Question

A 50 kg woman stands at the rim of a horizontal turntable having
a moment of inertia of 560 kg·m^{2} and a radius of 2.0 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 1.5 m/s relative to the Earth.

(a) In what direction and with what angular speed does the turntable rotate?

clockwise or counterclockwise

rad/s

(b) How much work does the woman do to set herself and the
turntable into motion?

J

Answer #1

(a) We know that, L_{initial} = L_{final}

(L_{1} + L_{2})_{i} = (L_{1} +
L_{2})_{f}

since, initially the system is at rest.

0 = (L_{1} + L_{2})_{f}

0 = I_{1}_{1} +
I_{2}_{2}

_{1} = -
I_{2}_{2} /
I_{1} { eq.1 }

_{1} = -
(m r^{2}) (v/r) / I_{1}

_{1} = -
m r v / I_{1} = - [(50 kg) (2 m) (1.5 m/s)] / (560
kg.m^{2})

_{1} =
0.267 rad/s

{ it's direction will be *counter clockwise*
}

(b) How much work does the woman do to set herself & the turntable into motion?

we know that, W = K.E

W =
(K.E_{final} - K.E_{initial})

W = [(1/2)
I_{1}_{1}^{2}
+ (1/2) m v^{2}] - (0 J)

W = [(0.5) (560
kg.m^{2}) (-0.267 rad/s)^{2} + (0.5) (50 kg) (-1.5
m/s)^{2}]

**W = 76.2
J**

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