A 10 m platform diver during his diving can increase his rotation rate from an initial rate of 0.8 rev/s to a maximum of 3.3 rev/s. If his initial moment of inertia was 10 kg·m2, what is his moment of inertia at his maximum rotation rate?
To find the moment of inertia at the maximum rotation rate, we need to use the conservation of angular momentum. Angular momentum is given by the formula:
L = Iω
where L represents the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Since angular momentum is conserved, we can say:
L1 = L2
Where L1 is the initial angular momentum and L2 is the angular momentum at the maximum rotation rate.
Given:
Initial angular velocity, ω1 = 0.8 rev/s
Initial moment of inertia, I1 = 10 kg·m2
Maximum angular velocity, ω2 = 3.3 rev/s
Moment of inertia at maximum rotation rate, I2 = ?
Let's substitute the values into the equation:
I1 * ω1 = I2 * ω2
10 kg·m2 * 0.8 rev/s = I2 * 3.3 rev/s
Now we can solve for I2:
I2 = (10 kg·m2 * 0.8 rev/s) / 3.3 rev/s
Simplifying:
I2 = 8 kg·m2 / 3.3
Calculating:
I2 ≈ 2.4242 kg·m2
Therefore, the moment of inertia at the maximum rotation rate is approximately 2.4242 kg·m2.