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

In summary, the conversation discusses a problem involving a child jumping onto a stationary merry-go-round and using conservation of momentum to find the final angular speed. The next part of the question asks about the amount of mechanical energy lost to friction during the jump. The suggested calculation involves using the change in kinetic energy, but the correct expression for the rotational kinetic energy of the merry-go-round should also include the concept of "moment of inertia."
  • #1
jybe
41
1

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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  • #2
You have the right idea. Make sure you include all the final kinetic energy.
 
  • #3
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?
 
  • #4
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").
 

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

1. What is mechanical energy?

Mechanical energy is the energy associated with an object's motion or position. It can be classified into two types: kinetic energy, which is the energy an object possesses due to its motion, and potential energy, which is the energy an object possesses due to its position or state.

2. How is mechanical energy lost when a child jumps onto a merry-go-round?

When a child jumps onto a merry-go-round, the mechanical energy of the system (child + merry-go-round) is converted into different forms. Some of the mechanical energy is lost due to friction, air resistance, and sound. This results in a decrease in the overall mechanical energy of the system.

3. Why does the child's jump affect the mechanical energy of the merry-go-round?

The child's jump adds additional energy to the system, which causes the merry-go-round to speed up. This increase in speed results in a change in the kinetic energy of the system, as well as an increase in the amount of energy lost due to friction and other factors.

4. Can the mechanical energy lost be recovered?

No, the mechanical energy lost due to the child's jump cannot be recovered. It is converted into other forms of energy, such as heat, sound, and motion. This is due to the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred or transformed.

5. How can the amount of mechanical energy lost be reduced?

The amount of mechanical energy lost can be reduced by minimizing factors that contribute to energy loss, such as reducing friction by using lubricants, or making the merry-go-round more aerodynamic to decrease air resistance. However, some amount of energy loss is inevitable and cannot be completely eliminated.

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