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

For a rolling solid sphere, the acceleration \(a\) down an inclined plane can be calculated using the formula that accounts for both the translation and rotation of the sphere:
\[a = \frac{5}{7} g \sin(\theta)\]
where \(\theta\) is the angle of inclination and \(g\) is the acceleration due to gravity. Here, \(\theta = 30^\circ\) and \(g = 9.8 \, \text{m/s}^2\). Plugging in the values:
\[a = \frac{5}{7} \times 9.8 \times \sin(30^\circ) = \frac{5}{7} \times 9.8 \times 0.5 = 3.5 \, \text{m/s}^2\]
This result shows that the rolling motion includes not just the translational kinetic energy but also rotational kinetic energy, which slows the acceleration compared to sliding without rotation.