Energy of translation and energy of rotation

AI Thread Summary
To determine the ratio of translational to rotational kinetic energy, the total kinetic energy is expressed as KEtotal = KEr + KEtr. The equations for kinetic energy are KEr = 1/2 Iω² and KEtr = 1/2 mv², with I for a disk being I = 1/2 mr². By substituting and simplifying, the ratio of KEtr to KEr results in 2:1 for a uniform disk rolling without slipping. This indicates that the translational kinetic energy is twice that of the rotational kinetic energy.
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Homework Statement


How would you go about figuring out the ratio of components (rotational and translational energy) of total kinetic energy?


Homework Equations





The Attempt at a Solution


i started with KEtotal=KEr+KEtr. KEr=1/2Iw^2; KEtr=1/2mv^2
use I for a disk-I=1/2mr^2
I put KEtr over KEr KEtr/KEr for a ratio

(1/2 mv^2)/1/2Iw^2
substitute in 1/mr^2 for I and v/r for w

(1/2 mv^2)/(1/2)(1/2mr^2)(v/r)^2

cancel m, v and r and 1/2

left with (1)/(1/2)

ratio is 2:1 KEtr:KEr ??
 
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That's right. For a uniform disk that is rolling without slipping, the translational KE is twice the rotational KE. Good!
 
YAY! Thanks for confirming. I am struggling with physics.
 
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