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NEET PHYSICSRotational motionMedium

Question

A solid cylinder of mass 2 kg2 \text{ kg} and radius 4 cm4 \text{ cm} is rotating about its axis at the rate of 3 rpm3 \text{ rpm}. The torque required to stop after 2π2\pi revolutions is:

A

2×106 N-m2 \times 10^6 \text{ N-m}

B

2×10−6 N-m2 \times 10^{-6} \text{ N-m}

C

2×10−3 N-m2 \times 10^{-3} \text{ N-m}

D

12×10−4 N-m12 \times 10^{-4} \text{ N-m}

Step-by-Step Solution

Given: Mass of the solid cylinder, m=2 kgm = 2 \text{ kg} Radius, r=4 cm=0.04 mr = 4 \text{ cm} = 0.04 \text{ m} Initial angular velocity, ω0=3 rpm=3×2π60 rad/s=π10 rad/s\omega_0 = 3 \text{ rpm} = 3 \times \frac{2\pi}{60} \text{ rad/s} = \frac{\pi}{10} \text{ rad/s} Final angular velocity, ω=0\omega = 0 Angular displacement, θ=2π revolutions=2π×2π rad=4π2 rad\theta = 2\pi \text{ revolutions} = 2\pi \times 2\pi \text{ rad} = 4\pi^2 \text{ rad}

Using the equation of rotational kinematics: ω2=ω02+2αθ\omega^2 = \omega_0^2 + 2\alpha\theta 0=(π10)2+2α(4π2)0 = \left(\frac{\pi}{10}\right)^2 + 2\alpha(4\pi^2) 2α(4π2)=−π21002\alpha(4\pi^2) = -\frac{\pi^2}{100} α=−1800 rad/s2\alpha = -\frac{1}{800} \text{ rad/s}^2

The moment of inertia of the solid cylinder about its axis: I=12mr2=12×2×(0.04)2=0.0016 kg m2=16×10−4 kg m2I = \frac{1}{2}mr^2 = \frac{1}{2} \times 2 \times (0.04)^2 = 0.0016 \text{ kg m}^2 = 16 \times 10^{-4} \text{ kg m}^2

The magnitude of the required torque is: τ=I∣α∣=(16×10−4 kg m2)×(1800 rad/s2)=2×10−6 N-m\tau = I|\alpha| = (16 \times 10^{-4} \text{ kg m}^2) \times \left(\frac{1}{800} \text{ rad/s}^2\right) = 2 \times 10^{-6} \text{ N-m}

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 motioncylinderradiusrotatingtorquerequired

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