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Question #42353

Difficulty: 8/10 | 🟢 Free

The Question

A system consists of a heavy base of mass MM resting on the horizontal floor of an elevator. A motor rigidly attached to the base drives a massless rod of length L=0.5 mL = 0.5 \text{ m} with a point mass m=2 kgm = 2 \text{ kg} at its free end. The rod rotates in a vertical plane with a constant angular velocity ω=10 rad/s\omega = 10 \text{ rad/s}. The elevator accelerates upwards with a constant acceleration a=5 m/s2a = 5 \text{ m/s}^2. Assume g=10 m/s2g = 10 \text{ m/s}^2. What is the maximum internal torque τmax\tau_{\text{max}} the motor must provide to maintain the constant angular velocity, and what is the minimum mass MminM_{\text{min}} of the base required so that it never loses contact with the elevator floor?

Options

  • A.

    15 N⋅m,143 kg15 \text{ N}\cdot\text{m}, \frac{14}{3} \text{ kg}

  • B.

    10 N⋅m,8 kg10 \text{ N}\cdot\text{m}, 8 \text{ kg}

  • C.

    10 N⋅m,143 kg10 \text{ N}\cdot\text{m}, \frac{14}{3} \text{ kg}

  • D.

    15 N⋅m,203 kg15 \text{ N}\cdot\text{m}, \frac{20}{3} \text{ kg}

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

Center of Mass (COM) Frame|Apparent Weight in Accelerating Elevators|Vector Cross-Products

Why This Question Is Tricky

Common Mistakes

This question tests the ability to analyze a multi-particle system with internal moving parts inside a non-inertial reference frame.

The primary trap lies in calculating the normal force for the lift-off condition.

Students often incorrectly equate the normal force to just the weight of the base minus the centrifugal force (N=Mgeff−mLω2N = Mg_{\text{eff}} - mL\omega^2), forgetting that the mass mm also contributes to the static weight of the system (M+m)geff(M+m)g_{\text{eff}}.

This specific omission leads directly to the distractor 20/320/3 kg.

Additionally, failing to adjust the gravity vector to geff=g+ag_{\text{eff}} = g+a for the internal torque calculation traps students into choosing 10 N⋅m10 \text{ N}\cdot\text{m}.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterSystem of Particles and Rotational Motion - Class 11 Part 1
ConceptCenter of Mass (COM) Frame
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree