मराठी

The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is 'I'. It is rotating with angular velocity 'ω'. -

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प्रश्न

The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is 'I'. It is rotating with angular velocity 'ω'. Another identical ring is gently placed on it so that their centres coincide. If both the ring are rotating about the same axis, then loss in kinetic energy is ______.

पर्याय

  • `(Iomega^2)/3`

  • `(Iomega^2)/4`

  • 2

  • `(Iomega^2)/2`

MCQ
रिकाम्या जागा भरा

उत्तर

The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is 'I'. It is rotating with angular velocity 'ω'. Another identical ring is gently placed on it so that their centres coincide. If both the ring are rotating about the same axis, then loss in kinetic energy is `underline((Iomega^2)/4)`.

Explanation:

By law of conservation of angular momentum

I1ω1 = I2ω2

∴ Iω = 2Iω

`thereforeomega'=omega/2`

Initial kinetic energy `K_1=1/2Iomega^2`

Final kinetic energy `K_2=1/2(2I)omega'^2`

`=1/2xx2Ixxomega^2/4=(Iomega^2)/4`

`K_1-K_2=1/2Iomega^2-1/4Iomega^2=(Iomega^2)/4`

shaalaa.com
Moment of Inertia as an Analogous Quantity for Mass
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