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Rotational Motion question

2014 · Q146
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Rotational Motion question

2014 · Q146

NEETPhysicsRotational MotionMCQ+4 / −1
A solid cylinder of mass 50 kg and radius 0.5 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 revolutions s−-−2 is
  1. A
    25 N
  2. B
    50 N
  3. C
    78.5 N
  4. D
    157 N
View written solutionFree

Correct answer: D

Here α\alpha α = 2 revolutions/s2 = 4π\pi π rad/s2 (given)

Icylinder=12MR2=12(50)(0.5)2=254Kg−m2{I_{cylinder}} = {1 \over 2}M{R^2} = {1 \over 2}\left( {50} \right){\left( {0.5} \right)^2} = {{25} \over 4}{\rm{Kg - }}{{\rm{m}}^{\rm{2}}}Icylinder​=21​MR2=21​(50)(0.5)2=425​Kg−m2

As τ=Iα\tau = I\alpha τ=Iα so TR=IαTR = I\alpha TR=Iα

⇒T=IαR=(254)(4π)0.5N \Rightarrow T = {{I\alpha } \over R} = {{\left( {{{25} \over 4}} \right)\left( {4\pi } \right)} \over {0.5}}N⇒T=RIα​=0.5(425​)(4π)​N

= 50 π\pi πN = 157 N

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