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. 18 J
C. 27 J
D. 36 J

Explanation

To find the total kinetic energy of the wheel, we need to consider both its translational and rotational kinetic energy. Here’s a step-by-step solution:

### Step 1: Identify the mass and speed of the wheel
- The mass \( m \) of the wheel is given as \( 1.0 \, \text{kg} \).
- The speed \( v \) of the wheel is given as \( 6.0 \, \text{m/s} \).

### Step 2: Calculate the translational kinetic energy
The translational kinetic energy \( KE_{trans} \) is given by the formula:
\[ KE_{trans} = \frac{1}{2} m v^2 \]
Substituting the values:
\[ KE_{trans} = \frac{1}{2} \times 1.0 \, \text{kg} \times (6.0 \, \text{m/s})^2 \]
\[ KE_{trans} = \frac{1}{2} \times 1.0 \times 36 \]
\[ KE_{trans} = 18 \, \text{J} \]

### Step 3: Calculate the moment of inertia for a solid disk
The moment of inertia \( I \) for a solid disk is given by:
\[ I = \frac{1}{2} m r^2 \]
Since the radius \( r \) is not provided, we will keep it as \( r \).

### Step 4: Relate linear velocity and angular velocity
For rolling without slipping, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ v = r \omega \implies \omega = \frac{v}{r} \]
Substituting \( v = 6.0 \, \text{m/s} \):
\[ \omega = \frac{6.0}{r} \]

### Step 5: Calculate the rotational kinetic energy
The rotational kinetic energy \( KE_{rot} \) is given by:
\[ KE_{rot} = \frac{1}{2} I \omega^2 \]
Substituting the expression for \( I \) and \( \omega \):
\[ KE_{rot} = \frac{1}{2} \left( \frac{1}{2} m r^2 \right) \left( \frac{v}{r} \right)^2 \]
\[ KE_{rot} = \frac{1}{2} \left( \frac{1}{2} \times 1.0 \, \text{kg} \times r^2 \right) \left( \frac{6.0 \, \text{m/s}}{r} \right)^2 \]
\[ KE_{rot} = \frac{1}{2} \left( \frac{1}{2} \times 1.0 \times r^2 \times \frac{36}{r^2} \right) \]
\[ KE_{rot} = \frac{1}{2} \times \frac{1}{2} \times 1.0 \times 36 = \frac{18}{2} = 9 \, \text{J} \]

### Step 6: Calculate the total kinetic energy
Now, we can find the total kinetic energy \( KE_{total} \):
\[ KE_{total} = KE_{trans} + KE_{rot} \]
\[ KE_{total} = 18 \, \text{J} + 9 \, \text{J} = 27 \, \text{J} \]

### Final Answer
The total kinetic energy of the wheel is \( 27 \, \text{J} \).

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