back to directory
NEET PHYSICSRotational motionMedium

Question

A circular platform is mounted on a frictionless vertical axle. Its radius R=2 mR=2\text{ m} and moment of inertia about the axle is 200 kg m2200\text{ kg m}^2. It is initially at rest. A 50 kg50\text{ kg} man stands on the edge of the platform and begins to walk along the edge at the speed of 1 ms−11\text{ ms}^{-1} relative to the ground. Time taken by the man to complete one revolution is:

A

π s\pi\text{ s}

B

3π2 s\frac{3\pi}{2}\text{ s}

C

2π s2\pi\text{ s}

D

π2 s\frac{\pi}{2}\text{ s}

Step-by-Step Solution

Since no external torque acts on the system (man + platform), the total angular momentum is conserved. The initial angular momentum is zero. Let the angular velocity of the platform be ωp\omega_p. The velocity of the man relative to the ground is v=1 m/sv = 1\text{ m/s}. His angular velocity relative to the ground is ωm=vR=12=0.5 rad/s\omega_m = \frac{v}{R} = \frac{1}{2} = 0.5\text{ rad/s}. According to the law of conservation of angular momentum: Lman+Lplatform=0L_{\text{man}} + L_{\text{platform}} = 0 mvR+Iωp=0m v R + I \omega_p = 0 (50)(1)(2)+200ωp=0  ⟹  100+200ωp=0  ⟹  ωp=−0.5 rad/s(50)(1)(2) + 200 \omega_p = 0 \implies 100 + 200 \omega_p = 0 \implies \omega_p = -0.5\text{ rad/s}. The negative sign indicates that the platform rotates in the opposite direction to the man's motion. The angular velocity of the man relative to the platform is: ωrel=ωm−ωp=0.5−(−0.5)=1.0 rad/s\omega_{\text{rel}} = \omega_m - \omega_p = 0.5 - (-0.5) = 1.0\text{ rad/s}. The time taken by the man to complete one revolution on the platform is: T=2πωrel=2π1.0=2π sT = \frac{2\pi}{\omega_{\text{rel}}} = \frac{2\pi}{1.0} = 2\pi\text{ s}.

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 motioncircularplatformmountedfrictionlessvertical

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