A ring of mass 1kg and diameter 1m is rolling on a plane road with a speed 2m/s. Its kinetic energy would be -

Options

A. 1 Joule
B. 4 Joule
C. 2 Joule
D. 0.5 Joule

Answer

To solve the problem, we need to calculate the total kinetic energy of a ring that is rolling on a plane road. The total kinetic energy will consist of both translational and rotational kinetic energy.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass (m) of the ring = 1 kg
- Diameter of the ring = 1 m, therefore, radius (r) = 0.5 m
- Speed (v) of the ring = 2 m/s

2. **Understand the types of kinetic energy:**
- The ring has both translational kinetic energy (due to its linear motion) and rotational kinetic energy (due to its rotation about its center).
- The formula for translational kinetic energy (TKE) is:
\[ TKE = \frac{1}{2} m v^2 \]
- The formula for rotational kinetic energy (RKE) is:
\[ RKE = \frac{1}{2} I \omega^2 \]
- For a ring, the moment of inertia (I) is given by:
\[ I = m r^2 \]
- The angular velocity (\(\omega\)) can be related to linear velocity (v) using:
\[ \omega = \frac{v}{r} \]

3. **Calculate the translational kinetic energy:**
\[ TKE = \frac{1}{2} \times 1 \, \text{kg} \times (2 \, \text{m/s})^2 = \frac{1}{2} \times 1 \times 4 = 2 \, \text{J} \]

4. **Calculate the rotational kinetic energy:**
- First, calculate the moment of inertia (I):
\[ I = m r^2 = 1 \, \text{kg} \times (0.5 \, \text{m})^2 = 1 \times 0.25 = 0.25 \, \text{kg m}^2 \]
- Next, calculate \(\omega\):
\[ \omega = \frac{v}{r} = \frac{2 \, \text{m/s}}{0.5 \, \text{m}} = 4 \, \text{rad/s} \]
- Now, substitute I and \(\omega\) into the RKE formula:
\[ RKE = \frac{1}{2} I \omega^2 = \frac{1}{2} \times 0.25 \, \text{kg m}^2 \times (4 \, \text{rad/s})^2 = \frac{1}{2} \times 0.25 \times 16 = 2 \, \text{J} \]

5. **Calculate the total kinetic energy:**
\[ \text{Total Kinetic Energy} = TKE + RKE = 2 \, \text{J} + 2 \, \text{J} = 4 \, \text{J} \]

### Final Answer:
The total kinetic energy of the ring is **4 Joules**.

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