Question

si un disco que gira a 800 rpm se acopla a uno cuyo momento de inercia es del triple, ¿que rapidez angular tendra la combinacion?

Answer 1

Answer:

the final angular velocity of the system would be 400 rpm

Explanation:

Recall that the kinetic energy of a rotating object is defined as:

$K=\frac{1}{2} \,I\,*\,\omega^2$

where I is the object's moment of inertia, and $\omega$ its angular velocity.

If the kinetic energy is conserved in the coupling of the two discs, then the final kinetic energy of the new system should be the same:

$K=\frac{1}{2} \,I\,*\,\omega^2=\frac{1}{2} (I+3I)\,*\,\omega_f^2\\\omega_f^2=\frac{\omega^2}{4}$

therefore the new angular velocity will be:

$\sqrt{\omega_f^2}=\sqrt{\frac{\omega_i^2}{4}} \\\omega_f=\frac{\omega_i}{2}$

half of the initial angular velocity. That is :

the final angular velocity of the system would be 400 rpm