A 55.0-kg woman stands at the rim of a horizontal turntable having a moment of inertia of 540 kg · m2 and a radius of 2.00 m. The turntable is initially at rest and is free to rotate about a frictionless vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.50 m/s relative to the Earth.
(a) In what direction does the turntable rotate? counterclockwise or clockwise
With what angular speed does the turntable rotate? ? rad/s
(b) How much work does the woman do to set herself and the turntable into motion? ? J

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lcmendozaf lcmendozaf · Answer 1 · 2019-11-10T13:20:46+01:00

Answer:

a) ω = 0.305 rad/s counterclockwise

b) W = 61.875J

Explanation:

The angular speed of the woman is:

[tex]\omega_w=V_W/R = -1.5/2=-0.75rad/s[/tex]

By conservation of the angular momentum:

[tex](I_t+I_w)*\omega_o=I_t*\omega_t-I_w*\omega_w[/tex] where ωo = 0

Solving for [tex]\omega_w[/tex]:

[tex]\omega_w=I_w*\omega_w/I_t=m_w*R^2*\omega_w/I_t=0.305rad/s[/tex]

Work done by the woman is given by the variation on her kinetic energy:

[tex]W=\Delta K = 1/2*m*V_w^2-1/2*m*Vo^2=1/2*55*1.5^2=61.875J[/tex]