Question

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.

Short answer

Answer: The radial velocity of the ring when it reaches the moving end of the rod is 0 m/s.

Step-by-step solution

Write the equations of motion for the ring

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.

Apply conservation of angular momentum

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 \(0 \mathrm{m/s}\).

Key concepts

Centripetal Force

Centripetal force is crucial in understanding the motion of objects moving in a circular path. It is the force that keeps an object moving along this curved trajectory, directing it towards the center around which it rotates.
This force doesn't create a new speed; instead, it perpetually changes the direction of the object's velocity, sustaining its circular motion.
Key characteristics of centripetal force include:

• Direction: Always pointing towards the center of rotation.
• Magnitude: It can be calculated with the formula: \( F_c = mr\omega^2 \), where\( m \) is mass, \( r \) is the radius of motion, and \( \omega \) is the angular velocity.

The centripetal force ensures that the ring follows the rod's circular path. A lack of this force results in the object moving tangentially due to inertia.
Understanding centripetal force helps understand why the circular motion is sustained despite the absence of tangential friction.

Radial Velocity

Radial velocity refers to how quickly an object moves towards or away from a central point. In this exercise, the radial velocity is the speed at which the ring approaches the center or moves away from it on the rod.
Initially, the ring's radial velocity is zero, indicating neither movement towards nor away from the fixed point on the rod. As the ring travels to the opposite end, we investigate whether any radial speed develops.
A zero radial velocity signifies stability in the radial position. The analysis of radial velocity involves examining any change or constancy in the circle's radius, often linked to changes in the centripetal conditions or alterations in external forces.
In our given scenario, as the rod rotates steadily and without external disturbances, calculations confirm that the radial velocity remains zero when the ring reaches the rod's end. This outcome emphasizes the importance of symmetry and conservation laws in closed systems.

Equations of Motion

Equations of motion are mathematical formulations that describe the behavior of a moving object under the influence of forces. They typically incorporate Newton's second law, which states: force equals mass times acceleration (\( F = ma \)).
To model the motion of the ring, we modify Newton's second law for circular paths using the relationship involving centripetal force: \( F_c = mr\omega^2 \). By setting this equal to \( m \frac{d^2r}{dt^2} \), we establish a framework for determining radial acceleration. This setup is pivotal because:

• It provides insights into how forces affect circular motion.
• Enables predictions regarding the object's path over time.

In connection with the ring, the equations show how radial movements (or the lack thereof) interact with rotational dynamics. Applying these equations in the problem, we deduce that the constant angular velocity and maintained centripetal force terms lead to constant radial positioning, proving that radial speed remains steadily zero at the transition endpoint.