Answer:
[tex]v_{\theta\text{-}f} = 12.49{\rm\dfrac{[rad]}{s}}[/tex]
Explanation:
We are given the following information:
We are solving for the final angular velocity. We can find this using the definition of rotational inertia:
[tex]I_\theta = \dfrac{L}{v_\theta}[/tex]
where:
Plugging in the known values, we can solve for the initial angular momentum:
[tex]5.33 \text{ kg}\cdot\text{m}^2= \dfrac{L_i}{3.75 \text{ s}^{-1}}[/tex]
[tex]L_i = 19.99\ \dfrac{\text{ kg}\cdot\text{m}^2}{\text{s}}[/tex]
Then, we can solve for the final angular velocity using conservation of angular momentum:
[tex]L_i = L_f[/tex]
[tex]19.99\ \dfrac{\text{ kg}\cdot\text{m}^2}{\text{s}} = I_{\theta\text{-}f}\,v_{\theta\text{-}f}[/tex]
[tex]19.99\ \dfrac{\text{ kg}\cdot\text{m}^2}{\text{s}} = (1.60\text{ kg}\cdot\text{m}^2)\,v_{\theta\text{-}f}[/tex]
[tex]\boxed{v_{\theta\text{-}f} = 12.49{\rm\dfrac{[rad]}{s}}}[/tex]