A uniform thin ring of mass 0.4kg rolls without slipping on a horizontal surface with a linear velocity of 10 cm/s. The kinetic energy of the ring is

Options

A. `4xx10^(-3)` joule
B. `4xx10^(-2)` joule
C. `2xx10^(-3)` joule
D. `2xx10^(-2)` joule

Answer

To find the kinetic energy of a uniform thin ring rolling without slipping on a horizontal surface, we need to consider both its translational and rotational kinetic energy. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Mass of the ring (m) = 0.4 kg
- Linear velocity (v) = 10 cm/s = 0.1 m/s (converting cm/s to m/s)

### Step 2: Write the formula for total kinetic energy
The total kinetic energy (KE) of the ring is the sum of its translational kinetic energy (TKE) and rotational kinetic energy (RKE):
\[ KE = TKE + RKE \]
Where:
- Translational Kinetic Energy (TKE) = \(\frac{1}{2} m v^2\)
- Rotational Kinetic Energy (RKE) = \(\frac{1}{2} I \omega^2\)

### Step 3: Find the moment of inertia (I) of the ring
For a thin ring, the moment of inertia about its central axis is given by:
\[ I = m r^2 \]
Where \(r\) is the radius of the ring. However, we can express the rotational kinetic energy in terms of linear velocity.

### Step 4: Relate angular velocity (ω) to linear velocity (v)
Using the condition for rolling without slipping:
\[ v = \omega r \implies \omega = \frac{v}{r} \]

### Step 5: Substitute ω in the RKE formula
Substituting \(\omega\) into the RKE formula:
\[ RKE = \frac{1}{2} I \omega^2 = \frac{1}{2} (m r^2) \left(\frac{v}{r}\right)^2 = \frac{1}{2} m r^2 \cdot \frac{v^2}{r^2} = \frac{1}{2} m v^2 \]

### Step 6: Combine TKE and RKE
Now, substituting TKE and RKE into the total kinetic energy formula:
\[ KE = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 = m v^2 \]

### Step 7: Calculate the kinetic energy
Now we can substitute the values of mass and velocity:
\[ KE = m v^2 = 0.4 \, \text{kg} \times (0.1 \, \text{m/s})^2 \]
Calculating \(v^2\):
\[ (0.1)^2 = 0.01 \, \text{m}^2/\text{s}^2 \]
Now substituting:
\[ KE = 0.4 \times 0.01 = 0.004 \, \text{J} = 4 \times 10^{-3} \, \text{J} \]

### Final Answer
The kinetic energy of the ring is:
\[ KE = 4 \times 10^{-3} \, \text{J} \]

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