Question

A uniform, 0.0300-kg rod of length 0.400 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.0200 kg, are mounted so that they can slide along the rod. They are initially held by catches at positions 0.0500 m on each side of the center of the rod, and the system is rotating at 48.0 rev/min. With no other changes in the system, the catches are released, and the rings slide outward along the rod and fly off at the ends. What is the angular speed (a) of the system at the instant when the rings reach the ends of the rod; (b) of the rod after the rings leave it?

Short answer

(a) 1.257 rad/s; (b) 6.285 rad/s.

Step-by-step solution

Convert Rotational Speed to Radians per Second

First, we need to convert the given angular velocity from revolutions per minute to radians per second. The initial angular velocity is \(48.0 \text{ rev/min}\). Therefore,\[\omega_i = 48.0 \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = 5.027 \text{ rad/s}.\]

Calculate Initial Moment of Inertia of the System

Next, find the initial moment of inertia \(I_i\) of the system when the rings are held at 0.0500 m from the center.The moment of inertia for the rod rotating about its center is \(I_{\text{rod}} = \frac{1}{12} m L^2 = \frac{1}{12} (0.0300 \text{ kg})(0.400 \text{ m})^2 = 4.0 \times 10^{-4} \text{ kg}\cdot\text{m}^2\).The moment of inertia for each ring is \(I_{\text{ring}} = mr^2\), where \(r = 0.0500 \text{ m}\). So for both rings,\[ I_{\text{rings,i}} = 2(0.0200 \text{ kg})(0.0500 \text{ m})^2 = 1.0 \times 10^{-4} \text{ kg}\cdot\text{m}^2. \]Thus, the initial total moment of inertia is \(I_i = I_{\text{rod}} + I_{\text{rings,i}} = 5.0 \times 10^{-4} \text{ kg}\cdot\text{m}^2.\)

Calculate Final Moment of Inertia of the System

When the rings reach the ends, the moment of inertia changes. Now, \(r = 0.200 \text{ m}\) (i.e., half the length of the rod).The moment of inertia for both rings now is\[ I_{\text{rings,f}} = 2(0.0200 \text{ kg})(0.200 \text{ m})^2 = 1.6 \times 10^{-3} \text{ kg}\cdot\text{m}^2. \]The final total moment of inertia is then\[ I_f = I_{\text{rod}} + I_{\text{rings,f}} = 4.0 \times 10^{-4} + 1.6 \times 10^{-3} = 2.0 \times 10^{-3} \text{ kg}\cdot\text{m}^2. \]

Use Conservation of Angular Momentum to Find Final Angular Velocity

Since no external torques act on the system, angular momentum is conserved. Thus,\[ I_i \omega_i = I_f \omega_f. \]Plugging in the known values:\[ (5.0 \times 10^{-4} \text{ kg}\cdot\text{m}^2)(5.027 \text{ rad/s}) = (2.0 \times 10^{-3} \text{ kg}\cdot\text{m}^2) \omega_f. \]Solving for \(\omega_f\):\[ \omega_f = \frac{(5.0 \times 10^{-4})(5.027)}{2.0 \times 10^{-3}} = 1.257\text{ rad/s}. \]

Calculate Angular Speed of the Rod After Rings Leave

After the rings leave, only the rod remains in rotation. The moment of inertia is just that of the rod, which is \(I_{\text{rod}}\).Using conservation of angular momentum at the instant before and after the rings leave:\[ I_f \omega_f = I_{\text{rod}} \omega_{\text{rod}} \]\[ (2.0 \times 10^{-3} \text{ kg}\cdot\text{m}^2)(1.257 \text{ rad/s}) = (4.0 \times 10^{-4} \text{ kg}\cdot\text{m}^2) \omega_{\text{rod}}. \]Solving for \(\omega_{\text{rod}}\):\[ \omega_{\text{rod}} = \frac{(2.0 \times 10^{-3})(1.257)}{4.0 \times 10^{-4}} = 6.285 \text{ rad/s}. \]

Conclusion

The angular speed of the system when the rings reach the ends is \(1.257 \text{ rad/s}\). After the rings leave, the angular speed of the rod is \(6.285 \text{ rad/s}\).

Key concepts

Moment of Inertia

The moment of inertia is a fundamental concept in physics that describes how mass is distributed in a rotating system. It acts as the rotational counterpart to mass in linear motion. The greater the moment of inertia, the harder it is to change the object's rotational speed. To compute the moment of inertia of a system, you sum the moments of inertia for individual components. To find the moment of inertia of the rod in the exercise, you use the formula for a rod rotating about its center: \[I_{\text{rod}} = \frac{1}{12} m L^2\] where \(m\) is the mass and \(L\) is the length of the rod. In our case, this becomes \(4.0 \times 10^{-4}\, \text{kg}\cdot\text{m}^2\).
For the rings, the moment of inertia is calculated using \(mr^2\), where \(r\) is the distance from the rotation axis. Initial placement at \(0.050\, \text{m}\) gives \(1.0 \times 10^{-4}\, \text{kg}\cdot\text{m}^2\) for both rings together, and when they slide to \(0.200\, \text{m}\), it becomes \(1.6 \times 10^{-3}\, \text{kg}\cdot\text{m}^2\).
Adding these computes the system's overall moment of inertia at different stages, influencing its rotation.

Rotational Motion

Rotational motion refers to objects spinning around a fixed axis. It follows rules akin to linear motion but considers additional parameters like angular displacement, velocity, and acceleration. In the context of our exercise, rotational motion involves the changes in angular velocity as the system's configuration changes.
The initial rotational speed is 48 revolutions per minute, but to work with standard physics equations, we convert this to radians per second. This is calculated by multiplying by \(\frac{2\pi}{60}\), resulting in \(5.027\, \text{rad/s}\). Understanding rotational dynamics is crucial here:

• Angular velocity represents how fast something spins, in radians per second.
• Changes in moment of inertia affect angular speed due to the inverse relationship between them.

As the rings move outward, the moment of inertia increases, thereby decreasing angular speed, illustrating the practical application of rotational motion theory.

Conservation of Angular Momentum

The principle of conservation of angular momentum states that if no external torques act on a system, its angular momentum remains constant. This concept helps predict rotational behavior when the system's configuration changes, such as the rings in the exercise.
Angular momentum \(L\) is the product of moment of inertia \(I\) and angular velocity \(\omega\): \[L = I \cdot \omega\]Initially, the system has distinct values for \(I\) and \(\omega\). As the rings slide, \(I\) changes. No external force affects the system, so angular momentum is the constant determinant of changes in \(\omega\).

• Start: \((I_i)(\omega_i)\) with low \(I_i\) and high \(\omega_i\)
• End: \((I_f)(\omega_f)\) with higher \(I_f\) due to rings moving outwards, lowering \(\omega_f\)

After the rings depart, leaving only the rotating rod, \(\omega_{\text{rod}}\) increases since \(L\) remains but \(I\) reduces back to \(I_{\text{rod}}\). This illustrates how angular momentum principles govern rotational behavior in systems lacking external disturbance.