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Question #42383

Difficulty: 8/10 | 🟢 Free

The Question

A solid cylinder of mass $m$ and radius $R$ is rolling without slipping down a rough inclined plane of angle $\theta = 30^\circ$ with the horizontal. It encounters a small fixed step of height $h = R/2$ perpendicular to the incline. Assuming the cylinder pivots about the edge of the step without slipping or bouncing during and after the impact, what is the minimum velocity $v_0$ of the center of mass just before impact so that the cylinder can just climb over the step? (Let $g$ be the acceleration due to gravity)

Options

A.
$\sqrt{ \frac{3}{2} gR (2 - \sqrt{3}) }$
B.
$\sqrt{ \frac{2}{3} gR (2 - \sqrt{3}) }$
C.
$\sqrt{ \frac{3}{2} gR }$
D.
$\sqrt{ 6 gR (2 - \sqrt{3}) }$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Rolling on Inclines|Collision Impulses|Integration

Why This Question Is Tricky

Common Mistakes

This problem tests the application of the impulse-momentum theorem during collisions involving rigid bodies.

The primary trap is assuming energy is conserved during the impact or incorrectly conserving angular momentum about the center of mass instead of the point of impact.

The impulsive normal force from the step exerts a massive torque about the CoM, invalidating angular momentum conservation about it.

Furthermore, the inclusion of the incline introduces a vector projection trap: students often fail to correctly identify the angle between the position vector of the CoM and the true vertical ( $\alpha - \theta$ ), leading to an incorrect calculation of the potential energy required to climb the step.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	System of Particles and Rotational Motion - Class 11 Part 1
Concept	Rolling on Inclines
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free