A 1.0 kg wheel in the form of a solid disk rolls along a horizontal surface with a speed of 6.0 m/s. What is the total kinetic energy of the wheel?

Options

A. 9.0 J
B. 27 J
C. 18 J
D. 36 J

Answer

To find the total kinetic energy of a 1.0 kg wheel in the form of a solid disk rolling along a horizontal surface with a speed of 6.0 m/s, we need to consider both translational and rotational kinetic energy.

### Step-by-Step Solution:

1. **Identify the mass and speed of the wheel:**
- Mass (m) = 1.0 kg
- Speed (v) = 6.0 m/s

2. **Calculate the translational kinetic energy (TKE):**
- The formula for translational kinetic energy is:
\[ TKE = \frac{1}{2} mv^2 \]
- Substituting the values:
\[ TKE = \frac{1}{2} \times 1.0 \, \text{kg} \times (6.0 \, \text{m/s})^2 \]
- Calculating:
\[ TKE = \frac{1}{2} \times 1.0 \times 36 = 18 \, \text{J} \]

3. **Calculate the moment of inertia (I) for a solid disk:**
- The moment of inertia for a solid disk about its center is given by:
\[ I = \frac{1}{2} m r^2 \]
- However, we do not need the radius (r) to find the total kinetic energy because we will use the relationship between linear speed (v) and angular speed (ω).

4. **Relate angular speed (ω) to linear speed (v):**
- The relationship is given by:
\[ v = \omega r \quad \Rightarrow \quad \omega = \frac{v}{r} \]

5. **Calculate the rotational kinetic energy (RKE):**
- The formula for rotational kinetic energy is:
\[ RKE = \frac{1}{2} I \omega^2 \]
- Substituting the moment of inertia and the expression for ω:
\[ RKE = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v^2}{r^2}\right) \]
- Simplifying:
\[ RKE = \frac{1}{4} mv^2 \]
- Substituting the values:
\[ RKE = \frac{1}{4} \times 1.0 \, \text{kg} \times (6.0 \, \text{m/s})^2 \]
- Calculating:
\[ RKE = \frac{1}{4} \times 1.0 \times 36 = 9 \, \text{J} \]

6. **Calculate the total kinetic energy (TKE + RKE):**
- Total kinetic energy (KE) is the sum of translational and rotational kinetic energy:
\[ KE = TKE + RKE = 18 \, \text{J} + 9 \, \text{J} = 27 \, \text{J} \]

### Final Answer:
The total kinetic energy of the wheel is **27 Joules**.