A ring is fitted loosely (with no friction) around a long, smooth rod of length \(L=0.50 \mathrm{~m} .\) The rod is fixed at one end, and the other end is spun in a horizontal circle at a constant angular velocity of \(\omega=4.0 \mathrm{rad} / \mathrm{s} .\) The ring has zero radial velocity at its initial position, a distance of \(r_{0}=0.30 \mathrm{~m}\) from the fixed end. Determine the radial velocity of the ring as it reaches the moving end of the rod.

Since there's no friction between the ring and the rod, and because the motion is circular, centripetal force is the only relevant force. With the help of the centripetal force equation and Newton's second law, we’ll create an equation for the ring's motion. The centripetal force is given by: \(\mathrm{F_c} =mr \omega^2\) We can write the equation of motion in terms of radial acceleration, \(\mathrm{F_c} =m \frac{\mathrm{d^2}r}{\mathrm{d}t^2}\) Combining these two equations, we get: \(m \frac{\mathrm{d^2}r}{\mathrm{d}t^2} = mr \omega^2\) This equation of motion will be used to find the radial velocity of the ring.

Since there is no external torque applied on the ring & rod system, the angular momentum of the system is conserved. The initial angular momentum of the system is equal to the final angular momentum. Initial angular momentum of the ring is: \(\mathrm{L_{initial}} = I_{\mathrm{initial}}\omega_{\mathrm{initial}} = m r_0^2 \omega\) Final angular momentum of the ring when it reaches the moving end of the rod is: \(\mathrm{L_{final}} = I_{\mathrm{final}}\omega_{\mathrm{final}} = m r_f^2 \omega\) Since the angular momentum is conserved, \(\mathrm{L_{initial}} = \mathrm{L_{final}}\). Therefore, we can write: \(m r_0^2 \omega = m r_f^2 \omega\) Divide both sides by m and ω, we get: \(r_0^2 = r_f^2\) As the final position is at the moving end of the rod, \(r_f = L\). Therefore: \(r_0^2 = L^2\) Now we can find the radial velocity by taking the derivative of the radius with respect to time: \(\frac{\mathrm{d}r }{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} (r_0^2)^{1/2}\) Using the chain rule, we have: \(\frac{\mathrm{d}r }{\mathrm{d}t} = \frac{1}{2}(r_0^2)^{-1/2} \frac{\mathrm{d}(r_0^2)}{\mathrm{d}t}\) Since \(r_0^2 = L^2\), we have: \(\frac{\mathrm{d}r }{\mathrm{d}t} = \frac{1}{2}(L^2)^{-1/2} \frac{\mathrm{d}(L^2)}{\mathrm{d}t}\) But, as the rod's length doesn't change, \(\frac{\mathrm{d}(L^2)}{\mathrm{d}t} = 0\) and thus, the radial velocity of the ring as it reaches the moving end of the rod is: \(\frac{\mathrm{d}r }{\mathrm{d}t} = 0\) So, the radial velocity of the ring when it reaches the moving end of the rod is \(0 \mathrm{m/s}\).