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 ________.
Correct Answer: 2
Approach Solution - 1
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.
Approach Solution -2
A hollow sphere is rolling on a plane surface about its axis of symmetry.
We need to find the ratio of rotational kinetic energy to total kinetic energy.
Step 2: Moment of inertia for a hollow sphere.
For a hollow sphere (thin spherical shell), the moment of inertia about its diameter is:
\[ I = \frac{2}{3} m r^2 \]
Step 3: Kinetic energy expressions.
For rolling motion, total kinetic energy is the sum of translational and rotational kinetic energies:
\[ K_{total} = K_{trans} + K_{rot} \] \[ K_{total} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \] For rolling without slipping, \( v = r\omega \).
Substitute \( I = \frac{2}{3} m r^2 \):
\[ K_{rot} = \frac{1}{2} \times \frac{2}{3} m r^2 \omega^2 = \frac{1}{3} m v^2 \] \[ K_{trans} = \frac{1}{2} m v^2 \] \[ K_{total} = \frac{1}{3} m v^2 + \frac{1}{2} m v^2 = \frac{5}{6} m v^2 \]
Step 4: Ratio of rotational kinetic energy to total kinetic energy.
\[ \frac{K_{rot}}{K_{total}} = \frac{(1/3) m v^2}{(5/6) m v^2} = \frac{2}{5} \]
Step 5: Final Answer.
The ratio is \( \frac{x}{5} = \frac{2}{5} \), hence \( x = 2 \).
Final Answer: 2
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