A physics student is standing on an initially motionless,frictionless turntable

Question

A physics student is standing on an initially motionless,frictionless turntable with rotational inertia .31kg*m^2. She'sholding a wheel of rotational inertia .22 kg*m^2 spinning at 130rpm about a vertical axis. When she turns the wheel upside down,student and turntable begin rotating at 70 rpm. A) What's the students mass, considering her to be a cylinder30cm in diameter>
B) How much work did she do in turning the wheel upside down?Neglect the distance between the axes of the turntable andwheel. A physics student is standing on an initially motionless,frictionless turntable with rotational inertia .31kg*m^2. She'sholding a wheel of rotational inertia .22 kg*m^2 spinning at 130rpm about a vertical axis. When she turns the wheel upside down,student and turntable begin rotating at 70 rpm. A) What's the students mass, considering her to be a cylinder30cm in diameter>
B) How much work did she do in turning the wheel upside down?Neglect the distance between the axes of the turntable andwheel.

Explanation / Answer

            Given that the rotational inertia of the table is Itable = 0.31 kg.m2             rotational inertia of the wheel is Iwheel = 0.22kg.m2             Initial angular velocity is 1 = 130 rpm = 130*2 /60rad/s = 13.60 rad/s              Finalangular velocity is 2 = 70 rpm = 7.32 rad/s             Radius of cyllinder is R = 0.30 m --------------------------------------------------------------------------------------------------------          Since there is noexternal force acting on the system then the angular momentum isconserved.                              (Iwheel )*1= ( Istudent +I table -Iwheel)2            then we get      Istudent= ------------ kg.m2                                    (1/2)MR2 =  ------------kg.m2                              slove for M = ------ kg    The work done = change in rotationalkinetic energy                             = (1/2)( Istudent +I table -Iwheel)22 -(1/2)(Iwheel )*12                              =------------ J                                  slove for M = ------ kg    The work done = change in rotationalkinetic energy                             = (1/2)( Istudent +I table -Iwheel)22 -(1/2)(Iwheel )*12                              =------------ J    

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.