A disc of the moment of inertia I 1 is rotating in a horizontal plane about an axis passing through a centre and perpendicular to its plane with constant angular speed ω 1 Another disc of the moment o

Question

A disc of the moment of inertia I 1 is rotating in a horizontal plane about an axis passing through a centre and perpendicular to its plane with constant angular speed ω 1 Another disc of the moment of inertia I 2 having zero angular speed is placed coaxially on a rotating disc. Now both the disc are rotating with the constant angular speed ω 2 . The energy lost by the initial rotating disc is

TestHub · Accepted Answer

I 1 ω 1 = I 1 + I 2 ω 2 ω 2 ω 1 = I 1 I 1 + I 2 E 1 - E 2 = 1 2 I 1 ω 1 2 - 1 2 ( I 1 + I 2 ) ω 2 2 = 1 2 ω 1 2 I 1 - I 1 + I 2 ω 2 2 ω 1 2 = 1 2 ω 1 2 I 1 - I 1 + I 2 I 1 2 I 1 + I 2 2 = 1 2 ω 1 2 I 1 2 + I 1 I 2 - I 1 2 I 1 + I 2 = 1 2 I 1 I 2 I 1 + I 2 ω 1 2

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