Question

A ring and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of both bodies are identical and the ratio of their kinetic energies is \( \frac{7}{x} \) where \( x \) is \(\_\_\_\_\_\_\_\).

Answer 1

Correct Answer: 7

To determine the ratio of kinetic energies for a ring and a solid sphere rolling without slipping down an inclined plane from rest, with identical radii, we first express the kinetic energy for each. Kinetic energy (\(KE\)) for an object rolling without slipping comprises translational (\(KE_t\)) and rotational (\(KE_r\)) components.
1. Ring Kinetic Energy:
Total kinetic energy is calculated as:
\[ KE_{\text{ring}} = KE_t + KE_r = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \]
For a ring, \(I = mr^2\) and \(\omega = \frac{v}{r}\). Substituting these yields:
\[ KE_{\text{ring}} = \frac{1}{2}mv^2 + \frac{1}{2}mr^2\left(\frac{v}{r}\right)^2 = mv^2 \]
2. Solid Sphere Kinetic Energy:
For a solid sphere, the moment of inertia is \(I = \frac{2}{5}mr^2\). The kinetic energy is:
\[ KE_{\text{sphere}} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ KE_{\text{sphere}} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2 \]
3. Kinetic Energy Ratio:
The ratio of the ring's kinetic energy to the sphere's kinetic energy is:
\[ \text{Ratio} = \frac{KE_{\text{ring}}}{KE_{\text{sphere}}} = \frac{mv^2}{\frac{7}{10}mv^2} = \frac{10}{7} \]
Thus, in the ratio \(\frac{7}{x}\), where the comparison is \(\frac{7}{x} = \frac{7}{10/7} = 7\), the value of \(x\) is confirmed to be 7.