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NEET PHYSICSRotational motionMedium

Question

A solid cylinder of mass 50 kg50 \text{ kg} and radius 0.5 m0.5 \text{ m} is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of 2 rev/s22 \text{ rev/s}^2 is:

A

25 N25 \text{ N}

B

50 N50 \text{ N}

C

78.5 N78.5 \text{ N}

D

157 N157 \text{ N}

Step-by-Step Solution

Given: Mass of the cylinder, M=50 kgM = 50 \text{ kg} Radius of the cylinder, R=0.5 mR = 0.5 \text{ m} Angular acceleration, α=2 rev/s2=2×2π rad/s2=4π rad/s2\alpha = 2 \text{ rev/s}^2 = 2 \times 2\pi \text{ rad/s}^2 = 4\pi \text{ rad/s}^2

The moment of inertia of a solid cylinder about its central axis is given by: I=12MR2I = \frac{1}{2}MR^2

The torque τ\tau produced by the tension TT in the string acting at the rim of the cylinder is: τ=T×R\tau = T \times R

According to Newton's second law for rotational motion, we know that: τ=Iα\tau = I\alpha

Equating the two expressions for torque: T×R=(12MR2)αT \times R = \left(\frac{1}{2}MR^2\right)\alpha T=12MRαT = \frac{1}{2}MR\alpha

Substituting the given values into the equation: T=12×50×0.5×4πT = \frac{1}{2} \times 50 \times 0.5 \times 4\pi T=12.5×4π=50π NT = 12.5 \times 4\pi = 50\pi \text{ N}

Taking π≈3.14\pi \approx 3.14: T=50×3.14=157 NT = 50 \times 3.14 = 157 \text{ N}

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 motioncylinderradiusrotatehorizontalmassless

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