Question

A hollow cylinder and a solid cylinder initially at rest at the top of an inclined plane are rolling down without slipping. If the time taken by the hollow cylinder to reach the bottom of the inclined plane is $2$ s, the time taken by the solid cylinder to reach the bottom of the inclined plane is?

• A. $2$ s
• B. $1.414$ s
• C. $1$ s
• D. $1.732$ s

Hint:

For rolling motion, different objects have different accelerations due to their moments of inertia. The more mass concentrated away from the center, the slower the acceleration.

Answer 1

The Correct Option is D

Step 1: Understanding the motion of rolling objects 
When a rolling object moves down an inclined plane without slipping, its acceleration is given by: $$a = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}}$$ where $K$ is the radius of gyration and $R$ is the radius of the object. For a hollow cylinder, the moment of inertia about its central axis is: $$I_{\text{hollow}} = MR^2 \Rightarrow K^2 = R^2$$ Thus, its acceleration is: $$a_{\text{hollow}} = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2}$$ For a solid cylinder, the moment of inertia about its central axis is: $$I_{\text{solid}} = \frac{1}{2} M R^2 \Rightarrow K^2 = \frac{R^2}{2}$$ Thus, its acceleration is: $$a_{\text{solid}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{1.5} = \frac{2g \sin \theta}{3}$$ 

Step 2: Relating acceleration to time 
Using the equation of motion for rolling motion: $$s = \frac{1}{2} a t^2$$ Since the same distance $s$ is covered by both cylinders, the time ratio is: $$\frac{t_{\text{solid}}}{t_{\text{hollow}}} = \sqrt{\frac{a_{\text{hollow}}}{a_{\text{solid}}}} = \sqrt{\frac{\frac{g \sin \theta}{2}}{\frac{2g \sin \theta}{3}}}$$ $$= \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$ Given $t_{\text{hollow}} = 2$ s: $$t_{\text{solid}} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \approx 1.732 \text{ s}$$ Thus, the time taken by the solid cylinder to reach the bottom is $1.732$ s.