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, we need to consider both translational and rotational kinetic energy. Here’s a step-by-step solution:

### Step 1: Identify the formulas for kinetic energy
The total kinetic energy (KE) of a rolling object is the sum of its translational kinetic energy (KE_trans) and rotational kinetic energy (KE_rot):
\[ KE_{total} = KE_{trans} + KE_{rot} \]
Where:
\[ KE_{trans} = \frac{1}{2} mv^2 \]
\[ KE_{rot} = \frac{1}{2} I \omega^2 \]

### Step 2: Determine the moment of inertia and angular velocity
For a uniform thin ring, the moment of inertia (I) is given by:
\[ I = mr^2 \]
Where:
- \(m\) is the mass of the ring
- \(r\) is the radius of the ring

Since the ring rolls without slipping, the relationship between linear velocity (v) and angular velocity (ω) is:
\[ v = r\omega \quad \Rightarrow \quad \omega = \frac{v}{r} \]

### Step 3: Substitute the values into the kinetic energy formulas
Substituting the expression for angular velocity into the rotational kinetic energy formula:
\[ KE_{rot} = \frac{1}{2} I \left(\frac{v}{r}\right)^2 \]
Substituting the moment of inertia:
\[ KE_{rot} = \frac{1}{2} (mr^2) \left(\frac{v}{r}\right)^2 = \frac{1}{2} m \frac{v^2}{r^2} r^2 = \frac{1}{2} mv^2 \]

### Step 4: Combine the kinetic energy components
Now we can combine the translational and rotational kinetic energy:
\[ KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 = mv^2 \]

### Step 5: Substitute the values for mass and velocity
Given:
- Mass \(m = 0.4 \, \text{kg}\)
- Velocity \(v = 10 \, \text{cm/s} = 0.1 \, \text{m/s}\)

Now, substituting these values into the total kinetic energy formula:
\[ KE_{total} = mv^2 = 0.4 \times (0.1)^2 = 0.4 \times 0.01 = 0.004 \, \text{J} = 4 \times 10^{-3} \, \text{J} \]

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

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