An acrobatic physics professor stands at the center of a turntable holding his a

Question

An acrobatic physics professor stands at the center of a turntable holding his arms extended horizontally with a 5.0 kg dumbbell in each hand He is set rotating about a vertical axis, making one revolution in 2.0 s Find the profs new angular velocity if he pulls the dumbbell in to his stomach His moment of inertia (without the dumbbells) is 3.0 kgm^2 when his arms out-stretched, dropping to 2.2 kg m^2 when his hands are at his stomach The dumbbells are 1.0 m from the axis initially and 0.20 m from it at the end. Treat the dumbbells as particles.

Explanation / Answer

Part a)

Wa.Ia=Wf.If

Moment of inertia of system: I=I prof+ I bells

Each dumbbell of mass m contributes mr^2 to Ibells

Ia=3+2(5).1^2

=13kgm^2

Wa=Irw/25

=0.5 rev/s

If=2.2+2(5)(0.2)^2=2.6kgm^2

Wa.Ia=Wf.If

Wf=Wa.Ia/If

=(0.5)13/2.6

=2.5 rev/s

the angular velocity increases by a factor of 5 while the angular momentum remains constant

b)

express Wa and Wf in rad/s

Ka=0.5Ia.Wa^2

=0.5x13[0.5 x2pi]^2

=64 J

Kf=0.5If.Wf^2

=0.5x2.6[2.5x2pi]^2

=320 J

The extra kinetic energy came from the work professor did to pull the arms and the dumbell inwards

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