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Sergio039 [100]
3 years ago
10

A skater extends her arms horizontally, holding a 5-kg mass in each hand. She is rotating about a vertical axis with an angular

velocity of one revolution per second. If she drops her hands to her sides, what will the final angular velocity (in rev/s) be if her moment of inertia remains approximately constant at 5 kg⋅m2, and the distance of the masses from the axis changes from 1 m to 0.1 m?

Physics
2 answers:
Crazy boy [7]3 years ago
7 0

The answer is attached

sp2606 [1]3 years ago
3 0

Answer:

1.98 rev/s

Explanation:

m  = mass attached to each hand = 5 kg

r_{i} = initial distance of masses in each hand = 1 m

r_{f} = final distance of masses in each hand = 0.1 m

I = moment of inertia of body = 5 kgm²

I_{i} = initial total moment of inertia = I + 2 mr_{i}^{2}

w_{i} = initial angular velocity = 1 rev/s

I_{f} = final total moment of inertia = I + 2 mr_{f}^{2}

w_{f} = final angular velocity = ?

Using conservation of angular momentum

I_{i} w_{i} = I_{f} w_{f}

(I + 2 mr_{i}^{2}) w_{i} = (I + 2 mr_{f}^{2}) w_{f}

(5 + 2 (5) (1)^{2}) (1) = (5 + 2 (5) (0.1)^{2}) w_{f}

w_{f} = 2.94 rev/s

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