Question

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

• A. \(\frac{14}{15}\)
• B. \(\frac{15}{14}\)
• C. \(\frac{13}{14}\)
• D. \(\frac{11}{15}\)

Hint:

Use a = g sin(theta) / (1 + K squared over R squared).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a body that rolls down an incline without slipping, the acceleration is \(a = \frac{g\sin\theta}{1 + K^2/R^2}\), where \(K\) is the radius of gyration.

Step 2: Key Formula or Approach:
Solid cylinder: \(\frac{K^2}{R^2} = \frac12\). Solid sphere: \(\frac{K^2}{R^2} = \frac25\).

Step 3: Detailed Explanation:
\(a_c = \frac{g\sin\theta}{1 + \frac12} = \frac{2}{3}g\sin\theta\).
\(a_s = \frac{g\sin\theta}{1 + \frac25} = \frac57 g\sin\theta\).
\[ \frac{a_c}{a_s} = \frac{2/3}{5/7} = \frac{14}{15} \]
The sphere is faster because it has less of its mass far from the axis. Option B, \(\frac{15}{14}\), is the inverse ratio.

Final Answer:
The ratio \(a_c : a_s\) is \(\frac{14}{15}\), option (A). \[ \boxed{\frac{14}{15}} \]