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