A circular disc of mass 2 kg and radius 0.1 m is rotating at an angular speed of 2 rad/s, about an axis passing through its centre and perpendicular to its plane. What is its rotational kinetic energy?

Options

A. 0.1 J
B. 0.2 J
C. 0.02 J
D. 0.05 J

Explanation

To solve the problem of finding the rotational kinetic energy of a circular disc, we can follow these steps:

### Step 1: Understand the formula for rotational kinetic energy
The rotational kinetic energy (K.E.) of a rotating object is given by the formula:
\[ K.E. = \frac{1}{2} I \omega^2 \]
where \(I\) is the moment of inertia and \(\omega\) is the angular speed.

### Step 2: Calculate the moment of inertia for the disc
For a circular disc rotating about an axis through its center and perpendicular to its plane, the moment of inertia \(I\) is given by:
\[ I = \frac{1}{2} m r^2 \]
where \(m\) is the mass of the disc and \(r\) is its radius.

### Step 3: Substitute the values into the moment of inertia formula
Given:
- Mass \(m = 2 \, \text{kg}\)
- Radius \(r = 0.1 \, \text{m}\)

Substituting these values into the moment of inertia formula:
\[ I = \frac{1}{2} \times 2 \, \text{kg} \times (0.1 \, \text{m})^2 \]
Calculating this:
\[ I = \frac{1}{2} \times 2 \times 0.01 = 0.01 \, \text{kg m}^2 \]

### Step 4: Substitute the values into the kinetic energy formula
Now we need to substitute the moment of inertia \(I\) and the angular speed \(\omega\) into the kinetic energy formula. Given:
- Angular speed \(\omega = 2 \, \text{rad/s}\)

Calculating \(\omega^2\):
\[ \omega^2 = (2 \, \text{rad/s})^2 = 4 \, \text{rad}^2/\text{s}^2 \]

Now substituting \(I\) and \(\omega^2\) into the kinetic energy formula:
\[ K.E. = \frac{1}{2} \times 0.01 \, \text{kg m}^2 \times 4 \, \text{rad}^2/\text{s}^2 \]
Calculating this:
\[ K.E. = \frac{1}{2} \times 0.01 \times 4 = 0.02 \, \text{J} \]

### Final Answer
The rotational kinetic energy of the circular disc is:
\[ K.E. = 0.02 \, \text{J} \]

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