For a hollow sphere rolling on a plane surface without slipping, its total kinetic energy (K.E.) is the sum of its translational kinetic energy and rotational kinetic energy. The translational kinetic energy results from the motion of the center of mass of the sphere, and the rotational kinetic energy is due to its rotation about an axis through its center of mass (in this case, the axis of symmetry).

The translational kinetic energy (TKE) can be expressed as:

$$\text{TKE} = \frac{1}{2}mv^2$$

Where:

• m is the mass of the sphere,


• v is the linear velocity of the center of mass.

The rotational kinetic energy (RKE) for a rolling object can be given by:

$$\text{RKE} = \frac{1}{2}I\omega^2$$

For a hollow sphere, the moment of inertia (I) about its axis of symmetry is:

$$I = \frac{2}{3}mr^2$$

where r is the radius of the sphere. The angular velocity, $$\omega$$, can be related to the linear velocity, v, by the relation $$v = r\omega$$, for an object rolling without slipping. We substitute $$\omega = \frac{v}{r}$$ into the expression for RKE:

$$\text{RKE} = \frac{1}{2} \cdot \frac{2}{3}mr^2 \cdot \left(\frac{v}{r}\right)^2$$

This simplifies to:

$$\text{RKE} = \frac{1}{3}mv^2$$

The total kinetic energy (Total K.E.) of the rolling hollow sphere is the sum of its translational and rotational kinetic energies:

$$\text{Total K.E.} = \text{TKE} + \text{RKE} = \frac{1}{2}mv^2 + \frac{1}{3}mv^2 = \frac{5}{6}mv^2$$

Now, we want to find the ratio of the rotational kinetic energy to the total kinetic energy:

$$\frac{\text{RKE}}{\text{Total K.E.}} = \frac{\frac{1}{3}mv^2}{\frac{5}{6}mv^2}$$

Since the mass and velocity are common in both the numerator and the denominator, they will cancel out, leaving:

$$\frac{\frac{1}{3}}{\frac{5}{6}} = \frac{1}{3} \cdot \frac{6}{5} = \frac{2}{5}$$

Therefore, the value of $$x$$, representing the ratio of the rotational kinetic energy to the total kinetic energy for a hollow sphere rolling on a plane surface about its axis of symmetry, is $$2$$. Thus, $$x = 2$$.