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?
Answers
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}=>Iω−(2I)(3ω)=(3I)ωfinal
= > I \omega - 6I \omega = (3I) \omega_{final}=>Iω−6Iω=(3I)ωfinal
= > - 5I \omega = (3I) \omega_{final}=>−5Iω=(3I)ωfinal
= > \omega_{final} = - \dfrac{5}{3} \omega=>ωfinal=−35ω
Negative sign denotes opposite Angular velocity as compared to the initial angular velocity of smaller disc .
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Answer:
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
negative sign represents direction of rotation.