Step 1: Understanding the Concept:
When a rigid body rolls down an incline without slipping, its potential energy is converted into both translational kinetic energy and rotational kinetic energy.
Because some energy goes into rotation, its linear acceleration will be less than an object simply sliding down a frictionless incline ($g\sin\theta$).
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:
\[ a = \frac{g \sin\theta}{1 + \frac{K^2}{R^2}} \]
where $g$ is gravity, $\theta$ is the angle of incline, and $K^2/R^2$ is a shape factor derived from the moment of inertia $I = MK^2$.
For a solid sphere, the moment of inertia is $I = \frac{2}{5}MR^2$, so its shape factor is $\frac{K^2}{R^2} = \frac{2}{5}$.
Step 3: Detailed Explanation:
Given values:
Incline angle, $\theta = 30^\circ$
$\sin 30^\circ = 0.5 = \frac{1}{2}$
Shape factor for solid sphere, $\frac{K^2}{R^2} = \frac{2}{5}$
Substitute these values into the acceleration formula:
\[ a = \frac{g \sin(30^\circ)}{1 + \frac{2}{5}} \]
\[ a = \frac{g \left(\frac{1}{2}\right)}{\frac{5}{5} + \frac{2}{5}} \]
\[ a = \frac{\frac{g}{2}}{\frac{7}{5}} \]
\[ a = \frac{g}{2} \times \frac{5}{7} \]
\[ a = \frac{5g}{14} \]
Step 4: Final Answer:
The linear acceleration is $\frac{5\text{ g}}{14}$.