A figure skater is spinning with an initial angular speed of 5.6rad/s. He then pulls his arms in, reducing his moment of inertia to 0.31 times its original value. What is his angular speed after pulling in his arms?
conservation of momentum
I*5.6=(.31)I wf
wf=5.6/.31 rad/sec
To find the angular speed of the figure skater after pulling in his arms, we can use the principle of conservation of angular momentum. The angular momentum of an object remains constant when no external torques act 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.
Initially, the skater's moment of inertia is I1 and angular speed is ω1.
After pulling in his arms, the skater's moment of inertia becomes 0.31 times its original value, or I2 = 0.31 * I1.
Since angular momentum is conserved, we can write: L1 = L2.
L1 = I1 * ω1
L2 = I2 * ω2
Since L1 = L2, we can set the two equations equal to each other and solve for ω2:
I1 * ω1 = I2 * ω2
Substituting the values we know:
I1 * (5.6 rad/s) = (0.31 * I1) * ω2
Simplifying the equation:
5.6 rad/s = 0.31 * ω2
Dividing both sides by 0.31:
ω2 = (5.6 rad/s) / 0.31
Calculating the result:
ω2 ≈ 18.06 rad/s
Therefore, the figure skater's angular speed after pulling in his arms is approximately 18.06 rad/s.