Question

For a rolling solid sphere on an incline of $30^\circ$, what is its acceleration $a$ in terms of gravity $g$?

• A. $\frac{g}{2}$
• B. $\frac{5g}{7}$
• C. $\frac{5g}{7} \sin 30^\circ$
• D. $\frac{2g}{7} \sin 30^\circ$

Hint:

Remember: Solid sphere → $a = \frac{5}{7} g \sin \theta$.

Answer 1

The Correct Option is C

Concept: For a body rolling without slipping down an incline, acceleration depends on both translational and rotational motion.
Step 1: General formula.
For rolling motion: \[ a = \frac{g \sin \theta}{1 + \frac{I}{mR^2}} \]
Step 2: Moment of inertia of solid sphere.
\[ I = \frac{2}{5} mR^2 \]
Step 3: Substitute in formula.
\[ a = \frac{g \sin \theta}{1 + \frac{2}{5}} = \frac{g \sin \theta}{\frac{7}{5}} = \frac{5}{7} g \sin \theta \]
Step 4: For $\theta = 30^\circ$.
\[ a = \frac{5}{7} g \sin 30^\circ \]
Step 5: Evaluating the options.

• $\frac{g}{2}$ $\rightarrow$ Incorrect
• $\frac{5g}{7}$ $\rightarrow$ Missing $\sin \theta$ (incorrect)
• $\frac{5g}{7} \sin 30^\circ$ $\rightarrow$ Correct
• $\frac{2g}{7} \sin 30^\circ$ $\rightarrow$ Incorrect

Step 6: Conclusion.
Thus, acceleration is $\frac{5g}{7} \sin 30^\circ$.