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Question #42309

Difficulty: 8/10 | 🟢 Free

The Question

A particle of mass m=1 kgm = 1 \text{ kg} is initially at rest at the origin of a Cartesian coordinate system. It is acted upon by a time-varying force F⃗(t)=(2t)i^+(3t2)j^ N\vec{F}(t) = (2t)\hat{i} + (3t^2)\hat{j} \text{ N}, where tt is in seconds. What is the magnitude of the angular momentum of the particle about the point P(0,1 m)P(0, 1 \text{ m}) at t=2 st = 2 \text{ s}?

Options

  • A.

    283 kg m2/s\frac{28}{3} \text{ kg m}^2\text{/s}

  • B.

    163 kg m2/s\frac{16}{3} \text{ kg m}^2\text{/s}

  • C.

    40 kg m2/s40 \text{ kg m}^2\text{/s}

  • D.

    20 kg m2/s20 \text{ kg m}^2\text{/s}

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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

Variable Force Integration|Angular Momentum conservation|Vector Cross-Products

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle angular momentum for a particle subjected to a time-varying, non-central force.

The primary trap is the assumption of constant torque or using the final torque value multiplied by time (τ⃗final×t\vec{\tau}_{final} \times t), which incorrectly simplifies the dynamic angular impulse integral ∫τ⃗(t)dt\int \vec{\tau}(t) dt.

Furthermore, by asking for the angular momentum about a shifted reference point P(0,1)P(0,1) rather than the origin, it penalizes students who blindly apply r⃗×v⃗\vec{r} \times \vec{v} without adjusting the position vector to the correct reference frame.

The dual-path solution (kinematics vs. torque integration) ensures that students who rely on static formulas will fail, forcing a fundamental application of Newton's Second Law and the definition of angular momentum.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterWork, Energy, and Power - Class 11 Part 1
ConceptVariable Force Integration
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree