A solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about axis of rotation of the sphere to the total energy of moving sphere is $$\pi : 22$$ then, the value of its angular speed will be _____ rad s$$^{-1}$$.

A solid sphere rolls without slipping on a horizontal plane. The ratio of angular momentum about its axis of rotation to total energy is $$\pi : 22$$.

Angular momentum about the axis of rotation.

For a solid sphere: $$I = \frac{2}{5}mR^2$$

$$L = I\omega = \frac{2}{5}mR^2\omega$$

Total kinetic energy.

$$E = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}m(R\omega)^2 + \frac{1}{2} \cdot \frac{2}{5}mR^2\omega^2$$

$$= \frac{1}{2}mR^2\omega^2\left(1 + \frac{2}{5}\right) = \frac{7}{10}mR^2\omega^2$$

Form the ratio.

$$\frac{L}{E} = \frac{\frac{2}{5}mR^2\omega}{\frac{7}{10}mR^2\omega^2} = \frac{2/5}{7\omega/10} = \frac{4}{7\omega}$$

Apply the given ratio.

$$\frac{4}{7\omega} = \frac{\pi}{22}$$

$$\omega = \frac{4 \times 22}{7\pi} = \frac{88}{7\pi}$$

Using $$\pi = \frac{22}{7}$$:

$$\omega = \frac{88}{7 \times \frac{22}{7}} = \frac{88}{22} = 4 \text{ rad s}^{-1}$$

Therefore, the answer is $$\mathbf{4}$$.