A light-weight boy holds two heavy dumb-bells of equal mass without stretched arms while standing on a turn-table which is rotating at an angular frequency $ω_{1}$ when the dumb-bells are at a distance r, from the axis of rotation. The boy suddenly pulls the dumb-bells towards his chest until they are at a distance r₂, from the axis of rotation. The new angular frequency of rotation $ω_{2}$ of the turn-table will be

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$ω_{1} (r_{2} / r_{1})$

$ω_{1} (r_{1} / r_{2})$

$ω_{1} {(r_{1} / r_{2})}^{2}$

$ω_{1} {(r_{2} / r_{1})}^{2}$

answer is C.

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Detailed Solution

If m be the mass of each dumb-bell. then total initial angular momentum L₁, is given by
$L_{1} = {mr}_{1}^{2} ω_{1} + {mr}_{1}^{2} ω_{1} = 2 {mr}_{1}^{2} ω_{1}$
When the boy pulls the dumb-bells towards his chest, let $ω_{2}$ , be the angular frequency. In this case
$L_{2} = {mr}_{2}^{2} ω_{2} + {mr}_{2}^{2} ω_{2} = 2 {mr}_{2}^{2} ω_{2}$
Applying the law of conservation of angular momentum, we have
$\begin{array}{l} 2 {mr}_{1}^{2} ω_{2} = 2 {mr}_{2}^{2} ω_{2} \\ \frac{ω_{1}}{ω_{2}} = \frac{r_{2}^{2}}{r_{1}^{2}} \end{array}$
or $ω_{2} = ω_{1} {(r_{1} / r_{2})}^{2}$