Relationship between linear and rotational kinetic energy

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  • #1
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Homework Statement


Consider a solid disk of mass m and radius R rolling along a horizontal surface with the center of the disk moving horizontally with a speed v. The total kinetic energy of the disk is a combination of translational kinetic energy , KEt and rotational kinetic energy, KEr. Find the ration KEt/KEr



Homework Equations





The Attempt at a Solution


I basically broke down the equations into their simplest forms to try to find the relationship. So
E = 1/2mv^2 + 1/2IW^2
E = 1/2mv^2 + 1/2 (1/2mR^2)w^2
E = 1/2mv^2 + 1/4mR^2 * W^2
I am really not sure how to how about solving this problem. I know I could make up some imaginary numbers for the variables and just guess and check for the relationship but I would prefer to solve this algebraically. The correct answer is 2.
 

Answers and Replies

  • #2
Orodruin
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You are after the translational and rotational kinetic energies (or rather, their ratio), not the total kinetic energy.

What is the ratio? How can you simplify it?
 
  • #3
CAF123
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Using the fact that the disk does not slip or travels in a 'pure roll', you can substitute for ##\omega## in terms of other variables. Then take the ratio.
 
  • #4
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You are after the translational and rotational kinetic energies (or rather, their ratio), not the total kinetic energy.

What is the ratio? How can you simplify it?

OK so I figured it out by breaking down the velocity into its parts. So
E = 1/2mv^2 + 1/4mR^2 * W^2
E = 1/2m(Rw)2 + 1/4mR^2 *w^2
E = 1/2 mR^2 * w^2 + 1/4mR^2 * w^2
Since everything in the translational and rotational kinetic energies is the same except 1/2 and 1/4 I can conclude that the translational kinetic energy is twice as big as the rotational kinetic energy. What was your approach?
 
  • #5
Orodruin
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Correct. There really is not much more to it.
 

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