Rotational kinetic energy of ice skater

In summary: Quite so, so why did you plug in 10rad/s in both energy calculations?In summary, the skater executes a spin about a vertical axis with her feet on a frictionless ice surface. Her own moment of inertia can be taken as being 50kgm^2, independent of her arm position. The skater then moves her arms so that both masses are 0.5m from the rotation axis. The skaters total kinetic energy before and after the arm movement is 3000J and 2625J, respectively.
  • #1
Erenjaeger
141
6

Homework Statement

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An ice skater executes a spin about a vertical axis with her feet on a frictionless ice surface. In each hand she holds a small 5kg mass of which are both 1m from the rotation axis and the angular velocity of the skater is 10rad/s. The skater then moves her arms so that both masses are 0.5m from the rotation axis. The skaters own moment of intertia can be taken as being 50kgm^2, independent of her arm position.

Find the total kinetic energy of the skater and the masses both before and after the arm movement. Explain any difference.


Homework Equations


KErotational = 1/2 Iω2

3. The Attempt at a Solution
KErotational initial = 1/2 Iω2 = 1/2 ((10⋅(1)2)+50)⋅(10)2
= 3000J
Then using the same method to find the final rotational KE once the skater has moved her arms in
KErotational initial = 1/2 Iω2 = 1/2 ((10⋅(0.5)2)+50)⋅(10)2
= 2625J
which doesn't make intuitive sense to me, since moving her arms in will increase her speed and should in turn increase her kinetic rotational energy?? on the worked answers for the exam they use the equation
KE= L2/2I
where L is the angular momentum

can someone explain to me how the got that equation and why you can't use the other one, thanks.

 
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  • #2
You are right that moving her arms in will increase her angular velocity and her kinetic energy. But her angular momentum should remain the same.

So you wrote 2 equations for KE, both with a subscript of "rotational initial". I'm pretty sure that you intended the second equation to be "rotational final".
 
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  • #3
TomHart said:
You are right that moving her arms in will increase her angular velocity and her kinetic energy. But her angular momentum should remain the same.

So you wrote 2 equations for KE, both with a subscript of "rotational initial". I'm pretty sure that you intended the second equation to be "rotational final".
oh whoops yeah i meant the second equation to be sub final, i just copied the equation for the initial ke so i didnt have to rewrite it haha.
So how did they come to that second equation involving the angular momentum ?
thanks.
 
  • #4
I'm not familiar with that second equation, but I can substitute Iω for L into it and see that it is a valid equation. So maybe they calculated the final momentum and substituted that into that second equation to get the final kinetic energy.

L = Iω is easier for me to remember. For these kinds of problems, unless there is something that is dissipating energy, angular momentum will be conserved. Once you calculate the final angular velocity, you can calculate the final kinetic energy.
 
  • #5
The other equation is useful as it contains quantities that are constant(angular Momentum)
 
  • #6
The relevant equations should have included L = I ω. As already answered, KE = L = 1/2 I ω2 = (I ω)2 / 2.

explain any difference
How technical should this part be? The skater performs internal work to move the weights in. For more detail, as the weights are pulled in, their path is not circular, somewhat spiral like, and during the transition, a component of the tension in the skaters arms is in the direction of the inwards spiral like path of the weights, increasing their speed. The skater arms also oppose some of this tendency for the weights to speed up, otherwise the skater's arms would wrap around the skaters body, so the skater's body also experiences an increase in angular velocity.
 
  • #7
Erenjaeger said:
moving her arms in will increase her speed
Quite so, so why did you plug in 10rad/s in both energy calculations?
 

What is rotational kinetic energy of an ice skater?

Rotational kinetic energy is a type of energy that an object has due to its rotation around an axis. In the case of an ice skater, it refers to the energy they have due to their spinning motion on the ice.

How is the rotational kinetic energy of an ice skater calculated?

The rotational kinetic energy of an ice skater can be calculated using the formula KE = (1/2) * I * ω^2, where KE is the rotational kinetic energy, I is the moment of inertia, and ω is the angular velocity of the skater.

What factors affect the rotational kinetic energy of an ice skater?

The main factors that affect the rotational kinetic energy of an ice skater are their mass, the distribution of their mass, and their angular velocity. The greater the mass and distribution of mass, the greater the rotational kinetic energy. Similarly, the higher the angular velocity, the greater the rotational kinetic energy.

How does the rotational kinetic energy of an ice skater change during a spin?

As an ice skater begins to spin, their rotational kinetic energy increases due to an increase in angular velocity. However, as they spin for longer periods of time, their rotational kinetic energy may decrease due to the effects of friction and air resistance, which can slow down their spin.

Why is understanding the rotational kinetic energy of an ice skater important?

Understanding the rotational kinetic energy of an ice skater is important because it allows us to analyze and predict their movements and performances on the ice. It also helps us understand the physical principles involved in ice skating, which can be applied to other areas of science and engineering.

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