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Rotational Motion question

2004 · Q139
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Rotational Motion question

2004 · Q139

NEETPhysicsRotational MotionMCQ+4 / −1
A round disc of moment of inertia III2 about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia III1 rotating with an angular velocity ω\omegaω about the same axis. The final angular velocity of the combination of discs is
  1. A
    I2ωI1+I2{{{I_2}\omega } \over {{I_1} + {I_2}}}I1​+I2​I2​ω​
  2. B
    ω\omegaω
  3. C
    I1ωI1+I2{{{I_1}\omega } \over {{I_1} + I{}_2}}I1​+I2​I1​ω​
  4. D
    (I1+I2)ωI1{{\left( {{I_1} + {I_2}} \right)\omega } \over {{I_1}}}I1​(I1​+I2​)ω​
View written solutionFree

Correct answer: C

The initial angular momentum of the system is I1ω{I_1}\omega I1​ω and final angular momentum is (I1+I2)ω′\left( {{I_1} + {I_2}} \right)\omega '(I1​+I2​)ω′ where

ω′\omega 'ω′ = final angular velocity of combination of discs
On equating the initial and final angular momentum, we get

ω′=I1ω(I1+I2)\omega ' = {{{I_1}\omega } \over {\left( {{I_1} + {I_2}} \right)}}ω′=(I1​+I2​)I1​ω​

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