A 5.0 kg skater begins a spin with an angular speed of 4 rad/s. By changing the position of her arms, the skater decreases her moment of inertia to one half its initial value. What is the skater's final angular speed? I thank you in advance for your time. JL
I omega = constant angular momentum
I/2 (2 omega) = I omega
twice 4 = 8 radians/second
To determine the skater's final angular speed, we need to apply the conservation of angular momentum. Angular momentum (L) is given by the product of moment of inertia (I) and angular velocity (ω), so we have:
L = I * ω
According to the conservation of angular momentum, if the moment of inertia decreases, the angular velocity must increase to keep the angular momentum constant.
Let's denote the initial moment of inertia as Ii, the initial angular velocity as ωi, the final moment of inertia as If, and the final angular velocity as ωf.
Given:
Ii = initial moment of inertia = 5.0 kg
ωi = initial angular speed = 4 rad/s
If = final moment of inertia = 1/2 * Ii (since it decreases to one-half its initial value)
From the conservation of angular momentum, we have:
L_initial = L_final
Ii * ωi = If * ωf
Substituting the given values:
5.0 kg * 4 rad/s = (1/2 * 5.0 kg) * ωf
Now we can solve for ωf:
20 kg·rad/s = (2.5 kg) * ωf
ωf = 20 kg·rad/s / 2.5 kg
ωf = 8 rad/s
Therefore, the skater's final angular speed is 8 rad/s.