Question

A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is \(\frac{x}{5}\). The value of x is ________.

Answer 1

Correct Answer: 2

For a hollow sphere rolling on a plane surface, the total kinetic energy \( K_{\text{total}} \) consists of translational kinetic energy \( K_{\text{trans}} \) and rotational kinetic energy \( K_{\text{rot}} \).

Step 1: Expressions for Kinetic Energy
The translational kinetic energy is given by:

\[ K_{\text{trans}} = \frac{1}{2} mv^2, \]

where \( m \) is the mass and \( v \) is the linear velocity of the center of mass.
The rotational kinetic energy about the axis of symmetry is given by:

\[ K_{\text{rot}} = \frac{1}{2} I \omega^2, \]

where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
For a hollow sphere:

\[ I = \frac{2}{3} m R^2, \]

where \( R \) is the radius of the sphere.

Step 2: Relating Translational and Rotational Motion
Since the sphere rolls without slipping:

\[ v = R \omega. \]

Step 3: Substituting the Moment of Inertia
The rotational kinetic energy becomes:

\[ K_{\text{rot}} = \frac{1}{2} \left( \frac{2}{3} m R^2 \right) \omega^2 = \frac{1}{3} m R^2 \omega^2. \]

Using \( v = R \omega \):

\[ K_{\text{rot}} = \frac{1}{3} m v^2. \]

Step 4: Calculating the Total Kinetic Energy
The total kinetic energy is:

\[ K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2} m v^2 + \frac{1}{3} m v^2. \]

Combining terms:

\[ K_{\text{total}} = \frac{5}{6} m v^2. \]

Step 5: Finding the Ratio of Kinetic Energies
The ratio of rotational kinetic energy to total kinetic energy is:

\[ \text{Ratio} = \frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{\frac{1}{3} m v^2}{\frac{5}{6} m v^2} = \frac{2}{5}. \]

Thus:

\[ \frac{x}{5} = \frac{2}{5} \implies x = 2. \]

Therefore, the value of \( x \) is 2.