An ice skater spins about a vertical axis at an angular speed of 15 rad s−1 when her arms are

outstretched. She then quickly pulls her arms into her sides in a very small time interval so
that the frictional forces due to ice are negligible. Her initial moment of inertia about the
axis of rotation is 1.72 kg m2 and her final moment of inertia is 0.61 kg m2. What is the
change in her angular speed? What is the change in her kinetic energy? Explain this change
in kinetic energy.

To answer the first question - what is the change in her angular speed - we can use the principle of conservation of angular momentum. According to this principle, the angular momentum of an object remains constant when no external torques are acting on it.

The formula for angular momentum is given by:

L = Iω

Where:
L is the angular momentum,
I is the moment of inertia,
and ω is the angular speed.

In this case, the ice skater's initial angular momentum is equal to her final angular momentum. Since the only change in the system is the moment of inertia, we can write:

I₁ω₁ = I₂ω₂

Where:
I₁ is the initial moment of inertia (1.72 kg m²),
ω₁ is the initial angular speed (15 rad/s),
I₂ is the final moment of inertia (0.61 kg m²),
and ω₂ is the final angular speed (which we need to find).

Rearranging the equation:

ω₂ = (I₁ω₁) / I₂

Substituting the given values:

ω₂ = (1.72 kg m² * 15 rad/s) / 0.61 kg m²

ω₂ = 42.295 rad/s

So, the change in her angular speed is 42.295 rad/s - 15 rad/s = 27.295 rad/s.

Now, moving on to the second question - what is the change in her kinetic energy. The kinetic energy of a rotating object is given by the formula:

KE = (1/2) * I * ω²

Where:
KE is the kinetic energy,
I is the moment of inertia,
and ω is the angular speed.

To find the change in kinetic energy, we can subtract the initial kinetic energy from the final kinetic energy:

ΔKE = KE₂ - KE₁
ΔKE = (1/2) * I₂ * ω₂² - (1/2) * I₁ * ω₁²

Substituting the given values:

ΔKE = (1/2) * 0.61 kg m² * (42.295 rad/s)² - (1/2) * 1.72 kg m² * (15 rad/s)²

ΔKE = 904.28 J - 191.25 J

ΔKE ≈ 713.03 J

Therefore, the change in her kinetic energy is approximately 713.03 J.

The change in kinetic energy is positive, indicating that her kinetic energy has increased. This increase in kinetic energy is due to the decrease in moment of inertia. When the skater pulls her arms into her sides, her moment of inertia decreases, allowing her to spin faster. Since kinetic energy is directly proportional to the square of the angular speed, the increase in speed leads to an increase in kinetic energy.

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