Final answer:
The moment of inertia of the ring is 0.2 kg*m^2, the angular momentum is 2 kg*m^2/s, and the rotational kinetic energy is 10 J.
Step-by-step explanation:
To calculate the moment of inertia, angular momentum, and rotational kinetic energy of a ring, we need to use the appropriate formulas.
First, we will calculate the moment of inertia. The moment of inertia for a ring rotating about its axis is given by:
I = 1/2 * m * r^2
where m is the mass of the ring and r is the radius.
Using the values given in the question, the moment of inertia is:
I = 1/2 * 10 kg * (0.2 m)^2 = 0.2 kg * m^2.
Next, we can calculate the angular momentum. The angular momentum of an object rotating about its axis is given by:
L = I * ω
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity in radians per second.
Using the given angular velocity of the ring, we have:
L = 0.2 kg * m^2 * 10 rad/s = 2 kg * m^2/s.
Finally, we can calculate the rotational kinetic energy. The rotational kinetic energy of an object rotating about its axis is given by:
KE = 1/2 * I * ω^2
where KE is the rotational kinetic energy.
Plugging in the values, we get:
KE = 1/2 * 0.2 kg * m^2 * (10 rad/s)^2 = 10 J.