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

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Homework Help Overview

The problem involves a child on a cart starting from rest at the top of a frictionless ramp and entering a frictionless circular loop. The objective is to determine the required height of the ramp for the cart to successfully traverse the loop, with discussions centered around the application of conservation of energy versus kinematic equations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the use of conservation of energy to find the height of the ramp, with one participant attempting to derive the height using energy equations and another questioning the validity of this approach compared to kinematics. There is also a debate over the correctness of the derived answers.

Discussion Status

There is an ongoing discussion regarding the appropriateness of using conservation of energy versus kinematics to solve the problem. Some participants express confusion over the differing answers and seek clarification on the conditions under which kinematic methods might be applicable.

Contextual Notes

Participants note that the problem may involve assumptions about energy conservation and the relationship between speed at different points in the motion. There is also mention of the need for additional information to potentially apply kinematic equations effectively.

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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