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Rotational Motion question

2019 · Q125
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Rotational Motion question

2019 · Q125

NEETPhysicsRotational MotionMCQ+4 / −1
A disc of radius 2 m and mass 100 kg rolls on a horizontal floor. Its centre of mass has speed of 20 cm/s. How much work is needed to stop it?
  1. A
    1 J
  2. B
    2 J
  3. C
    3 J
  4. D
    30 kJ
View written solutionFree

Correct answer: C

ω=VR=0.22=\omega = {V \over R} = {{0.2} \over 2} = ω=RV​=20.2​= 0.1 rad/sec

Apply the law of conservation of energy,

Work required = change in kinetic energy

Since, final KE = 0

And, initial KE = 12mv2+12Iω2{1 \over 2}m{v^2} + {1 \over 2}I{\omega ^2}21​mv2+21​Iω2

= 12mv2+12Ml22ω2{1 \over 2}m{v^2} + {1 \over 2}{{M{l^2}} \over 2}{\omega ^2}21​mv2+21​2Ml2​ω2

= 12(100)(0.2)2+12100×(2)22(0.1)2{1 \over 2}\left( {100} \right){\left( {0.2} \right)^2} + {1 \over 2}{{100 \times {{\left( 2 \right)}^2}} \over 2}{\left( {0.1} \right)^2}21​(100)(0.2)2+21​2100×(2)2​(0.1)2

= 1 + 12×4{1 \over 2} \times 421​×4 = 3 J

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