A physics student is standing on an initially motionless, frictionless turntable with rotational inertia 0.31?

A physics student is standing on an initially motionless, frictionless turntable with rotational inertia 0.31 kg·m2. He's holding a wheel of rotational inertia 0.22 kg·m2 spining at 124 rpm about a vertical axis, as we showed in Fig. 11.8. When he turns the wheel upside down, student and turntable begin rotating at 61 rpm.





Figure 11.8


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


kg





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


J

A physics student is standing on an initially motionless, frictionless turntable with rotational inertia 0.31?
I have the same problem and got part a but can't get part b.





This problem is also using the conservation of angular momentum, which is I0ω0=I1ω1.





Looking at the diagram, L(total)=L(wheel) before he turns the wheel, and afterwards L(total)=L(student + table)+L(wheel). We know that L=I*ω, so just start plugging things into the formula above...





Since the total angular momentum is conserved,


L(wheel) = L(student + table) - L(wheel)





L(wheel) = (.22 kg*m^2)*(124) = 27.28


L(student + table) - L(wheel) = (I(student) + .31 kg*m^2)(61) - 27.28





So...





27.28 = (I(student) + .31)(61) - 27.28


((27.28+27.28)/61)-.31 = I(student)





Since the student is to be considered a cylinder, the rotational inertia will be I = 1/2*MR^2, so...





1/2*M*(.15^2) = .588443





Then just solve for mass...





M = .58443/(.5*.15^2) = 51.95 kg