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Rotational Motion question

2003 · Q162
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Rotational Motion question

2003 · Q162

NEETPhysicsRotational MotionMCQ+4 / −1
A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity ω\omegaω. Four objects each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be :
  1. A
    Mω4m{{M\omega } \over {4m}}4mMω​
  2. B
    MωM+4m{{M\omega } \over {M + 4m}}M+4mMω​
  3. C
    (M+4m)ωM{{\left( {M + 4m} \right)\omega } \over M}M(M+4m)ω​
  4. D
    (M−4m)ωM+4m{{\left( {M - 4m} \right)\omega } \over {M + 4m}}M+4m(M−4m)ω​
View written solutionFree

Correct answer: B

Applying conservation law of angular momentum, I1ω1=I2ω2{I_1}{\omega _1} = {I_2}{\omega _2}I1​ω1​=I2​ω2​

I2=(Mr2)+4(m)(r2)=(M+4m)r2{I_2} = \left( {M{r^2}} \right) + 4\left( m \right)\left( {{r^2}} \right) = \left( {M + 4m} \right){r^2}I2​=(Mr2)+4(m)(r2)=(M+4m)r2

(Taking ω1=ω{\omega _1} = \omega ω1​=ω and ω1=ω1{\omega _1} = \omega _1ω1​=ω1​)

⇒Mr2ω=(M+4m)r2ω1 \Rightarrow M{r^2}\omega = \left( {M + 4m} \right){r^2}{\omega _1}⇒Mr2ω=(M+4m)r2ω1​

⇒ω1=MωM+4m \Rightarrow {\omega _1} = {{M\omega } \over {M + 4m}}⇒ω1​=M+4mMω​

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