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

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.