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 know the answer to be 8 rad/s, however, I thank you in advance for a detailed response. JL
if angular momentum is conserved
Initialangulmomentum=I*w
= I*4
final angular momentum=
1/2 I * wf=I*4
wf=8 rad/sec
To solve this problem, we will use the principle of conservation of angular momentum. According to this principle, the initial angular momentum of a system remains constant unless an external torque acts on it.
The angular momentum (L) of an object is given by the product of its moment of inertia (I) and its angular velocity (ω): L = Iω.
Initially, the skater has a certain angular momentum (L1) with a moment of inertia (I1) and an angular speed (ω1). We can write this as L1 = I1ω1.
When the skater changes the position of her arms, the moment of inertia of her body decreases to one half its initial value, i.e., I2 = 0.5I1.
According to the conservation of angular momentum, L1 = L2. Rewriting this equation in terms of moment of inertia and angular velocity, we get I1ω1 = I2ω2.
Substituting the values, we have:
I1ω1 = (0.5I1)ω2.
Simplifying the equation, I1 cancels out:
ω1 = 0.5ω2.
Now we can solve for ω2, which is the final angular speed:
ω2 = 2ω1.
Given that ω1 = 4 rad/s, we can substitute this value to find ω2:
ω2 = 2(4) = 8 rad/s.
Therefore, the skater's final angular speed is 8 rad/s.