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

Options

A. `2:1`
B. `1:2`
C. `1:1`
D. `4:1`

Answer

To find the ratio of linear kinetic energy (K_T) to rotational kinetic energy (K_R) for a ring rolling down an inclined plane, we can follow these steps:

### Step 1: Understand the Kinetic Energies
The total kinetic energy of a rolling ring consists of two components:
1. **Linear Kinetic Energy (K_T)**: This is due to the translational motion of the center of mass of the ring.
\[ K_T = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the ring and \( v \) is the linear velocity of the center of the ring.

2. **Rotational Kinetic Energy (K_R)**: This is due to the rotation of the ring about its center.
\[ K_R = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the ring and \( \omega \) is the angular velocity.

### Step 2: Moment of Inertia of the Ring
For a ring of mass \( m \) and radius \( r \), the moment of inertia about its central axis is given by:
\[ I = m r^2 \]

### Step 3: Relationship Between Linear and Angular Velocity
For a ring rolling without slipping, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ v = r \omega \quad \Rightarrow \quad \omega = \frac{v}{r} \]

### Step 4: Substitute \( \omega \) in the Rotational Kinetic Energy
Substituting \( \omega \) in the expression for \( K_R \):
\[ K_R = \frac{1}{2} I \omega^2 = \frac{1}{2} (m r^2) \left(\frac{v}{r}\right)^2 \]
\[ K_R = \frac{1}{2} m r^2 \cdot \frac{v^2}{r^2} = \frac{1}{2} m v^2 \]

### Step 5: Calculate the Ratio of Kinetic Energies
Now we can find the ratio of linear kinetic energy to rotational kinetic energy:
\[ \text{Ratio} = \frac{K_T}{K_R} = \frac{\frac{1}{2} m v^2}{\frac{1}{2} m v^2} \]
\[ \text{Ratio} = \frac{1}{1} = 1 \]

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