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Two discs of same moment of inertia rota...

Two discs of same moment of inertia rotating their regular axis passing through centre and perpendicular to the plane of disc with angular velocities `omega_(1)` and `omega_(2)`. They are brought into contact face to the face coinciding the axis of rotation. The expression for loss of enregy during this process is :

A

`1/2 l (omega_(1)+omega_(2))^(2)`

B

`1/4 l(omega_(1)-omega_(2))^(2)`

C

`l (omega_(1)-omega_(2))^(2)`

D

`l/8 (omega_(1)-omega_(2))^(2)`

Text Solution

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The correct Answer is:
b
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Knowledge Check

  • Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities omega_(1) and omega_(2) . They are brought into contact face to face coinciding the axis of rotation . The expression for loss of energy during this process is

    A
    `(1)/(4) I (omega_1 - omega_2)^(2)`
    B
    `I (omega_(1) - omega_(2))^(2)`
    C
    `(1)/(8) I (omega_(1) - omega_(2))^(2)`
    D
    `(1)/(2) I (omega_(1) + omega_(2))^(2)`

  • The moment of inertia of a copper disc, rotating about an axis passing through its centre and perpendicular to its plane

    A
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    B
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    C
    increases if its angular velocity is increased
    D
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  • Two discs of moment of inertia I_(1) and I_(2) about their respectively axes (normal) to the disc and passing through the centre) , and rotating with angular speed omega_(1) and omega_(2) are brought into contact face to face with their axes of rotation coincident . What is the loss in kinetic energy of the system in the process ?

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    `(I_(1) I_(2) (omega_(1) - omega_(2))^(2))/(2 (I_(1) + I_(2)))`
    B
    `(I_(1) I_(2) (omega_(1) - omega_2)^(2))/((I_(1) + I_(2)))`
    C
    `(I_(1) I_(2) (omega_(1) + omega_(2))^(2))/((I_(1) - I_(2)))`
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