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 solve this problem, we need to determine the ratio of the kinetic energies of a ring and a solid sphere rolling down an inclined plane without slipping. Both start from rest, and their radii are identical. We can start by expressing the kinetic energy for both objects.
Kinetic energy (\(KE\)) for an object rolling without slipping consists of translational kinetic energy (\(KE_t\)) and rotational kinetic energy (\(KE_r\)).
1. Kinetic Energy of the Ring: 
The total kinetic energy is given by:
\[ KE_{\text{ring}} = KE_t + KE_r = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \]
For a ring, \(I = mr^2\) and \(\omega = \frac{v}{r}\):
\[ KE_{\text{ring}} = \frac{1}{2}mv^2 + \frac{1}{2}mr^2\left(\frac{v}{r}\right)^2 = mv^2 \]
2. Kinetic Energy of the Solid Sphere:
For a solid sphere, \(I = \frac{2}{5}mr^2\):
\[ KE_{\text{sphere}} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2 \]
\[ KE_{\text{sphere}} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2 \]
3. Ratio of Kinetic Energies:
The ratio of the kinetic energies of the ring to the sphere is determined as follows:
\[ \text{Ratio} = \frac{KE_{\text{ring}}}{KE_{\text{sphere}}} = \frac{mv^2}{\frac{7}{10}mv^2} = \frac{10}{7} \]
Therefore, the value of \(x\) in the ratio \(\frac{7}{x}\) where the kinetic energies are compared as \(\frac{7}{x} = \frac{7}{10/7} = 7\) is confirmed to be 7. 

Answer 2

In pure rolling motion, the work done by friction is zero. Hence, the potential energy is completely converted into kinetic energy. Since the ring and the sphere initially have the same potential energy, they will also have the same total kinetic energy at the end.

- Ratio of kinetic energies = 1.

Given:

\(\frac{7}{x} = 1 \implies x = 7.\)

The Correct answer is: 7