An ice skater has a moment of inertia of 2.3 kg * m^2 when his arms are outstretched and a moment of inertia of 1.0 kg * m^2 when his arms are tucked in close to his chest. If he starts to spin at an angular speed of 2.0 rps (revolutions per second) with his arms outstretched, what will his angular speed be when they are tucked in?
To solve this problem, we can apply the law of conservation of angular momentum. According to this law, the initial angular momentum of the skater with his arms outstretched should be equal to the final angular momentum of the skater when his arms are tucked in.
The formula for angular momentum is given by L = I * ω, where L represents angular momentum, I represents moment of inertia, and ω represents angular velocity.
Given that the initial angular momentum with arms outstretched is L1 = I1 * ω1, and the final moment of inertia with arms tucked in is I2, we can set up an equation:
L1 (initial) = L2 (final)
I1 * ω1 = I2 * ω2
Plugging in the values we have:
I1 = 2.3 kg * m^2 (moment of inertia with arms outstretched)
ω1 = 2.0 rps (angular speed with arms outstretched)
I2 = 1.0 kg * m^2 (moment of inertia with arms tucked in)
ω2 = unknown (angular speed with arms tucked in)
We can rearrange the equation to solve for ω2:
ω2 = (I1 * ω1) / I2
Substituting the given values:
ω2 = (2.3 kg * m^2 * 2.0 rps) / 1.0 kg * m^2
= 4.6 rps
Therefore, the skater's angular speed when his arms are tucked in will be 4.6 rps.