A disk rotates freely on a vertical axis with an angular velocity of 30 rpm. An identical disk rotates above it in the same direction about the same axis, but without touching the lower disk, at 20 rpm. The upper disk then drops onto the lower disk. After a short time, because of friction, they rotate together. The final angular velocity of the disks is
A. 50 rpm
B. 40 rpm
C. 25 rpm
D. 20 rpm
E. 10 rpm

Respuesta :

Answer:

c. 25 rpm

Explanation:

The inertia of both disk is the same so

[tex]I_1=I_2[/tex]

[tex]I_1*w_{1i}+I_2*w_{2i}=I_1*w_{1f}+I_2*w_{2f}[/tex]

[tex]I*(w_{1i}+w_{2i})=I*(w_{1f}+w_{2f})[/tex]

[tex]w_{1f}=w_{2f}=w_f[/tex]

[tex]w_{1f}+w_{2f}=2w_f[/tex]

[tex]w_f=\frac{w_{1i}+w_{i2}}{2}[/tex]

[tex]w_f=\frac{30rpm+20rpm}{2}[/tex]

[tex]w_f=\frac{50rpm}{2}[/tex]

[tex]w_f=25rpm[/tex]

he final angular momentum is equal to 25rpm

Data:

  • ω1 = 30rmp
  • ω2 = 20 rmp

Angular Momentum

From the conservation of angular momentum, the initial angular momentum is equal to the final angular momentum.

[tex]L_1 = L_2\\I_1\omega_1 + I_2 \omega_2 = I' \omega'\\I(\omega_1 + \omega_2) = 2I\omega'\\\omega' = \frac{\omega_1 + \omega_2}{2}[/tex]

Let's substitute the values

[tex]\omega ' = \frac{30+20}{2} = 25rpm[/tex]

The final angular momentum is equal to 25rpm

Learn more on conservation of angular momentum here;

https://brainly.com/question/10124070