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Who spent hours watching the recent winter olympics? During the spin sequence, a skater starts spinning at 16.1 rad/s with their arms extended, producing a moment of inertia of 2.75 kg·m2. Then, the skaters pulls their arms in close to their body, so their moment of inertia is reduced to 1.70 kg·m2. What is their final angular speed? (Friction may be ignored).

Respuesta :

Answer:

The final angular speed is 26.0 rad/s.

Explanation:

Given that,

Initial angular velocity = 161.1 rad/s

Moment of inertia = 2.75 kg m²

Reduced moment of inertia = 1.70 kg m²

We need to calculate the angular velocity

Using conservation of momentum

[tex]I_{i}\omega_{i}=I_{f}\omega_{f}[/tex]

[tex]\omega_{f}=\dfrac{I_{i}\omega_{i}}{I_{f}}[/tex]

Put the value into the formula

[tex]\omega_{f}=\dfrac{2.75\times16.1}{1.70}[/tex]

[tex]\omega_{f}=26.0\ rad/s[/tex]

Hence, The final angular speed of skater is 26.0 rad/s.

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