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A man stands on a rotating platform with his arms stretched holding a $5 kg$ weight in each hand. The angular speed of the platform is $1.2 {revs}^{- 1}$ . The moment of inertia of the man together with the platform may be taken to be constant and equal to $6 kg m^{2}$ . If the man brings his arms close to his chest with the distance of each weight from the axis changing from $100 cm$ to $20 cm$ . The new angular speed of the platform is

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$2 {revs}^{- 1}$

$3 {revs}^{- 1}$

5 ${revs}^{- 1}$

$6 {revs}^{- 1}$

answer is B.

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Detailed Solution

Initial moment of inertia, $I_{1} = 6 + [2 \times 5 \times (1)^{2}] = 16 kg m^{2}$

Initial angular velocity, $ω_{1} = 1.2 {revs}^{- 1}$

Initial angular momentum, $L_{1} = I_{1} ω_{1}$

Final moment of inertia, $I_{2} = 6 + [2 \times 5 \times (0.2)^{2}] = 6.4 kg m^{2}$

Final angular speed = $ω_{2}$

Final angular momentum, $L_{2} = I_{2} ω_{2}$

According to law of conservation of angular momentum,

$L_{1} = L_{2} or I_{1} ω_{1} = I_{2} ω_{2} ω_{2} = \frac{I_{1} ω_{1}}{I_{2}} = \frac{(16 kg m^{2}) (1.2 {revs}^{- 1})}{(6.4 kg m^{2})} = 3 {revs}^{- 1}$

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