Question

The mass of a flywheel is \(5.6 \times 10^{4} \mathrm{kg} .\) This particular flywheel has its mass concentrated at the rim of the wheel. If the radius of the wheel is \(2.6 \mathrm{m}\) and it is rotating at 350 rpm, what is the magnitude of its angular momentum?

Short answer

Answer: The magnitude of the flywheel's angular momentum is approximately 1.3837 * 10^7 kg*m^2/s.

Step-by-step solution

Determine the moment of inertia of the flywheel

Using the formula for a hoop's moment of inertia, calculate the moment of inertia (I) of the flywheel: I = MR^2 M = 5.6 * 10^4 kg (mass of the flywheel) R = 2.6 m (radius of the flywheel) I = (5.6 * 10^4 kg) * (2.6 m)^2 I = 3.7776 * 10^5 kg * m^2 (moment of inertia)

Convert rotational speed to angular velocity

The given rotational speed of the flywheel is 350 rpm. To convert rpm to radians per second, we can use the formula: ω = (2 * π radians * rotational_speed) / 60 seconds rotational_speed = 350 rpm ω = (2 * π * 350) / 60 ω ≈ 36.65 rad/s (angular velocity)

Calculate the angular momentum

Now that we have the moment of inertia and angular velocity, we can calculate the angular momentum (L) of the flywheel using the formula: L = Iω L = (3.7776 * 10^5 kg*m^2) * (36.65 rad/s) L ≈ 1.3837 * 10^7 kg*m^2/s The magnitude of the flywheel's angular momentum is approximately 1.3837 * 10^7 kg*m^2/s.