Question

A sphere rolls down an inclined plane of inclination \( \theta \). What is the acceleration as the sphere reaches the bottom?

• A. \( \frac{5g}{7} \sin \theta \)
• B. \( \frac{3g}{5} \sin \theta \)
• C. \( \frac{7g}{5} \sin \theta \)
• D. \( \frac{5g}{3} \sin \theta \)

Hint:

For a rolling body, the acceleration is affected by the moment of inertia. A solid sphere has \( k^2/r^2 = 2/5 \).

Answer 1

The Correct Option is A

Step 1: Rolling motion equation.
For a rolling sphere, the acceleration \( a \) is given by: \[ a = \frac{g \sin \theta}{1 + k^2/r^2} \] where \( k \) is the radius of gyration and \( r \) is the radius of the sphere. For a solid sphere, \( k^2/r^2 = 2/5 \).

Step 2: Substitute values.
Substituting \( k^2/r^2 = 2/5 \) into the equation, we get the acceleration as: \[ a = \frac{5g}{7} \sin \theta \]