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A horizontal platform in the shape of a circular disc rotates on a frictionless bearing about a vertical axle through the centre of the platform. The platform has a mass of 150 kg, a radius of 2.0 m, and a rotational inertia of 300 kg $m^{2}$ about the axis of rotation. A 60 kg student walks slowly from the rim of the platform toward the centre. If the angular speed of the system is 1.5 rad/s when the student starts at the rim, what is the angular speed when she is 0.50 m from the centre?

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1.2 rad/s

2.6 rad/s

1.5 rad/s

3.6 rad/s

answer is B.

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Detailed Solution

The initial rotational inertia of the system is $I_{i} = I_{disk}$ $+ I_{student,} where I_{disk} = 300 kg m^{2}$ and $I_{student} = {mR}^{2}$ where m = 60 kg and R = 2.0 m.

The rotational inertia when the student reaches r = 0.5 m is $I_{f} = I_{disk} + {mr}^{2}$ . Angular momentum conservation leads to

$I_{i} ω_{i} = I_{f} ω_{f} \Rightarrow ω_{f} = ω_{i} (\frac{I_{disk} + {mR}^{2}}{I_{disk} + {mr}^{2}})$

which yields, for $ω_{i} = 1.5 rad / s, a final angular velocity of ω_{f} = 2.6 rad / s .$

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