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Question #42025

Difficulty: 8/10 | 🟢 Free

The Question

A massive string of length $L$ and total mass $M$ has a non-uniform linear mass density $\lambda(x) = kx$ , where $x$ is the distance from the fixed pivot. The string is given an initial angular velocity at its lowest position ( $\theta = 0$ ) such that the tension at the pivot is exactly zero when the string reaches the highest point ( $\theta = \pi$ ). Find the tension in the string at its midpoint ( $x = L/2$ ) when it is at an angle $\theta = 2\pi/3$ from the downward vertical.

Options

A.
$\frac{17}{6} Mg$
B.
$\frac{43}{12} Mg$
C.
$\frac{23}{18} Mg$
D.
$\frac{35}{16} Mg$

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

The Directional Sign Convention|Vertical Circular Motion|Turning Point Identification

Why This Question Is Tricky

Common Mistakes

This question tests the dynamics of a non-uniform massive string in vertical circular motion.

The primary trap lies in the 'Directional Sign Convention' at obtuse angles ( $\theta = 2\pi/3$ ), where the radial component of gravity points inward and assists the required centripetal force, making the term $g\cos\theta$ negative.

Students often misapply the sign, treating gravity as purely resistive, or incorrectly use the center of mass of the entire string for local tension calculations.

Furthermore, identifying the correct boundary condition for the string not going slack ( $T(0)=0$ at the top) requires integrating the non-uniform density rather than relying on standard point-mass formulas.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Straight Line - Class 11 Part 1
Concept	The Directional Sign Convention
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free