Mechanical energy lost when child jumps onto merry-go-round

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Homework Help Overview

The problem involves a child jumping tangentially onto a stationary merry-go-round, requiring the application of conservation of momentum to determine the final angular speed. Additionally, the question addresses the mechanical energy lost to friction during this interaction.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the mechanical energy lost using the change in kinetic energy but struggles to arrive at the correct answer. Some participants suggest ensuring all final kinetic energy is accounted for in the calculations.

Discussion Status

Participants are engaged in clarifying the approach to calculating mechanical energy loss. There is an acknowledgment of the need to include the correct expressions for kinetic energy, particularly regarding the rotational kinetic energy of the merry-go-round. Multiple interpretations of the problem are being explored, with no explicit consensus reached yet.

Contextual Notes

There are indications that the original poster may be missing key concepts such as the moment of inertia in their calculations. The discussion reflects uncertainty about the correct expressions and calculations needed to solve the problem.

jybe
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Homework Statement


I have a basic problem where a child jumps tangentially onto the outer edge of a stationary merry-go-round, and you have to use conservation of momentum to find the final angular speed of the merry-go-round.

But the next part of the question asks "how much mechanical energy was lost to friction as the child jumped onto the merry-go-round?"

I tried using change in kinetic energy -- 1/2(m*r(omega)) - 1/2(mv2) = 0 but didn't get the right answer. I really can't think of how else I would find this, because the only thing that's changing is the kinetic energy. I would really appreciate it if somebody could help me out with this
 
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You have the right idea. Make sure you include all the final kinetic energy.
 
TSny said:
You have the right idea. Make sure you include all the final kinetic energy.

1/2 * (mass of child + mass of merry-go-round) * (r*omega)2 - 1/2mv2 = 0

Does this look better? I'm still not getting the right answer but it's much closer...maybe I messed up the calculation or the answer is wrong...but is this correct?
 
jybe said:
1/2 * (mass of child + mass of merry-go-round) * (r*omega)2 - 1/2mv2 = 0

Does this look better? I'm still not getting the right answer but it's much closer...maybe I messed up the calculation or the answer is wrong...but is this correct?
You are not using the correct expression for the rotational KE of the merry-go-round. It will involve the concept of "moment of inertia" (sometimes called "rotational inertia").
 

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