A ring and a disc having the same mass, roll without slipping with the same linear velocity. If the kinetic energy of the ring is 8 j , Find the kinetic energy of disc (in J)

Answer

To find the kinetic energy of the disc given that the kinetic energy of the ring is 8 joules, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Kinetic Energy Formula**:
The total kinetic energy (KE) of a rolling object can be expressed as the sum of translational and rotational kinetic energy. For a rolling object, the formula is:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
where \(m\) is the mass, \(v\) is the linear velocity, \(I\) is the moment of inertia, and \(\omega\) is the angular velocity.

2. **Relate Linear and Angular Velocity**:
Since the ring and disc roll without slipping, we have the relation:
\[ \omega = \frac{v}{R} \]
where \(R\) is the radius of the object.

3. **Kinetic Energy of the Ring**:
The moment of inertia \(I\) for a ring is given by:
\[ I_{ring} = mR^2 \]
Therefore, the kinetic energy of the ring can be expressed as:
\[ KE_{ring} = \frac{1}{2} mv^2 + \frac{1}{2} (mR^2) \left(\frac{v}{R}\right)^2 \]
Simplifying this, we find:
\[ KE_{ring} = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 = mv^2 \]
Given that \(KE_{ring} = 8 \, J\), we have:
\[ mv^2 = 8 \, J \]

4. **Kinetic Energy of the Disc**:
The moment of inertia \(I\) for a disc is given by:
\[ I_{disc} = \frac{1}{2} mR^2 \]
Thus, the kinetic energy of the disc can be expressed as:
\[ KE_{disc} = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} mR^2\right) \left(\frac{v}{R}\right)^2 \]
Simplifying this, we find:
\[ KE_{disc} = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]

5. **Finding the Kinetic Energy of the Disc**:
Now we can substitute \(mv^2\) from the kinetic energy of the ring:
\[ KE_{disc} = \frac{3}{4} (8 \, J) = 6 \, J \]

### Final Answer:
The kinetic energy of the disc is \(6 \, J\).