Question

A ring is fitted loosely (with no friction) around a long, smooth rod of length \(L=0.500 \mathrm{~m} .\) The rod is fixed at one end, and the other end is spun in a horizontal circle at a constant angular velocity of \(\omega=4.00 \mathrm{rad} / \mathrm{s} .\) The ring has zero radial velocity at its initial position, a distance of \(r_{0}=0.300 \mathrm{~m}\) from the fixed end. Determine the radial velocity of the ring as it reaches the moving end of the rod.

Short answer

Answer: The radial velocity of the ring when it reaches the moving end of the smooth rod is approximately 2.26 m/s.

Step-by-step solution

Find Initial Conditions

Initially, the ring is at a distance \(r_0=0.300 \mathrm{~m}\) from the fixed end, and it has zero radial velocity. The rod is spinning at a constant angular velocity of \(\omega=4.00 \mathrm{rad} / \mathrm{s} .\)

Determine the centripetal force acting on the ring

The centripetal force acting on the ring is given by the equation: \(F_c = m\omega^2 r\) where \(m\) is the mass of the ring and \(r\) is the distance of the ring from the fixed end. Note that the mass and moment of inertia of the ring are not given. However, their ratio is constant, and we are only interested in the change in radial velocity, which will depend on this constant.

Determine the radial velocity of the ring as a function of distance

In order to determine the radial velocity of the ring as it moves along the rod, we need to determine the radial acceleration of the ring at each position \(r\). The radial acceleration is given by the centripetal acceleration, which is given by \(a_r = \omega^2 r\) Using this equation, we can find the relationship between radial velocity \(v_r\) and distance \(r\), and then integrate over the distance from the initial position to the moving end of the rod. \(v_r^2 = v_{r0}^2 + 2\int_{r_0}^r a_r dr\) Since the initial radial velocity \(v_{r0}=0\), this simplifies to: \(v_r^2 = 2\int_{r_0}^r a_r dr = 2\int_{r_0}^r \omega^2 r dr\)

Calculate the radial velocity when the ring reaches the moving end of the rod

Now we can calculate the radial velocity as the ring reaches the moving end of the rod (when r=L) using the relation obtained in Step 3. \(v_r^2 = 2\int_{r_0}^{L} \omega^2 r dr = 2\omega^2\int_{r_0}^{L} r dr\) We can now perform the integration, and then find \(v_r\): \(v_r^2 = 2\omega^2\left[\frac{1}{2}(L^2 - r_0^2)\right]\) Finally, we can plug in the given values for \(\omega\), \(L\), and \(r_0\): \(v_r^2 = 2(4.00 \mathrm{rad} / \mathrm{s})^2\left[\frac{1}{2}(0.500 \mathrm{~m}^2 - 0.300 \mathrm{~m}^2)\right]\) \(v_r^2 \approx 5.12\) And taking the square root, we find the radial velocity: \(v_r \approx \pm \sqrt{5.12} \mathrm{m/s} \approx \pm 2.26 \mathrm{m/s}\) Since the ring can only move in a positive direction along the rod, we choose the positive solution: \(v_r \approx 2.26 \mathrm{m/s}\)