A disc of mass 2 kg is rolling on a horizontal surface without slipping with a velocity of 0.1 m/s. What is its rotational kinetic energy?

Options

A. `5xx10^(-3)J`
B. `2.5xx10^(-3)J`
C. `15xx10^(-3)J`
D. `8xx10^(-3)J`

Explanation

To find the rotational kinetic energy of a disc rolling on a horizontal surface without slipping, we can follow these steps:

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

### Step 2: Determine the moment of inertia for the disc
For a solid disc of mass \(m\) and radius \(r\), the moment of inertia about its central axis is:
\[ I = \frac{1}{2} m r^2 \]
Given that the mass \(m = 2 \, \text{kg}\), we can substitute this value into the equation later.

### Step 3: Relate angular velocity to linear velocity
Since the disc is rolling without slipping, the relationship between linear velocity \(v\) and angular velocity \(\omega\) is given by:
\[ \omega = \frac{v}{r} \]
Given that the linear velocity \(v = 0.1 \, \text{m/s}\), we can express \(\omega\) in terms of \(r\):
\[ \omega = \frac{0.1}{r} \]

### Step 4: Substitute \(I\) and \(\omega\) into the rotational kinetic energy formula
Substituting \(I\) and \(\omega\) into the rotational kinetic energy formula:
\[ K_{\text{rot}} = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{0.1}{r}\right)^2 \]
This simplifies to:
\[ K_{\text{rot}} = \frac{1}{4} m \cdot \frac{0.01}{r^2} \cdot r^2 \]
The \(r^2\) terms cancel out:
\[ K_{\text{rot}} = \frac{1}{4} m \cdot 0.01 \]

### Step 5: Substitute the mass into the equation
Now, substituting \(m = 2 \, \text{kg}\):
\[ K_{\text{rot}} = \frac{1}{4} \cdot 2 \cdot 0.01 \]
Calculating this gives:
\[ K_{\text{rot}} = \frac{2 \cdot 0.01}{4} = \frac{0.02}{4} = 0.005 \, \text{J} \]

### Step 6: Convert to standard scientific notation
The value \(0.005 \, \text{J}\) can be expressed in scientific notation as:
\[ K_{\text{rot}} = 5 \times 10^{-3} \, \text{J} \]

### Final Answer
The rotational kinetic energy of the disc is:
\[ \boxed{5 \times 10^{-3} \, \text{J}} \]