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A disk of mass M and radius R rotates at angular velocity ω0. Another disk of mass M and radius r is dropped on top of the rotating disk such that their centers coincide. Both disks now spin at a new angular velocity ω. What is ω?

Respuesta :

Answer:

[tex]\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}[/tex]

Explanation:

As we know that there is no external torque on the system of two disc

then the angular momentum of the system will remains conserved

So we will have

[tex]L_i = L_f[/tex]

now we have

[tex]L_i = (\frac{1}{2}MR^2)\omega_o[/tex]

also we have

[tex]L_f = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega[/tex]

now from above equation we have

[tex](\frac{1}{2}MR^2)\omega_o  = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega[/tex]

now we have

[tex]\omega = \frac{MR^2\omega_o}{(MR^2 + Mr^2)}[/tex]

[tex]\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}[/tex]

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