A figure skater is spinning with an initial angular speed of 6.9rad/s. She then pulls her arms in, reducing her moment of inertia to 0.38 times its original value. What is her angular speed after pulling in her arms?

To solve this problem, we can use the conservation of angular momentum.

Angular momentum (L) is the product of moment of inertia (I) and angular speed (ω). According to the conservation of angular momentum, the initial angular momentum should be equal to the final angular momentum.

The formula for angular momentum is:

L = I * ω,

where L is angular momentum, I is moment of inertia, and ω is angular speed.

Initially, the skater's angular speed is 6.9 rad/s. Let's denote this as ω1, and the initial moment of inertia as I1.

After pulling her arms in, her moment of inertia becomes 0.38 times its original value, denoted as I2 = 0.38 * I1. We need to find the final angular speed, denoted as ω2.

According to the conservation of angular momentum:

I1 * ω1 = I2 * ω2.

Plugging in the given values, we have:

I1 * 6.9 rad/s = 0.38 * I1 * ω2.

Divide both sides by I1:

6.9 rad/s = 0.38 * ω2.

Now, solve for ω2:

ω2 = 6.9 rad/s / 0.38.

Evaluating the expression, we find:

ω2 ≈ 18.16 rad/s.

Therefore, the skater's angular speed after pulling in her arms is approximately 18.16 rad/s.

conservation of momentum:

I1*W1=I2*w2