The final rotational kinetic energy, [tex]K_f[/tex], of the two (2) spinning disks is equal to [tex]K_f = K_i[\frac{I_t }{I_r + I_t}][/tex].
When the surfaces of two (2) spinning disks slip over each other, rotational kinetic energy is not conserved as a result of the effect of friction. Thus, the two (2) spinning disks would rotate with the same angular velocity.
To find the final rotational kinetic energy, [tex]K_f[/tex], of the two (2) spinning disks:
Mathematically, rotational kinetic energy is given by the formula:
[tex]K.E_R = \frac{1}{2} I\omega^2[/tex] ....equation 1.
Where:
- I is the moment of inertia.
- [tex]\omega[/tex] is the angular velocity.
For final rotational kinetic energy, [tex]K_f[/tex]:
[tex]K_f = \frac{1}{2} (I_r + I_t)\omega_f^2[/tex] ...equation 2.
But, the final angular velocity is given by:
[tex]\omega_f = \frac{I_t \omega_i}{I_r + I_t}[/tex] ...equation 3.
Substituting eqn. 3 into eqn. 2, we have:
[tex]K_f = \frac{1}{2} (I_r + I_t)[ \frac{I_t \omega_i}{I_r + I_t}]^2\\\\K_f = \frac{1}{2}I_t \omega_i^2[\frac{I_t }{I_r + I_t}]\\\\K_f = K_i[\frac{I_t }{I_r + I_t}][/tex]
Note: [tex]K_i = \frac{1}{2}I_t \omega_i^2[/tex]
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