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Rotational Motion question

2021 · 26 Feb · Shift 2 · Q63
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  5. /2021 · 26 Feb · Shift 2 · Q63

Rotational Motion question

2021 · 26 Feb · Shift 2 · Q63

JEE MainPhysicsRotational MotionMCQ+4 / −1
A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance 'h', the square of angular velocity of wheel will be :
  1. A
    2mghI+2mr2{{2mgh} \over {I + 2m{r^2}}}I+2mr22mgh​
  2. B
    2mghI+mr2{{2mgh} \over {I + m{r^2}}}I+mr22mgh​
  3. C
    2gh
  4. D
    2ghI+mr2{{2gh} \over {I + m{r^2}}}I+mr22gh​
View written solutionFree

Correct answer: B

  1. Set up the energy equation

As the mass mmm falls through a distance hhh, its loss in gravitational potential energy becomes:

  • translational kinetic energy of the mass,
  • rotational kinetic energy of the wheel.

So, mgh=12mv2+12Iω2mgh = \frac12 m v^2 + \frac12 I\omega^2mgh=21​mv2+21​Iω2

  1. Use the no-slip condition

Since the cord unwinds without slipping, v=rωv = r\omegav=rω

Substitute into the energy equation: mgh=12m(rω)2+12Iω2mgh = \frac12 m (r\omega)^2 + \frac12 I\omega^2mgh=21​m(rω)2+21​Iω2

mgh=12(mr2+I)ω2mgh = \frac12 \left(mr^2 + I\right)\omega^2mgh=21​(mr2+I)ω2

  1. Solve for ω2\omega^2ω2

ω2=2mghI+mr2\omega^2 = \frac{2mgh}{I + mr^2}ω2=I+mr22mgh​

  1. Match with options

This matches Option B: 2mghI+mr2\boxed{\frac{2mgh}{I+mr^2}}I+mr22mgh​​

  1. Verification with stored answer

Stored correct answer: B

Our derived answer is also B, so they agree.

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