Please explain how to work problem in detail so that it\'s easy to understand wh

Question

Please explain how to work problem in detailso that it's easy to understand what you did. && pleaseprovide answer. Thank you!
A playground merry-go-round of radius 2.00 m has a moment ofinertia I = 295kg·m2 and is rotating about a frictionlessvertical axle. As a child of mass 25.0 kg stands at a distance of1.00 m from the axle, the system (merry-go-round and child) rotatesat the rate of 15.0 rev/min. The childthen proceeds to walk toward the edge of the merry-go-round. Whatis the angular speed of the system when the child reaches theedge? Please explain how to work problem in detailso that it's easy to understand what you did. && pleaseprovide answer. Thank you!
A playground merry-go-round of radius 2.00 m has a moment ofinertia I = 295kg·m2 and is rotating about a frictionlessvertical axle. As a child of mass 25.0 kg stands at a distance of1.00 m from the axle, the system (merry-go-round and child) rotatesat the rate of 15.0 rev/min. The childthen proceeds to walk toward the edge of the merry-go-round. Whatis the angular speed of the system when the child reaches theedge?

Explanation / Answer

I1 1 = I22      conservation ofangular momentum I1 = 295 + 25 * 12 = 320kg-m2 I2 = 295 + 25 * 22 = 395kg-m2 2 = 320 / 395 * 1 = 320 /395 * 2 f = 5.09 f f = 15 / min / (60 sec / min) = .25 /sec 2 = .25 * 5.09 = 1.27 /sec (As a check f2 = 1.27 / ( 2 ) = .202 / sec =12.1 / min) 12.1 / 15 = .807 = 320 / 395 2 = 320 / 395 * 1 = 320 /395 * 2 f = 5.09 f f = 15 / min / (60 sec / min) = .25 /sec 2 = .25 * 5.09 = 1.27 /sec (As a check f2 = 1.27 / ( 2 ) = .202 / sec =12.1 / min) 12.1 / 15 = .807 = 320 / 395

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