Final answer:
To find the moment of inertia for the ring, multiply the mass and square of the radius. Angular momentum is the product of moment of inertia and angular velocity. Rotational kinetic energy is half the moment of inertia multiplied by the square of angular velocity, resulting in 0.4 kg*m², 88 kg*m²/s, and 9680 J respectively for the provided ring.
Step-by-step explanation:
To calculate the moment of inertia (I), angular momentum (L), and rotational kinetic energy (KErot) for a ring rotating about its axis, we use the following formulas: Moment of inertia for a thin ring: I = m * r2, where m is the mass of the ring and r is the radius.Angular momentum: L = I * ω, where ω is the angular velocity in rad/s, which can be calculated from the rpm (revolutions per minute) by ω = rpm * 2π / 60.Rotational kinetic energy: KErot = (1/2) * I * ω2.
For a ring with a diameter of 0.4 m (radius of 0.2 m) and a mass of 10 kg rotating at 2100 rpm:Moment of inertia: I = 10 kg * (0.2 m)2 = 0.4 kg*m2.Converting rpm to rad/s: ω = 2100 * 2π / 60 ≈ 220 rad/s.Angular momentum: L = 0.4 kg*m2 * 220 rad/s = 88 kg*m2/s.Rotational kinetic energy: KErot = (1/2) * 0.4 kg*m2 * (220 rad/s)2 = 9680 J.