A solid sphere rolls without slipping along a horizontal surface. What percentage of its total kinetic energy is rotational kinetic energy?

Options

A. 0.33
B. 0.29
C. 0.5
D. 0.75

Answer

To find the percentage of the total kinetic energy that is rotational kinetic energy for a solid sphere rolling without slipping, we can follow these steps:

### Step 1: Understand the Kinetic Energy Components
A rolling solid sphere has two components of kinetic energy:
1. **Translational Kinetic Energy (KT)**: Due to the motion of the center of mass.
2. **Rotational Kinetic Energy (KR)**: Due to the rotation about its own axis.

### Step 2: Write the Formulas for Kinetic Energies
- The translational kinetic energy (KT) is given by:
\[ K_T = \frac{1}{2} M v^2 \]
- The rotational kinetic energy (KR) is given by:
\[ K_R = \frac{1}{2} I \omega^2 \]
where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.

### Step 3: Find the Moment of Inertia for a Solid Sphere
The moment of inertia \(I\) of a solid sphere about its center is:
\[ I = \frac{2}{5} M r^2 \]

### Step 4: Relate Angular Velocity to Linear Velocity
For a sphere rolling without slipping, the relationship between linear velocity \(v\) and angular velocity \(\omega\) is:
\[ v = r \omega \implies \omega = \frac{v}{r} \]

### Step 5: Substitute \(\omega\) into the Rotational Kinetic Energy Formula
Substituting \(\omega\) into the formula for \(K_R\):
\[ K_R = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{2}{5} M r^2\right) \left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ K_R = \frac{1}{2} \cdot \frac{2}{5} M r^2 \cdot \frac{v^2}{r^2} = \frac{1}{5} M v^2 \]

### Step 6: Calculate the Total Kinetic Energy
The total kinetic energy \(K\) is the sum of translational and rotational kinetic energies:
\[ K = K_T + K_R = \frac{1}{2} M v^2 + \frac{1}{5} M v^2 \]
Finding a common denominator:
\[ K = \frac{5}{10} M v^2 + \frac{2}{10} M v^2 = \frac{7}{10} M v^2 \]

### Step 7: Find the Percentage of Rotational Kinetic Energy
Now, we can find the percentage of the total kinetic energy that is rotational:
\[ \text{Percentage} = \left(\frac{K_R}{K}\right) \times 100 = \left(\frac{\frac{1}{5} M v^2}{\frac{7}{10} M v^2}\right) \times 100 \]
The \(M v^2\) cancels out:
\[ \text{Percentage} = \left(\frac{1/5}{7/10}\right) \times 100 = \left(\frac{1 \times 10}{5 \times 7}\right) \times 100 = \left(\frac{10}{35}\right) \times 100 = \frac{2}{7} \times 100 \approx 28.57\% \]

### Final Answer
Approximately **28.57%** of the total kinetic energy of a solid sphere rolling without slipping is rotational kinetic energy.

---