Answer:
C) ωf = 2ωi
Explanation:
To solve this problem, we can apply the principle of conservation of angular momentum. When the second disk is dropped onto the first disk, the total angular momentum of the system is conserved.
The initial angular momentum of the system is given by the angular momentum of the first disk, which is rotating with angular velocity
[tex]\( \omega_i \).[/tex]
Since the second disk is initially at rest, its angular momentum is zero. Therefore, the initial total angular momentum
[tex]\( L_i \)[/tex]
of the system is:
[tex]\[ L_i = I_1 \omega_i + I_2 \cdot 0 \][/tex]
Where:
[tex]- \( I_1 \) is the moment of inertia of the first disk ( \( I_1 = \frac{1}{2} M R^2 \) for a disk rotating about its center).[/tex]
[tex]- \( I_2 \) is the moment of inertia of the second disk, which is identical to \( I_1 \).[/tex]
[tex]- \( \omega_i \) is the initial angular velocity of the first disk.[/tex]
Since the two disks eventually rotate together, they act as one combined system with a total moment of inertia equal to the sum of their moments of inertia
[tex]( \( I_{\text{total}} = I_1 + I_2 = 2I_1 \) ).[/tex]
[tex]Therefore, the final angular momentum \( L_f \) of the system is:[/tex]
[tex]\[ L_f = I_{\text{total}} \omega_f \][/tex]
According to the principle of conservation of angular momentum,
[tex]\( L_i = L_f \).[/tex]
So, we can equate the initial and final angular momentum expressions:
[tex]\[ I_1 \omega_i = I_{\text{total}} \omega_f \][/tex]
Substituting the expressions for
[tex]\( I_1 \) and \( I_{\text{total}} \):[/tex]
[tex]\[ \frac{1}{2} M R^2 \omega_i = 2 \left( \frac{1}{2} M R^2 \right) \omega_f \][/tex]
[tex]Solving for \( \omega_f \):[/tex]
[tex]\[ \omega_f = \frac{\omega_i}{2} \][/tex]
So, the correct answer is c) C) ωf = 2ωi
[tex]\[ \omega_f = \frac{\omega_i}{2} \][/tex]
