A hoop having a mass of 1kg and a diameter of 1 meter rolls along a level road at 2m/sec. Its total K.E. would be -

Options

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

Answer

To find the total kinetic energy of the hoop, 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 hoop (m) = 1 kg
- Diameter of the hoop = 1 meter, thus the radius (r) = 0.5 meters (since radius is half of the diameter)
- Speed of the hoop (v) = 2 m/s

### Step 2: Calculate the translational kinetic energy (TKE)
The translational kinetic energy (TKE) is given by the formula:
\[ \text{TKE} = \frac{1}{2} m v^2 \]
Substituting the values:
\[ \text{TKE} = \frac{1}{2} \times 1 \, \text{kg} \times (2 \, \text{m/s})^2 \]
\[ \text{TKE} = \frac{1}{2} \times 1 \times 4 \]
\[ \text{TKE} = 2 \, \text{J} \]

### Step 3: Calculate the moment of inertia (I) of the hoop
The moment of inertia (I) for a hoop is given by:
\[ I = m r^2 \]
Substituting the values:
\[ I = 1 \, \text{kg} \times (0.5 \, \text{m})^2 \]
\[ I = 1 \times 0.25 \]
\[ I = 0.25 \, \text{kg m}^2 \]

### Step 4: Calculate the angular velocity (ω)
The angular velocity (ω) can be related to the linear velocity (v) by the formula:
\[ \omega = \frac{v}{r} \]
Substituting the values:
\[ \omega = \frac{2 \, \text{m/s}}{0.5 \, \text{m}} \]
\[ \omega = 4 \, \text{rad/s} \]

### Step 5: Calculate the rotational kinetic energy (RKE)
The rotational kinetic energy (RKE) is given by the formula:
\[ \text{RKE} = \frac{1}{2} I \omega^2 \]
Substituting the values:
\[ \text{RKE} = \frac{1}{2} \times 0.25 \, \text{kg m}^2 \times (4 \, \text{rad/s})^2 \]
\[ \text{RKE} = \frac{1}{2} \times 0.25 \times 16 \]
\[ \text{RKE} = \frac{1}{2} \times 4 \]
\[ \text{RKE} = 2 \, \text{J} \]

### Step 6: Calculate the total kinetic energy (TKE + RKE)
The total kinetic energy (TKE_total) is the sum of translational and rotational kinetic energy:
\[ \text{TKE}_{\text{total}} = \text{TKE} + \text{RKE} \]
\[ \text{TKE}_{\text{total}} = 2 \, \text{J} + 2 \, \text{J} \]
\[ \text{TKE}_{\text{total}} = 4 \, \text{J} \]

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

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