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  • Two discs of moments of inertia I_{1} and I_{2} about their respective 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

Two discs of moments of inertia I_{1} and I_{2} about their respective 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.
(a) Does the law of conservation of angular momentum apply to the situation? why?
(b) Find the angular speed of the two-disc system.
(c) Calculate the loss in kinetic energy of the system in the process.
(d) Account for this loss.

Answers (1)

(a) The law of conservation of angular momentum is applicable to this situation because there is no net external torque acting on the system. Gravitational and its normal reaction cancel out each other thus resulting in a 0 net torque.

(a) According to the law of conservation of angular momentum

L_{f}=L_{i}\\implies\\I{\omega }=I_{1}\omega _{1}+I_{2}\; \omega _{2}

\omega =\frac{I_{1}\omega _{1}+I_{2}\; \omega _{2}}{I_{1}+I_{2}}

(c) Final Kinetic Energy = (rotational +translational) Kinetic Energy

KE_{f}=KE_{R}+KE_{T}

In the absence of translation energy KE_{T}=0

KE_{f}=KE_{R}=\frac{1}{2}I\omega ^{2}=\frac{1}{2}(I_{1}+I_{2})\left [ \frac{I_{1}\omega _{1}+I_{2}\omega _{2}}{I_{1}+I_{2}} \right ]^{2}

KE_{f}=\frac{1}{2} \frac{(I_{1}\omega _{1}+I_{2}\omega _{2})^{2}}{(I_{1}+I_{2})}

KE_{i}=KE_{1R}+KE_{2R}+KE_{1T}+KE_{2T}

As discs don't have any translational motion KE_{1T}=KE_{2T}=0

KE_{i}=\frac{1}{2}I_{1}\omega _{1}^{2}+\frac{1}{2}I_{2}\omega _{2}^{2}=\frac{1}{2}(I_{1}\omega _{1}^{2}+I_{2}\omega _{2}^{2})

\Delta KE=KE_{f}-KE_{i}=\frac{1}{2}\frac{(I_{1}\omega _{1}+I_{2}\omega _{2})^{2}}{(I_{1}+I_{2})}-\frac{1}{2}(I_{1}\omega _{1}^{2}+I_{2}\omega _{2}^{2})

=\frac{1}{2}\left [ \frac{(I_{1}^{2}\omega _{1}^{2}+I_{2}^{2}\omega _{2}^{2}+2I_{1}I_{2}\omega _{1}\omega _{2}-\left [ (I_{1}+I_{2})(I_{1}\omega _{1}^{2}+I_{2}\omega _{2}^{2}) \right ])}{I_{1}+I_{2}} \right ]

=\left [ \frac{[I_{1}^{2}\omega _{1}^{2}+I_{2}^{2}\omega _{2}^{2}+2I_{1}I_{2}\omega _{1}\omega _{2}]-[I_{1}^{2}\omega _{1}^{2}+I_{2}^{2}\omega _{2}^{2}+I_{1}I_{2}\omega _{1}^{2}+I_{1}I_{2}\omega _{2}^{2}]}{2(I_{1}+I_{2})} \right ]

=-\frac{I_{1}I_{2}}{2(I_{1}+I_{2})}(-2\omega _{1}\omega _{2}+\omega _{2}^{2}+\omega _{1}^{2})

=-\frac{I_{1}I_{2}}{2(I_{1}+I_{2})}(\omega _{1}-\omega _{2})^{2}<0

The negative sign in the answer shows that final kinetic energy is less than the initial since energy is lost during friction between moving surfaces of the disc.

Posted by

infoexpert23

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