User6060 Posted October 27, 2005 Share Posted October 27, 2005 Can someone check my work or post corrections to calculations thanks. 1. A record player rotates at 33.33 revolutions per minute. If a speck of dust of mass 0.2 mg is to stay on the record (and rotate with the record) at a distance of 10 cm from the center, what is the force of friction? 33.33/60s= 0.5555 rev/s 10cm= 0.1 m , 0.20mg = 2x10^-7 kg circumfrence of circle made by dust = 2pi(0.1 m) = 0.63m angular velocity= (0.63m)(0.5555 rev/s) = 0.35m/s Fc = Ff Ff= (mv^2)/r = ((2x10^-7kg)(0.35m/s))/ 0.1 m = 7x10^-7 N Link to comment Share on other sites More sharing options...
darkpuma Posted October 27, 2005 Share Posted October 27, 2005 as i'm in aerospace engineering i'd love to help ya, but you caught me just as i was going to bed... if you still need help in the monring pm me ;) Link to comment Share on other sites More sharing options...
futb0l Posted October 27, 2005 Share Posted October 27, 2005 Can someone check my work or post corrections to calculations thanks.1. A record player rotates at 33.33 revolutions per minute. If a speck of dust of mass 0.2 mg is to stay on the record (and rotate with the record) at a distance of 10 cm from the center, what is the force of friction? 33.33/60s= 0.5555 rev/s 10cm= 0.1 m , 0.20mg = 2x10^-7 kg circumfrence of circle made by dust = 2pi(0.1 m) = 0.63m angular velocity= (0.63m)(0.5555 rev/s) = 0.35m/s Fc = Ff Ff= (mv^2)/r = ((2x10^-7kg)(0.35m/s))/ 0.1 m = 7x10^-7 N 586730300[/snapback] The method looks right to me. Link to comment Share on other sites More sharing options...
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