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Question #42064

Difficulty: 8/10 | 🟢 Free

The Question

A block of mass $M = 2m$ is free to slide on a smooth horizontal floor. A uniform rod of mass $m$ and length $L$ is smoothly hinged at its lower end to the block. Initially, the system is at rest with the rod hanging vertically downwards. A constant horizontal force $F$ ( $F > mg$ ) is applied to the free top end of the rod. Find the angular velocity $\omega$ of the rod at the instant it reaches the horizontal position.

Options

A.
$\sqrt{\frac{4(5F - 3mg)}{3mL}}$
B.
$\sqrt{\frac{5F - 3mg}{mL}}$
C.
$\sqrt{\frac{12(5F - 3mg)}{mL}}$
D.
$\sqrt{\frac{3(2F - mg)}{mL}}$

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

Work-Energy Theorem for Rotation|Non-Uniform Acceleration Constraint|Vector Decomposition Ambiguity

Why This Question Is Tricky

Common Mistakes

This question tests the candidate's ability to handle work-energy in non-inertial frames with non-uniform acceleration.

A common trap is to assume the block M has a constant acceleration a = F/(M+m), leading to an incorrect constant pseudo-force -mF/(M+m) in the pivot frame.

In reality, the block's acceleration fluctuates due to the rod's changing angular velocity and acceleration.

To avoid integrating a time-varying pseudo-force, one must use the ground frame or the center-of-mass frame, carefully separating the system's translational kinetic energy from its internal kinetic energy, which includes both rotational and vertical translational components of the rod.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Plane - Class 11 Part 1
Concept	Vector Decomposition Ambiguity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free