Question

Фигурист с расставленными руками вращается со скоростью 2 об/с, при этом его момент инерции равен 2,94 кг·м2. С какой скоростью он будет вращаться, если резко согнет руки, тем самым уменьшая момент инерции до 0,98 кг·м2?

Answer 1

Iw=const, где i - момент инерции, w - угловая скорость
6 об/с

Answer 2

Calculation of Final Angular Velocity

To calculate the final angular velocity of the figure skater after bending their arms and reducing the moment of inertia, we can use the principle of conservation of angular momentum. The initial angular momentum of the skater is equal to the final angular momentum after bending the arms.

The formula for angular momentum is:

L = Iω

Where: - L is the angular momentum - I is the moment of inertia - ω is the angular velocity

Initially, the skater has a moment of inertia of 2.94 kg·m^2 and an angular velocity of 2 revolutions per second. After bending the arms, the moment of inertia reduces to 0.98 kg·m^2.

Using the conservation of angular momentum, we can set up the equation:

I1ω1 = I2ω2

Where: - I1 is the initial moment of inertia (2.94 kg·m^2) - ω1 is the initial angular velocity (2 revolutions per second) - I2 is the final moment of inertia (0.98 kg·m^2) - ω2 is the final angular velocity (to be calculated)

Solving for ω2, we can rearrange the equation:

ω2 = (I1 * ω1) / I2

Substituting the given values:

ω2 = (2.94 kg·m^2 * 2 revolutions per second) / 0.98 kg·m^2

Calculating the result:

ω2 = 5.98 revolutions per second

Therefore, the figure skater will rotate with a final angular velocity of approximately 5.98 revolutions per second after bending their arms and reducing the moment of inertia to 0.98 kg·m^2.