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Question #42290

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The Question

A motorized gyroscope consists of a rotor of mass $M$ and moment of inertia $I$ about its spin axis. It is mounted on a light, rigid horizontal axle of total length $2l$ , with the rotor at the exact center. The ends of the axle rest on two identical digital weighing scales, $S_1$ and $S_2$ . The rotor spins at a constant angular speed $\omega$ such that its angular momentum vector $\vec{L}$ points exactly towards $S_1$ . The setup is located at a latitude $\lambda$ in the Northern Hemisphere, and the Earth rotates with a constant angular speed $\Omega$ . The axle is oriented at an angle $\theta$ North of East. Horizontal guides keep the axle stationary relative to the Earth without exerting any vertical forces. If $N_1$ and $N_2$ are the vertical normal forces exerted by scales $S_1$ and $S_2$ respectively, what is the value of $N_2 - N_1$ ?

Options

A.
$\frac{I \omega \Omega \sin\lambda}{l}$
B.
$\frac{I \omega \Omega \sin\lambda \cos\theta}{l}$
C.
$\frac{I \omega \Omega \cos\lambda \sin\theta}{l}$
D.
$\frac{I \omega \Omega \sqrt{\sin^2\lambda + \cos^2\lambda \sin^2\theta}}{l}$

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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

Effect of Earth's Rotation on g|Gyroscopic Stability|Vector Cross-Products

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to analyze rigid body dynamics in a non-inertial rotating frame (the Earth).

The primary trap lies in the intuitive assumption that the gyroscopic torque (or Coriolis effect) must depend on the azimuthal orientation angle $\theta$ between the spin axis and the Earth's rotation axis.

Students often incorrectly resolve the full $\vec{\Omega} \times \vec{L}$ cross product and assume the vertical normal forces must balance the entire magnitude of the torque.

However, the horizontal component of the Earth's rotation only produces a vertical torque (which is balanced by the horizontal guides), while the vertical component of Earth's rotation produces the horizontal torque balanced by the vertical scales.

The cross-product geometry elegantly cancels out the azimuthal angle $\theta$ , making the weight difference purely a function of latitude.

This forces students to rigorously separate the vector components of torque rather than relying on scalar magnitude formulas.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Gravitation - Class 11 Part 1
Concept	Effect of Earth's Rotation on g
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free