Question

In the case of a rolling solid sphere on an inclined plane making an angle of $30^\circ$ with the horizontal plane, the acceleration of the sphere rolling down the plane is (where acceleration due to gravity is $9.8 \, \text{m/s}^2$):

• A. $3.5 \, \text{m/s}^2$
• B. $35 \, \text{m/s}^2$
• C. $5.3 \, \text{m/s}^2$
• D. $0.53 \, \text{m/s}^2$

Hint:

Rolling motion on an incline demonstrates the conservation of energy and the interplay between rotational and translational kinetic energy. This is a fundamental principle in mechanics that shows how rotation affects acceleration on an incline.

Answer 1

The Correct Option is A

The acceleration \(a\) of a solid sphere rolling down an inclined plane, accounting for both translation and rotation, is determined by the formula:
\[a = \frac{5}{7} g \sin(\theta)\]
where \(\theta\) is the angle of inclination and \(g\) is the acceleration due to gravity. For \(\theta = 30^\circ\) and \(g = 9.8 \, \text{m/s}^2\), the calculation yields:
\[a = \frac{5}{7} \times 9.8 \times \sin(30^\circ) = \frac{5}{7} \times 9.8 \times 0.5 = 3.5 \, \text{m/s}^2\]
This result demonstrates that rolling motion involves both translational and rotational kinetic energy, which causes a reduction in acceleration compared to frictionless sliding.