A playground carousel is free to rotate about its center onfrictionless bearings

Question

A playground carousel is free to rotate about its center onfrictionless bearings, and air resistance is negligible. Thecarousel itself (without riders) has a moment of inertia of120kg·m2. When one person isstanding at a distance of 1.50 m from the center, the carousel hasan angular velocity of 0.500 rad/s.However, as this person moves inward to a point located 0.750 mfrom the center, the angular velocity increases to 0.600 rad/s. What is the person's mass?

*36.7kg, 37.1kg, and 38.1kg are incorrect answers.

Explanation / Answer

       Given that the moment ofinertia of the carousel is I = 120 kg.m2        The distance between thecenter of the carousel and man is d1 = 1.50 m         Initial angularvelocity of the carousel is 1 = 0.500 rad/s         Final distancebetween the center of the carousel and man is d2 = 0.750m         final angularvelocity of the carousel is 2 = 0.600rad/s       --------------------------------------------------------------------------------------------        since there is noexternal force acts on the system the angular momentum isconserved                        initial angular momentum = final angualr moemntum                                                   I1*1 = I2* 2                                         (I + md12)1 = ( I +md22)* 2                                        (I + md12) = ( I +md22)* 2/1                                                      I+ md12 = I*2/ 1 +md22* 2/1                                                    m = [ I - I*2/ 1 ] /( d12+ d22*2/ 1)                                                   = ---------- kg                The distance between thecenter of the carousel and man is d1 = 1.50 m         Initial angularvelocity of the carousel is 1 = 0.500 rad/s         Final distancebetween the center of the carousel and man is d2 = 0.750m         final angularvelocity of the carousel is 2 = 0.600rad/s       --------------------------------------------------------------------------------------------        since there is noexternal force acts on the system the angular momentum isconserved                        initial angular momentum = final angualr moemntum                                                   I1*1 = I2* 2                                         (I + md12)1 = ( I +md22)* 2                                        (I + md12) = ( I +md22)* 2/1                                                      I+ md12 = I*2/ 1 +md22* 2/1                                                    m = [ I - I*2/ 1 ] /( d12+ d22*2/ 1)                                                   = ---------- kg        

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