rg) stills hold.

vf^2 = vi^2 + 2ax

rg = 0 + 2gsin(theta)*(h/sin(theta)

rg = 2gh

r = 2h

h = r/2 (CORRECT answer)

I understand the mathematical process of the correct solution.
However, I don't understand why I can't use conservation of energy(gives me wrong answer) instead of kinematics.

 

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The second answer (r/2) is clearly wrong since it would not provide enough energy to reach the top of the loop even with no remaining KE.
The calculation goes wrong because it equates the speed at the bottom of the ramp to that required at the top of the loop.
The first answer is correct.
 
The solutions I have showed the second answer as the correct answer.
haruspex said:
The second answer (r/2) is clearly wrong since it would not provide enough energy to reach the top of the loop even with no remaining KE.
The calculation goes wrong because it equates the speed at the bottom of the ramp to that required at the top of the loop.
The first answer is correct.

Really? That's the solution provided to me.
Glad to know I wasn't paranoid or something.

Is there any additional information that would make solving this question using kinematic possible, then?
 
squirrelschaser said:
Is there any additional information that would make solving this question using kinematic possible, then?
The v2-u2=2as equation is effectively KE+PE constant. All that's different is factoring out the mass.
 

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