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Question

A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the centre of the disk. The platform has a mass of $150 k g,$ a radius of $2.0 m$ , and a rotational inertia of $300 k g m^{2}$ about the axis of rotation. A $60 k g$ student walks slowly from the rim of the platform toward the center. If the angular speed of the system is $1.5 r a d / s$ when the student starts at the rim, what is the angular speed when she is $0.5 m$ from the center?

1.2 r a d / s

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2.6 r a d / s

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3.6 r a d / s

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1.5 r a d / s

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Solution

The correct option is B $2.6 r a d / s$
Given: $m_{d i s c} = 150 k g, R = 2 m, I_{d i s c} = 300 k g m^{2},$ $m_{s t u d e n t} = 60 k g, ω_{i} = 1.5 r a d / s, r = 0.5 m$
The initial rotational inertia of the system is

$I_{i} = I_{d i s k} + I_{s t u d e n t}$

Where $I_{d i s k} = 300 K g m^{2} a n d R = 2 m$

The rotational inertia when the student reaches $r = 0.5 m$ is

$I_{f} = I_{d i s k} + m r^{2}$

Angular momentum conservation leads to

$I_{i} ω_{i} = I_{f} ω_{f}$

$ω_{f} = \frac{ω_{i} (I_{d i s k} + m R^{2})}{I_{d i s k} + m r^{2}} = \frac{1.5 (300 + 60 \times 2^{2})}{300 + 60 \times {0.5}^{2}}$

$ω_{f} = 2.6 r a d / s$

FINAL ANSWER: (b)

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