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Question #42098

Difficulty: 8/10 | 🟢 Free

The Question

A small block of mass $m$ is projected with an initial velocity $v_0$ along the inside of a fixed, rough vertical circular track of radius $R$ from its lowest point. The coefficient of kinetic friction between the block and the track varies with the angle $\theta$ (measured from the lowest point) as $\mu(\theta) = \mu_0 \sin\theta$ , where $\mu_0$ is a positive constant. Find the minimum value of $v_0^2$ required for the block to just complete the vertical circle.

Options

A.
$gR \left[ \frac{3}{2\mu_0} (e^{4\mu_0} - 1) - 1 \right]$
B.
$gR \left[ \frac{3}{2\mu_0} (e^{2\mu_0} - 1) + 1 \right]$
C.
$gR \left[ \frac{1}{\mu_0} (e^{4\mu_0} - 1) + 5 \right]$
D.
$gR \left[ \frac{3}{\mu_0} (e^{4\mu_0} - 1) - 5 \right]$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Minimum Speed for Looping|Work Done by Friction|Conservative Force Field|Calculus/Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle non-conservative forces that depend on the instantaneous state of the system.

A common trap is attempting to use a macroscopic algebraic energy conservation equation and approximating the work done by friction (e.g., assuming $N = mg\cos\theta$ or treating $\mu$ as constant).

Because the normal force $N$ depends on the velocity $v(\theta)$ , the work done by friction is path-dependent and coupled to the kinetic energy.

This forces the student to apply the Work-Energy theorem in its differential form, set up a first-order linear ordinary differential equation, and integrate it over the circular path to find the exact energy loss.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Plane - Class 11 Part 1
Concept	Minimum Speed for Looping
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free