back to directory
NEET PHYSICSRotational motionMedium

Question

A circular disk of moment of inertia ItI_t is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed ωi\omega_i. Another disk of moment of inertia IbI_b is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed ωf\omega_f. The energy lost by the initially rotating disc to friction is:

A

12Ib2(It+Ib)ωi2\frac{1}{2}\frac{I_b^2}{(I_t+I_b)}\omega_i^2

B

12It2(It+Ib)ωi2\frac{1}{2}\frac{I_t^2}{(I_t+I_b)}\omega_i^2

C

12Ib−It(It+Ib)ωi2\frac{1}{2}\frac{I_b-I_t}{(I_t+I_b)}\omega_i^2

D

12IbIt(It+Ib)ωi2\frac{1}{2}\frac{I_b I_t}{(I_t+I_b)}\omega_i^2

Step-by-Step Solution

Let the initial angular momentum of the system be LiL_i and the final angular momentum be LfL_f. Since there is no external torque on the system, angular momentum is conserved . By conservation of angular momentum: Li=LfL_i = L_f Itωi+Ib(0)=(It+Ib)ωfI_t \omega_i + I_b (0) = (I_t + I_b) \omega_f ωf=ItωiIt+Ib\omega_f = \frac{I_t \omega_i}{I_t + I_b}

Initial kinetic energy of the system : Ki=12Itωi2K_i = \frac{1}{2} I_t \omega_i^2

Final kinetic energy of the system: Kf=12(It+Ib)ωf2K_f = \frac{1}{2} (I_t + I_b) \omega_f^2 Kf=12(It+Ib)(ItωiIt+Ib)2=12It2ωi2It+IbK_f = \frac{1}{2} (I_t + I_b) \left(\frac{I_t \omega_i}{I_t + I_b}\right)^2 = \frac{1}{2} \frac{I_t^2 \omega_i^2}{I_t + I_b}

Energy lost due to friction is the difference between initial and final kinetic energy: ΔK=Ki−Kf\Delta K = K_i - K_f ΔK=12Itωi2−12It2ωi2It+Ib\Delta K = \frac{1}{2} I_t \omega_i^2 - \frac{1}{2} \frac{I_t^2 \omega_i^2}{I_t + I_b} ΔK=12Itωi2(1−ItIt+Ib)\Delta K = \frac{1}{2} I_t \omega_i^2 \left( 1 - \frac{I_t}{I_t + I_b} \right) ΔK=12Itωi2(It+Ib−ItIt+Ib)\Delta K = \frac{1}{2} I_t \omega_i^2 \left( \frac{I_t + I_b - I_t}{I_t + I_b} \right) ΔK=12ItIbIt+Ibωi2\Delta K = \frac{1}{2} \frac{I_t I_b}{I_t + I_b} \omega_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 motioncircularmomentinertiarotatinghorizontal

More Rotational motion Questions

View all

From a circular ring of mass $M$ and radius $R$, an arc corresponding to a $90^\circ$ sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is $K$ times $MR^2$. The value of $K$ will be:

A.$\frac{1}{4}$
B.$\frac{1}{8}$
C.$\frac{3}{4}$
D.$\frac{7}{8}$
MediumSolve

The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis touching the disc at its edge and normal to the disc is:

A.$MR^2$
B.$\frac{2}{5}MR^2$
C.$\frac{3}{2}MR^2$
D.$\frac{1}{2}MR^2$
EasySolve

A solid cylinder of mass $2 \text{ kg}$ and radius $4 \text{ cm}$ is rotating about its axis at the rate of $3 \text{ rpm}$. The torque required to stop after $2\pi$ revolutions is:

A.$2 \times 10^6 \text{ N-m}$
B.$2 \times 10^{-6} \text{ N-m}$
C.$2 \times 10^{-3} \text{ N-m}$
D.$12 \times 10^{-4} \text{ N-m}$
MediumSolve

Three identical spherical shells, each of mass $m$ and radius $r$ are placed as shown in the figure. Consider an axis $XX'$, which is touching two shells and passing through the diameter of the third shell. The moment of inertia of the system consisting of these three spherical shells about the $XX'$ axis is:

A.$\frac{11}{5}mr^2$
B.$3mr^2$
C.$\frac{16}{5}mr^2$
D.$4mr^2$
MediumSolve

A solid cylinder of mass $3 \text{ kg}$ is rolling on a horizontal surface with a velocity of $4 \text{ ms}^{-1}$. It collides with a horizontal spring of force constant $200 \text{ Nm}^{-1}$. The maximum compression produced in the spring will be:

A.0.5 m
B.0.6 m
C.0.7 m
D.0.2 m
MediumSolve

The moment of inertia of a thin rod about an axis passing through its mid-point and perpendicular to the rod is $2400\text{ g cm}^2$. The length of the $400\text{ g}$ rod is nearly:

A.$17.5\text{ cm}$
B.$20.7\text{ cm}$
C.$72.0\text{ cm}$
D.$8.5\text{ cm}$
EasySolve

A solid sphere is in rolling motion. In rolling motion, a body possesses translational kinetic energy ($K_t$) as well as rotational kinetic energy ($K_r$) simultaneously. The ratio $K_t : (K_t + K_r)$ for the sphere will be:

A.7:10
B.5:7
C.10:7
D.2:5
MediumSolve

An energy of $484 \text{ J}$ is spent in increasing the speed of a flywheel from $60 \text{ rpm}$ to $360 \text{ rpm}$. The moment of inertia of the flywheel is:

A.$0.7 \text{ kg-m}^2$
B.$3.22 \text{ kg-m}^2$
C.$30.8 \text{ kg-m}^2$
D.$0.07 \text{ kg-m}^2$
MediumSolve

Why 32,000+ students choose TopperSquare

Everything you need to crack NEET

15,000+ Chapter MCQs

Physics, Chemistry, Biology — chapter-wise, topic-wise practice

AI Performance Analytics

Know your weak topics. Get a personalised study plan.

Full NEET Mock Tests

180 questions, timed, with detailed score analysis

PYQ Sets 2010–2024

15 years of past papers with solved explanations

Start Practising Free

No credit card · 30-second signup · Free forever plan included

This neet physics practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice physics sets →

Explore more subjects:Physics MCQs•Chemistry MCQs•Biology MCQs•Botany MCQs•Zoology MCQs•NEET Mock Tests•NEET PYQs