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NEET PHYSICSRotational motionMedium

Question

A round disc of moment of inertia I2I_2 about its axis perpendicular to its plane and passing through its center is placed over another disc of moment of inertia I1I_1 rotating with an angular velocity ω\omega about the same axis. The final angular velocity of the combination of discs is:

A

I2ωI1+I2\frac{I_2\omega}{I_1+I_2}

B

ω\omega

C

I1ωI1+I2\frac{I_1\omega}{I_1+I_2}

D

(I1+I2)ωI1\frac{(I_1+I_2)\omega}{I_1}

Step-by-Step Solution

When the second disc is placed over the first disc, there is no external torque acting on the system about the axis of rotation. Therefore, the angular momentum of the system is conserved. Initial angular momentum, Li=I1ω1+I2ω2=I1ω+I2(0)=I1ωL_i = I_1\omega_1 + I_2\omega_2 = I_1\omega + I_2(0) = I_1\omega Let the common final angular velocity be ω′\omega'. The final moment of inertia of the system is (I1+I2)(I_1 + I_2). Final angular momentum, Lf=(I1+I2)ω′L_f = (I_1 + I_2)\omega' By conservation of angular momentum: Li=LfL_i = L_f I1ω=(I1+I2)ω′I_1\omega = (I_1 + I_2)\omega' ω′=I1ωI1+I2\omega' = \frac{I_1\omega}{I_1+I_2}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Rotational motion. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRotational motionmomentinertiaperpendicularpassingthrough

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