The moment of inertia of a ring about an axis passing through the 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 rings are rotating about the same axis then loss in…

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The moment of inertia of a ring about an axis passing through the 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 rings are rotating about the same axis then loss in kinetic energy is

Aicharya · Accepted Answer

From conservation of angular momentum I 1 &#969; 1 = I 2 &#969; 2 I &#969; = 2 I &#969; 2 &#969; 2 = &#969; 2 New KE = 1 2 . ( 2 I ) ( &#969; 2 ) 2 = I &#969; 2 4 Loss in KE = 1 2 I &#969; 2 &#8722; I &#969; 2 4 = I &#969; 2 4

The moment of inertia of a ring about an axis passing through the centre and perpendicular to its plane is I…

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