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QJEEM20Phc898: JEE Main 2020 Physics - Class 11 JEEM20Phc898 Solution: As shown in the figure, a bob of mass m is tied by a massless…

JEE Main 2020 (Online) 7th January Morning Slot Physics - Rotational Motion Question 143 English As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a fly wheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance of h, the angular speed of the wheel will be:

Multiple Choice Options

  • A r32ghr\sqrt {{3 \over {2gh}}}
  • B r34ghr\sqrt {{3 \over {4gh}}}
  • C 1r4gh3{1 \over r}\sqrt {{{4gh} \over 3}}
  • D 1r2gh3{1 \over r}\sqrt {{{2gh} \over 3}}

Step-by-Step Verified Solution

mgh = Δ\Delta KE

= 12mv2+12Iω2{1 \over 2}m{v^2} + {1 \over 2}I{\omega ^2}

For no slipping, v = ω\omega R

∴\therefore mgh = 12mω2R2+12mR22ω2{1 \over 2}m{\omega ^2}{R^2} + {1 \over 2}{{m{R^2}} \over 2}{\omega ^2}

⇒\Rightarrow mgh = 34mω2R2{3 \over 4}m{\omega ^2}{R^2}

⇒\Rightarrow ω\omega = 1r4gh3{1 \over r}\sqrt {{{4gh} \over 3}}

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