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प्रश्न
A disc of the moment of inertia Ia is rotating in a horizontal plane about its symmetry axis with a constant angular speed ω. Another disc initially at rest of moment of inertia Ib is dropped coaxially onto the rotating disc. Then, both the discs rotate with the same constant angular speed. The loss of kinetic energy due to friction in this process is, ______
विकल्प
`1/2(I_b^2)/((I_a + I_b))ω^2`
`(I_b^2)/((I_a + I_b))ω^2`
`(I_b - I_a)^2/((I_a + I_b))ω^2`
`1/2(I_bI_b)/((I_a + I_b))ω^2`
उत्तर
A disc of the moment of inertia Ia is rotating in a horizontal plane about its symmetry axis with a constant angular speed ω. Another disc initially at rest of moment of inertia Ib is dropped coaxially onto the rotating disc. Then, both the discs rotate with the same constant angular speed. The loss of kinetic energy due to friction in this process is, `underline(1/2(I_bI_b)/((I_a + I_b))ω^2)`.
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