A disc of mass 4 kg and of radius 1 m rolls on a horizontal surface without slipping such that the velocity of its centre of mass is `10" cm sec"^(-1)`, Its rotatonal kinetic energy is

Options

A. 0.005 J
B. 0.02 J
C. 0.03 J
D. 0.01 J

Answer

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

### Step 1: Identify the given values
- Mass of the disc (m) = 4 kg
- Radius of the disc (r) = 1 m
- Velocity of the center of mass (V_cm) = 10 cm/s = 0.1 m/s (conversion from cm/s to m/s)

### Step 2: Convert the velocity to meters per second
Since the velocity is given in cm/s, we convert it to m/s:
\[ V_{cm} = 10 \, \text{cm/s} = 10 \times 10^{-2} \, \text{m/s} = 0.1 \, \text{m/s} \]

### Step 3: Relate linear velocity to angular velocity
For a disc rolling without slipping, the relationship between linear velocity (V_cm) and angular velocity (ω) is given by:
\[ V_{cm} = r \cdot \omega \]
From this, we can solve for ω:
\[ \omega = \frac{V_{cm}}{r} = \frac{0.1 \, \text{m/s}}{1 \, \text{m}} = 0.1 \, \text{rad/s} \]

### Step 4: Calculate the moment of inertia (I) of the disc
The moment of inertia (I) for a solid disc about its central axis is given by:
\[ I = \frac{1}{2} m r^2 \]
Substituting the values:
\[ I = \frac{1}{2} \times 4 \, \text{kg} \times (1 \, \text{m})^2 = \frac{1}{2} \times 4 = 2 \, \text{kg m}^2 \]

### Step 5: Calculate the rotational kinetic energy (K.E._rotational)
The formula for rotational kinetic energy is:
\[ K.E._{rotational} = \frac{1}{2} I \omega^2 \]
Substituting the values we found:
\[ K.E._{rotational} = \frac{1}{2} \times 2 \, \text{kg m}^2 \times (0.1 \, \text{rad/s})^2 \]
\[ K.E._{rotational} = \frac{1}{2} \times 2 \times 0.01 \]
\[ K.E._{rotational} = 1 \times 0.01 = 0.01 \, \text{J} \]

### Final Answer
The rotational kinetic energy of the disc is:
\[ K.E._{rotational} = 0.01 \, \text{J} \]