A physics student is standing on an initiallymotionless, frictionless turntable

Question

A physics student is standing on an initiallymotionless, frictionless turntable with rotational inertia 0.31kg·m2. He's holding a wheel of rotational inertia0.22 kg·m2 spining at 123 rpm about a vertical axis, as we showed in Fig.11.8. When he turns the wheel upside down, student and turntablebegin rotating at 69 rpm.

(a) What is the student's mass, considering him to be a cylinder 30cm in diameter?

(b) How much work did he do in turning the wheel upside down?Neglect the distance between the axes of the turntable andwheel.

Explanation / Answer

Initial angular momentum of the wheel Li =0.22kgm2 * 123 rpm When wheel turns upside down the angular momentum (student+table) Lf =2Li            But Lf =Itotf                    ==>        Lf= Itotf                              2Li  =Itotf                               2(0.22kgm2 * 123 rpm ) =Itot(69rpm)          ==>        Itot = 0.7843kg.m2          Assumed Student as cylinder , then moment ofinertia Is = 0.5Mr2                                                                                      = 0.5M(0.15)2 Therefore total moment of inertia = moment of inertia ofstudent + moment of inertia of table                                          Itot = Is + It           ==>                         Is = Itot - It                          0.5M(0.15)2 = 0.7843 kg.m2  -0.31kg.m2                ==>                      M=42.16 Kg We know that workdone = KE =0.5Itotf2                                                     =0.5 * 0.7843kg.m2 (69rpm)2                                                    solve to get   
  

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