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Question #42405

Difficulty: 8/10 | 🟢 Free

The Question

A rigid physical pendulum consists of a uniform circular disc of mass $M$ and radius $R$ , pivoted freely at a point $P$ on its circumference. A point mass of mass $m = \frac{M}{2}$ is firmly attached to the disc at a point diametrically opposite to the pivot $P$ . The pendulum is held such that the diameter passing through the pivot and the point mass is horizontal, and then released from rest. What is the magnitude of the linear velocity of the point mass when the pendulum reaches its lowest vertical position?

Options

A.
$\sqrt{\frac{32gR}{7}}$
B.
$\sqrt{\frac{8gR}{7}}$
C.
$\sqrt{\frac{24gR}{7}}$
D.
$\sqrt{4gR}$

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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

Angular Velocity vs Linear Velocity|Physical Pendulum Inertia|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly apply energy conservation to a non-uniform physical pendulum.

The primary trap lies in the calculation of the moment of inertia (forgetting the parallel axis theorem for the disc) and the center of mass of a composite body.

Furthermore, students often incorrectly calculate the linear velocity by using $v = \omega R$ (the velocity of the disc's center) instead of $v = \omega (2R)$ for the point mass at the edge, failing to transition properly from rotational dynamics to point-mass kinematics in a non-uniform field.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Plane - Class 11 Part 1
Concept	Angular Velocity vs Linear Velocity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free