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Rotational Motion question

2007 · Q137
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Rotational Motion question

2007 · Q137

NEETPhysicsRotational MotionMCQ+4 / −1
A uniform rod AB of length lll and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is ml2/3, the initial angular acceleration of the rod will be

AIPMT 2007 Physics - Rotational Motion Question 46 English
  1. A
    mgl2{{mgl} \over 2}2mgl​
  2. B
    32gl{3 \over 2}gl23​gl
  3. C
    3g2l{{3g} \over {2l}}2l3g​
  4. D
    2g3l{{2g} \over {3l}}3l2g​
View written solutionFree

Correct answer: C

Torque about A,

τ=mg×l2=mgl2\tau = mg \times {l \over 2} = {{mgl} \over 2}τ=mg×2l​=2mgl​

τ=Iα\tau = I\alpha τ=Iα

∴\therefore∴ Angular acceleration,
α=τI=mgl/2ml2/3=3g2l\alpha = {\tau \over I} = {{mgl/2} \over {m{l^2}/3}} = {{3g} \over {2l}}α=Iτ​=ml2/3mgl/2​=2l3g​

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