Before You Begin

Question #42381

Difficulty: 8/10 | 🟢 Free

The Question

A solid uniform sphere of mass $M$ and radius $R$ is released from rest at a vertical height $h$ on a rough inclined plane. It rolls without slipping down the incline. At the exact instant it reaches the horizontal surface at the bottom, its center of mass is suddenly clamped by a fixed, frictionless pivot, instantly halting its translational motion but allowing it to rotate freely.

Simultaneously, a rigid, massless rod of length $R$ , which is attached to the pivot and has a point mass $m = \frac{2}{5}M$ at its free end, is coupled to the sphere via a clutch. Initially, the rod was at rest, hanging vertically downwards. The clutch instantly forces the rod and the sphere to rotate together with the same angular velocity.

Assuming no external torques act during the coupling process, what is the maximum vertical height $d$ reached by the point mass $m$ as the system swings upwards?

Options

A.
$\frac{5}{14} h$
B.
$\frac{5}{7} h$
C.
$\frac{35}{18} h$
D.
$\frac{5}{2} h$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Rolling on Inclines|Physical Pendulum Inertia|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to distinguish between the conservation of mechanical energy and the conservation of angular momentum across abrupt physical transitions.

The primary trap is assuming mechanical energy is conserved throughout the entire process, leading to $d = 5h/2$ .

A secondary trap involves correctly identifying the loss of translational kinetic energy when the CM is clamped, but failing to account for the energy lost during the inelastic rotational coupling, leading students to equate the initial rotational kinetic energy directly to the pendulum's potential energy ( $d = 5h/7$ ).

Another trap is incorrectly applying angular momentum conservation about the contact point instead of the CM ( $d = 35h/18$ ).

The correct approach requires isolating the impulsive event, recognizing that only angular momentum about the CM is conserved, and then applying energy conservation for the subsequent pendulum swing.

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