Question
A solid sphere rolls along a horizontal, smooth surface at a constant linear speed without slipping. What is the ratio between the rotational kinetic energy about the center of the sphere and the sphere’s total kinetic energy?
1. 3/7
2. 7/2
3. 5/3
4. None of these
5. 3/5
6. 2/5
7. 2/7
1. 3/7
2. 7/2
3. 5/3
4. None of these
5. 3/5
6. 2/5
7. 2/7
Answers
GPT-4o mini
To find the ratio between the rotational kinetic energy of a solid sphere about its center and the total kinetic energy, we start by defining the relevant quantities.
Let:
- \( m \) = mass of the sphere
- \( R \) = radius of the sphere
- \( v \) = linear speed of the sphere (the speed of its center of mass)
The total kinetic energy \( K_{\text{total}} \) of the sphere consists of both translational and rotational kinetic energy and is calculated as follows:
1. **Translational Kinetic Energy (\( K_{\text{trans}} \))**:
\[
K_{\text{trans}} = \frac{1}{2} m v^2
\]
2. **Rotational Kinetic Energy (\( K_{\text{rot}} \))**:
The moment of inertia \( I \) of a solid sphere about its center is given by:
\[
I = \frac{2}{5} m R^2
\]
The angular velocity \( \omega \) of the sphere is related to the linear velocity \( v \) through the rolling condition:
\[
\omega = \frac{v}{R}
\]
Thus, the rotational kinetic energy is:
\[
K_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{2}{5} m R^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{2} \cdot \frac{2}{5} m R^2 \cdot \frac{v^2}{R^2} = \frac{1}{5} m v^2
\]
3. **Total Kinetic Energy**:
Now, we combine the translational and rotational kinetic energy:
\[
K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2} m v^2 + \frac{1}{5} m v^2
\]
We need a common denominator to sum these:
\[
K_{\text{total}} = \frac{5}{10} m v^2 + \frac{2}{10} m v^2 = \frac{7}{10} m v^2
\]
4. **Ratio of Rotational Kinetic Energy to Total Kinetic Energy**:
Now, we find the ratio:
\[
\text{Ratio} = \frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{\frac{1}{5} m v^2}{\frac{7}{10} m v^2}
\]
Simplifying this:
\[
\text{Ratio} = \frac{\frac{1}{5}}{\frac{7}{10}} = \frac{1}{5} \cdot \frac{10}{7} = \frac{2}{7}
\]
Thus, the final answer is:
\[
\boxed{\frac{2}{7}}
\]
Let:
- \( m \) = mass of the sphere
- \( R \) = radius of the sphere
- \( v \) = linear speed of the sphere (the speed of its center of mass)
The total kinetic energy \( K_{\text{total}} \) of the sphere consists of both translational and rotational kinetic energy and is calculated as follows:
1. **Translational Kinetic Energy (\( K_{\text{trans}} \))**:
\[
K_{\text{trans}} = \frac{1}{2} m v^2
\]
2. **Rotational Kinetic Energy (\( K_{\text{rot}} \))**:
The moment of inertia \( I \) of a solid sphere about its center is given by:
\[
I = \frac{2}{5} m R^2
\]
The angular velocity \( \omega \) of the sphere is related to the linear velocity \( v \) through the rolling condition:
\[
\omega = \frac{v}{R}
\]
Thus, the rotational kinetic energy is:
\[
K_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{2}{5} m R^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{2} \cdot \frac{2}{5} m R^2 \cdot \frac{v^2}{R^2} = \frac{1}{5} m v^2
\]
3. **Total Kinetic Energy**:
Now, we combine the translational and rotational kinetic energy:
\[
K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2} m v^2 + \frac{1}{5} m v^2
\]
We need a common denominator to sum these:
\[
K_{\text{total}} = \frac{5}{10} m v^2 + \frac{2}{10} m v^2 = \frac{7}{10} m v^2
\]
4. **Ratio of Rotational Kinetic Energy to Total Kinetic Energy**:
Now, we find the ratio:
\[
\text{Ratio} = \frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{\frac{1}{5} m v^2}{\frac{7}{10} m v^2}
\]
Simplifying this:
\[
\text{Ratio} = \frac{\frac{1}{5}}{\frac{7}{10}} = \frac{1}{5} \cdot \frac{10}{7} = \frac{2}{7}
\]
Thus, the final answer is:
\[
\boxed{\frac{2}{7}}
\]