Question

Q. Two discs are mounted on frictionless bearings on  a common shaft. The first disc has rotational inertia I and is spinning with angular velocity ω. The second disc has rotational inertia 2I and is spinning in opposite direction with angular velocity 3ω  , as shown in figure. The two discs are slowly forced towards each other along the shaft until they couple and have a final common angular velocity of?

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Answer 1

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★ As we know angular momentum = I.w

(w= omega here),

(external torque is absent)

so, applying LAW OF CONSERVATION OF ANGULAR MOMENTUM we get

(I1+I2)wc = I1W1 - I2W2

substitute the given values of I and w

we get wc(omega common ) =

(3I)wc = IW - 6IW

wc = -5w/3

here, negative sign represents direction of rotation.

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Answer 2

Answer:

Given:

Smaller disc of Moment of Inertia I rotating with ω.

Larger disc of Moment of Inertia 2I rotating in the opposite direction with 3ω

To find:

Common angular Velocity

Concept:

Since no external torque is applied to the system, the Angular Momentum of the system shall remain constant.

Calculation:

Initial Angular Momentum = Final Angular Momentum

$= > I \omega - (2I)(3 \omega) = (3I) \omega_{final}$

$= > I \omega - 6I \omega = (3I) \omega_{final}$

$= > - 5I \omega = (3I) \omega_{final}$

$= > \omega_{final} = - \dfrac{5}{3} \omega$

Negative sign denotes opposite Angular velocity as compared to the initial angular velocity of smaller disc .