Question

A solid cylinder and a solid sphere having same mass and same radius roll down on the same inclined plane. The ratio of the acceleration of the cylinder '$a_c$' to that of sphere '$a_s$' is

• A. 11/15
• B. 13/14
• C. 15/14
• D. 14/15

Hint:

The acceleration of a rolling body is $a = \frac{g \sin \theta}{1 + \beta}$ where $\beta = \frac{I}{MR^2}$. For a cylinder $\beta = 0.5$ and for a sphere $\beta = 0.4$.

Answer 1

The Correct Option is A

Step 1: The formula for the acceleration of a rolling body on an inclined plane of angle $\theta$ is given by \[ a = \frac{g \sin \theta}{1 + \frac{I}{MR^2 \]
Step 2: For a solid cylinder, the moment of inertia about its axis is $I_c = \frac{1}{2} MR^2$. Substituting this into the acceleration formula: \[ a_c = \frac{g \sin \theta}{1 + \frac{1}{2 = \frac{2}{3} g \sin \theta \]
Step 3: For a solid sphere, the moment of inertia about its diameter is $I_s = \frac{2}{5} MR^2$. Substituting this into the acceleration formula: \[ a_s = \frac{g \sin \theta}{1 + \frac{2}{5 = \frac{5}{7} g \sin \theta \]
Step 4: The ratio of the acceleration of the cylinder to that of the sphere is calculated as follows: \[ \frac{a_c}{a_s} = \frac{\frac{2}{3} g \sin \theta}{\frac{5}{7} g \sin \theta} = \frac{2}{3} \times \frac{7}{5} = \frac{14}{15} \]