Step 1: Understanding the Concept:
When a rigid body rolls down an inclined plane without slipping, part of its potential energy is converted into translational kinetic energy and part into rotational kinetic energy.
Because of the rotational inertia, its downward acceleration is less than that of an object simply sliding down a frictionless incline (which would be \(g \sin \theta\)).
Step 2: Key Formula or Approach:
The linear acceleration \(a\) of a body rolling without slipping down an incline of angle \(\theta\) is given by:
\[ a = \frac{g \sin \theta}{1 + \frac{I}{mR^2}} \]
where \(I\) is the moment of inertia about the center of mass, \(m\) is the mass, and \(R\) is the radius.
We can also express this using the radius of gyration \(K\), where \(I = mK^2\), making the formula:
\[ a = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}} \]
Step 3: Detailed Explanation:
For a solid sphere, the moment of inertia is \(I = \frac{2}{5}mR^2\).
Therefore, the ratio \(\frac{I}{mR^2} = \frac{2}{5}\).
Substitute this ratio and the given angle \(\theta = 30^\circ\) into the acceleration formula:
\[ a = \frac{g \sin 30^\circ}{1 + \frac{2}{5}} \]
\[ a = \frac{g \sin 30^\circ}{\frac{7}{5}} \]
\[ a = \frac{5}{7} g \sin 30^\circ \]
(Note: Since \(\sin 30^\circ = \frac{1}{2}\), the numerical value is actually \(\frac{5}{14}g\), but the answer is kept in terms of \(\sin 30^\circ\) to match the given options.)
Step 4: Final Answer:
The acceleration of the rolling solid sphere is \(\frac{5g}{7} \sin 30^\circ\).