To find the linear acceleration of a solid sphere rolling down an inclined plane, we need to consider both the translational and rotational motion, as well as the forces acting on the sphere.
Given:
When a sphere rolls down an inclined plane without slipping, both translational (linear) and rotational motions are present. The forces acting on the sphere are:
For pure rolling motion, the linear acceleration \((a)\) and angular acceleration \((\alpha)\) are related by the equation: \(a = r \alpha\), where \(r\) is the radius of the sphere.
The moment of inertia \(I\) for a solid sphere is given by: \(\frac{2}{5}mr^2\)
The torque due to friction is related to the angular acceleration by: \(\tau = I \alpha\)
Using Newton's second law for linear and rotational motion, we have:
Substituting \(\alpha = \frac{a}{r}\) in the rotational equation gives: \(f \cdot r = \frac{2}{5}mr^2 \cdot \frac{a}{r}\)
Simplifying, we find: \(f = \frac{2}{5}ma\)
Substituting the value of \(f\) into the translational equation: \(mg \sin \theta - \frac{2}{5}ma = ma\)
Simplifying further: \(mg \sin \theta = ma + \frac{2}{5}ma\)
Combining terms gives: \(mg \sin \theta = \frac{7}{5}ma\)
Solving for \(a\): \(a = \frac{5}{7}g \sin \theta\)
With \(\theta = 30^\circ\), we have \(\sin(30^\circ) = \frac{1}{2}\): \(a = \frac{5}{7} \cdot g \cdot \frac{1}{2} = \frac{5g}{14}\)
Thus, the correct answer is: \(\frac{5g}{14}\), which matches option \(a\).