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Rotational Motion question

2017 · Q152
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Rotational Motion question

2017 · Q152

NEETPhysicsRotational MotionMCQ+4 / −1
Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities ω1{\omega _1}ω1​ and ω2{\omega _2}ω2​. They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is
  1. A
    14I(ω1−ω2)2{1 \over 4}I{\left( {{\omega _1} - {\omega _2}} \right)^2}41​I(ω1​−ω2​)2
  2. B
    I(ω1−ω2)2I{\left( {{\omega _1} - {\omega _2}} \right)^2}I(ω1​−ω2​)2
  3. C
    18I(ω1−ω2)2{1 \over 8}I{\left( {{\omega _1} - {\omega _2}} \right)^2}81​I(ω1​−ω2​)2
  4. D
    12I(ω1+ω2)2{1 \over 2}I{\left( {{\omega _1} + {\omega _2}} \right)^2}21​I(ω1​+ω2​)2
View written solutionFree

Correct answer: A

According to the problem,

Iω1{\omega _1}ω1​ + Iω2{\omega _2}ω2​ = 2Iω0{\omega _0}ω0​

ω0{\omega _0}ω0​ = (ω1−ω2)2{{\left( {{\omega _1} - {\omega _2}} \right)} \over 2}2(ω1​−ω2​)​

Ki=12I(ω12+ω22){K_i} = {1 \over 2}I\left( {{\omega _1}^2 + {\omega _2}^2} \right)Ki​=21​I(ω1​2+ω2​2)

Kf=14(ω1+ω2)2{K_f} = {1 \over 4}{\left( {{\omega _1} + {\omega _2}} \right)^2}Kf​=41​(ω1​+ω2​)2

Loss ΔK=I[ω122+ω222−ω124−ω224−2ω1ω24]\Delta K = I\left[ {{{{\omega _1}^2} \over 2} + {{{\omega _2}^2} \over 2} - {{{\omega _1}^2} \over 4} - {{{\omega _2}^2} \over 4} - {{2{\omega _1}{\omega _2}} \over 4}} \right]ΔK=I[2ω1​2​+2ω2​2​−4ω1​2​−4ω2​2​−42ω1​ω2​​]

=I[ω124+ω224−2ω1ω24] = I\left[ {{{{\omega _1}^2} \over 4} + {{{\omega _2}^2} \over 4} - {{2{\omega _1}{\omega _2}} \over 4}} \right]=I[4ω1​2​+4ω2​2​−42ω1​ω2​​]

=I4[ω1−ω2]2 = {I \over 4}{\left[ {{\omega _1} - {\omega _2}} \right]^2}=4I​[ω1​−ω2​]2

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