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Question #42366

Difficulty: 8/10 | 🟢 Free

The Question

A uniform solid cylinder of mass M=2mM = 2m and radius RR rests on its flat face on a smooth horizontal table. A uniform thick rope of mass mm and length LL is completely wrapped around the cylinder. A constant horizontal force FF is applied to the free end of the rope, causing it to unwrap. The unwrapped portion of the rope remains straight and tangent to the cylinder. At the instant when exactly half the length of the rope (L/2L/2) is unwrapped, the cylinder has an angular velocity ω\omega. What is the angular acceleration α\alpha of the cylinder at this instant?

Options

  • A.

    10FL+2mω2R223mRL\frac{10 F L + 2 m \omega^2 R^2}{23 m R L}

  • B.

    10F23mR\frac{10 F}{23 m R}

  • C.

    10FL−2mω2R223mRL\frac{10 F L - 2 m \omega^2 R^2}{23 m R L}

  • D.

    10FL+2mω2R217mRL\frac{10 F L + 2 m \omega^2 R^2}{17 m R L}

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

Torque about non-fixed axes|Tension in Massless vs Massive Strings|Vector Cross-Products

Why This Question Is Tricky

Common Mistakes

This question tests the dynamics of variable mass systems in rotational and translational motion.

The lethal trap lies in assuming the tension TT simply accelerates the cylinder's mass (T=MsysacT = M_{sys}a_c).

Students fail to account for the momentum flux (thrust force) caused by the massive string unwrapping.

As the string transitions from the rotating cylinder to the translating straight segment, it accelerates from vcv_c to vc+ωRv_c + \omega R.

This mass transfer requires a continuous force, manifesting as a backward thrust λω2R2\lambda \omega^2 R^2 on the cylinder.

Ignoring this thrust leads to the incorrect option B.

Furthermore, students often miscalculate the torque by failing to realize that the angular momentum flux exactly cancels the changing moment of inertia term, leaving τ=Iα\tau = I\alpha intact.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterSystem of Particles and Rotational Motion - Class 11 Part 1
ConceptTorque about non-fixed axes
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree