A figure skater is spinning at a rate of 1.7 rev/s with her arms outstretched. She then draws her arms in to her chest, reducing her rotational inertia to 64% of its original value. What is her new rate of rotation?

Use conservation of angular momentum

L1*W1 = L2*W2

We're solving for W2

From the information given in the question:

L2 = 64*(L1/100) = .64*L1

W2 = (L1*W1)/L2

So plugging this into the momentum equation:

W2 = (L1*1.7)/(.64*L1)

We can kill the L1 on top and bottom leaving us with:

W2 = 1.7/.64

W2 = ___ rev/s

To find the figure skater's new rate of rotation, we can use the principle of conservation of angular momentum.

The formula for angular momentum is:
L = I * ω

Where:
L is the angular momentum
I is the rotational inertia
ω (omega) is the angular velocity

Initially, the skater has an angular momentum (L1) with the initial rotational inertia (I1) and the initial angular velocity (ω1).

When the skater draws her arms in, the rotational inertia reduces to 64% of its original value, meaning the new rotational inertia (I2) is 0.64 times I1.

Since angular momentum is conserved, we have:
L1 = L2

Substituting the formula for angular momentum, we get:
I1 * ω1 = I2 * ω2

We are given that the initial angular velocity (ω1) is 1.7 rev/s and the ratio of the new rotational inertia to the initial rotational inertia is 0.64.

Substituting these values into the equation:
I1 * 1.7 rev/s = (0.64 * I1) * ω2

Simplifying the equation:
1.7 rev/s = 0.64 * ω2

Dividing both sides of the equation by 0.64:
ω2 = (1.7 rev/s) / 0.64

Calculating the value of ω2:
ω2 ≈ 2.656 rev/s

Therefore, the skater's new rate of rotation is approximately 2.656 revolutions per second.

To determine the new rate of rotation, we need to apply the law of conservation of angular momentum. According to this principle, the product of rotational inertia and angular velocity remains constant as long as no external torques act on the system.

The angular momentum (L) is given by the equation:

L = Iω

where L is the angular momentum, I is the rotational inertia, and ω is the angular velocity.

Initially, the skater is rotating with an angular velocity (ω_initial) of 1.7 rev/s. Now, when the skater draws her arms in, her rotational inertia (I) decreases to 64% of its original value. Let's denote the final angular velocity as ω_final.

Since angular momentum is conserved, we can equate the initial and final angular momenta:

I_initial * ω_initial = I_final * ω_final

Given that the final rotational inertia (I_final) is 64% of the initial rotational inertia (I_initial), we can express it as:

I_final = 0.64 * I_initial

Now, substituting the values into the equation:

I_initial * ω_initial = (0.64 * I_initial) * ω_final

Simplifying this equation, we find:

ω_final = (I_initial * ω_initial) / (0.64 * I_initial)

ω_final = ω_initial / 0.64

Now, we can substitute the given value of ω_initial = 1.7 rev/s into the equation:

ω_final = 1.7 rev/s / 0.64

Calculating this value, we get:

ω_final ≈ 2.656 rev/s

Therefore, the new rate of rotation for the figure skater after drawing her arms in is approximately 2.656 revolutions per second.