A ring of radius 0.5 m and mass 10 kg is rotating about its diameter with an angular velocity of 20 rad/s. Its kinetic energy is

To find the rotational kinetic energy of the ring, we use the formula for rotational kinetic energy:
$K_R=\frac{1}{2}I{\omega }^2$

where:
- $I$ is the moment of inertia of the ring about the axis of rotation.
- $\omega$ is the angular velocity of the ring.

First, we determine the moment of inertia of a ring of mass $M$ and radius $R$ when rotating about its diameter. The standard moment of inertia of a thin ring about its central axis perpendicular to its plane is $M{R}^2$. According to the perpendicular axis theorem, the moment of inertia about any diameter in the plane of the ring is:
$I=\frac{1}{2}M{R}^2$

Given the values from the problem statement:
- Mass of the ring, $M=10\text{kg}$
- Radius of the ring, $R=0.5\text{m}$
- Angular velocity, $\omega =20\text{rad/s}$

Now, let's calculate the moment of inertia $I$:
$I=\frac{1}{2}\times 10\text{kg}\times \left(0.5{\text{m}}^2\right)$
$I=5\times 0.25=1.25{\text{kg·m}}^2$

Next, we substitute $I$ and $\omega$ into the rotational kinetic energy formula:
$K_R=\frac{1}{2}\times 1.25{\text{kg·m}}^2\times \left(20{\text{rad/s}}^2\right)$
$K_R=0.5\times 1.25\times 400$
$K_R=1.25\times 200=250\text{J}$

Therefore, the kinetic energy of the rotating ring is 250 J.