Answer:
E). 7.097 rad/s
Explanation:
As we know that the turn table and the clay both are isolated from all external forces and torque so the angular momentum of the system must be conserved
so we will have
[tex]I_1\omega_1 = (I_1 + I_2)\omega_2[/tex]
[tex]\frac{1}{2}MR^2\omega_1 = (\frac{1}{2}MR^2 + mr^2)\omega_2[/tex]
now plug in all data in it
[tex]\frac{1}{2}(25)3^2 (7.9) = (\frac{1}{2}(25) 3^2 + 12 (1.1)^2) \omega[/tex]
[tex]888.75 = 127.02 \omega[/tex]
[tex]\omega = \frac{888.75}{127.02}[/tex]
[tex]\omega = 7.097 rad/s[/tex]
