A ring is rolling on an inclined plane. The ratio of the linear and rotational kinetic energies will be

Options

A. (a)`2:1`
B. (b)`1:2`
C. (c)`1:1`
D. (d)`4:1`

Answer

To find the ratio of the linear and rotational kinetic energies of a ring rolling down an inclined plane, we can follow these steps:

### Step 1: Define Linear Kinetic Energy
The linear kinetic energy (KE_linear) of a ring is given by the formula:
\[ KE_{\text{linear}} = \frac{1}{2} mv^2 \]
where \(m\) is the mass of the ring and \(v\) is its linear velocity.

### Step 2: Define Rotational Kinetic Energy
The rotational kinetic energy (KE_rotational) of a ring is given by the formula:
\[ KE_{\text{rotational}} = \frac{1}{2} I \omega^2 \]
where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.

### Step 3: Calculate Moment of Inertia for a Ring
The moment of inertia \(I\) for a ring about its central axis is:
\[ I = mr^2 \]
where \(r\) is the radius of the ring.

### Step 4: Relate Angular Velocity to Linear Velocity
For a ring rolling without slipping, the relationship between linear velocity \(v\) and angular velocity \(\omega\) is:
\[ \omega = \frac{v}{r} \]

### Step 5: Substitute Angular Velocity into Rotational Kinetic Energy
Substituting \(\omega\) into the rotational kinetic energy formula:
\[ KE_{\text{rotational}} = \frac{1}{2} (mr^2) \left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ KE_{\text{rotational}} = \frac{1}{2} m r^2 \cdot \frac{v^2}{r^2} = \frac{1}{2} mv^2 \]

### Step 6: Find the Ratio of Kinetic Energies
Now, we can find the ratio of the rotational kinetic energy to the linear kinetic energy:
\[ \text{Ratio} = \frac{KE_{\text{rotational}}}{KE_{\text{linear}}} = \frac{\frac{1}{2} mv^2}{\frac{1}{2} mv^2} = 1 \]

### Conclusion
Thus, the ratio of the linear kinetic energy to the rotational kinetic energy for a ring rolling down an inclined plane is:
\[ \text{Ratio} = 1:1 \]