First slide

Law of conservation of angular momentum

Question

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 rev $s^{- 1}$ . 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

Moderate

A

2 rev $s^{- 1}$
B

3 rev $s^{- 1}$
C

5 rev $s^{- 1}$
D

6 rev $s^{- 1}$

Solution

Initial moment of inertia,
   $I_{1} = 6 + [2 \times 5 \times {(1)}^{2}] = 16 kg m^{2}$
Initial angular velocity, $ω_{1} = 1.2 rev s^{- 1}$
Initial angular momentum is
   $L_{1} = I_{1} ω_{1}$
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 is
   $L_{2} = I_{2} ω_{2}$
According to law of conservation of angular momentum,
$L_{1} = L_{2} or I_{1} ω_{1} = I_{2} ω_{2}$
$ω_{2} = \frac{I_{1} ω_{1}}{I_{2}} = \frac{(16 kg m^{2}) (1.2 rev s^{- 1})}{(6.4 kg m^{2})} = 3 rev s^{- 1}$

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