A frictionless 10 kg disk with a radius of 0.5 meters is spinning at 2 radians/s

Question

A frictionless 10 kg disk with a radius of 0.5 meters is spinning at 2 radians/second with a 2.0 kg piece of spherical dry ice with a radius of 0.05 meters on the edge of the disk. (you can assume that the center of mass of the dry ice is 0.5 meters from the center of the disk). After the dry ice sublimes (i.e. vaporizes into carbon dioxide) and there isn't anymore dry ice on the spinning disk, what is the angular velocity of the disk?

I have nooooooo clue what I am doing and really need help understanding this! :)

Explanation / Answer

mass of the disk M = 10 kg radius r = 0.5 m Angular speed w ' = 2 rad / s Moment of inertia of the disk I = ( 1/ 2) Mr^ 2 I = 1.25 kg m^ 2 Mass of the dry ice m = 2 kg total dry ice before sublies the dry ice I ' = I + mr^ 2 I ' = 1.25 + 0.5 = 1.75 kg m^ 2 total dry ice after sublies the dryice I" = I = 1.25 kg m^ 2 from law of conservation of angular momentum , I' w ' = I" w" from this required angular velocity of the disk w " = I ' w' / I " w " = 2.8 rad / s

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