A ring of mass `3kg` is rolling Without Slipping with linear velocity `1 ms^(-1)` on a smooth horizontal surface. A rod of same mass is fitted along its one diameter. Find total kinetic energy of the system (in `J`).

Explanation

To find the total kinetic energy of the system consisting of a ring and a rod, we need to consider both the translational and rotational kinetic energies.

### Step 1: Identify the components of kinetic energy
The total kinetic energy (KE) of the system will be the sum of the translational kinetic energy (TKE) and the rotational kinetic energy (RKE).

### Step 2: Calculate the translational kinetic energy
The translational kinetic energy for the system is given by the formula:
\[ \text{TKE} = \frac{1}{2} m v^2 \]
where:
- \( m \) is the total mass of the system,
- \( v \) is the linear velocity.

Since both the ring and the rod have the same mass of \( 3 \, \text{kg} \), the total mass \( m \) is:
\[ m = 3 \, \text{kg} + 3 \, \text{kg} = 6 \, \text{kg} \]
Substituting the values:
\[ \text{TKE} = \frac{1}{2} \times 6 \, \text{kg} \times (1 \, \text{m/s})^2 = \frac{1}{2} \times 6 \times 1 = 3 \, \text{J} \]

### Step 3: Calculate the moment of inertia of the system
The moment of inertia \( I \) of the ring about its center is:
\[ I_{\text{ring}} = m r^2 = 3 \, \text{kg} \cdot r^2 \]
The moment of inertia \( I \) of the rod about its end (using the parallel axis theorem) is:
\[ I_{\text{rod}} = \frac{1}{3} m r^2 = \frac{1}{3} \cdot 3 \, \text{kg} \cdot r^2 = r^2 \]
Thus, the total moment of inertia \( I_{\text{total}} \) is:
\[ I_{\text{total}} = I_{\text{ring}} + I_{\text{rod}} = 3r^2 + r^2 = 4r^2 \]

### Step 4: Calculate the rotational kinetic energy
The rotational kinetic energy is given by:
\[ \text{RKE} = \frac{1}{2} I \omega^2 \]
We know that \( \omega \) can be related to \( v \) by the equation \( v = r \omega \), which gives:
\[ \omega = \frac{v}{r} \]
Substituting this into the RKE formula:
\[ \text{RKE} = \frac{1}{2} (4r^2) \left(\frac{v}{r}\right)^2 = \frac{1}{2} (4r^2) \left(\frac{1}{r^2}\right) = 2 \, \text{J} \]

### Step 5: Calculate the total kinetic energy
Now, we can find the total kinetic energy of the system:
\[ \text{Total KE} = \text{TKE} + \text{RKE} = 3 \, \text{J} + 2 \, \text{J} = 5 \, \text{J} \]

### Final Answer
The total kinetic energy of the system is \( 5 \, \text{J} \).

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