Problem 88 - University Physics with Modern Physics

Question

A uniform, \(0.0300-\mathrm{kg}\) rod of length 0.400 \(\mathrm{m}\) 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 \(\mathrm{kg}\) , are mounted so that they can slide along the rod. They are initially held by catches at positions 0.0500 \(\mathrm{m}\) on each side of the center of the rod, and the system is rotating at 30.0 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. (a) What is the angular speed of the system at the instant when the rings reach the ends of the rod? (b) What is the angular speed of the rod after the rings leave it?

Step-by-Step Solution

Answer

(a) 0.785 rad/s; (b) 1.57 rad/s.

Step 1: Determine initial moment of inertia

Initially, the system consists of the rod and the two rings held at a distance of 0.0500 m from the center. Calculate the initial moment of inertia, \(I_i\), taking into account the rod's and the rings' moments of inertia.For the rod: \( I_{rod} = \frac{1}{12} m_{rod} L^2 = \frac{1}{12} (0.0300\, \text{kg}) (0.400\, \text{m})^2 = 4.00 \times 10^{-4} \text{ kg}\cdot\text{m}^2 \).For each ring at a distance of 0.0500 m from the center: \( I_{ring} = m_{ring} r^2 = (0.0200\, \text{kg}) (0.0500\, \text{m})^2 = 5.00 \times 10^{-5} \text{ kg}\cdot\text{m}^2 \).There are two rings, so the total moment of inertia for the rings is: \( 2 \times 5.00 \times 10^{-5} \text{ kg}\cdot\text{m}^2 = 1.00 \times 10^{-4} \text{ kg}\cdot\text{m}^2 \).Adding them up: \( I_i = 4.00 \times 10^{-4} + 1.00 \times 10^{-4} = 5.00 \times 10^{-4} \text{ kg}\cdot\text{m}^2 \).

Step 2: Calculate initial angular velocity

Convert the initial angular velocity from revolutions per minute to radians per second. Given \( \, \omega_i = 30.0 \, \text{rev/min}\),Convert to \( \text{rad/s} \):\[ \omega_i = 30.0 \, \frac{2\pi\, \text{rad}}{1\, \text{rev}} \cdot \frac{1}{60\text{ s}} = \pi \, \text{rad/s} \].

Step 3: Apply conservation of angular momentum for system with rings reaching the end

According to the conservation of angular momentum, the initial angular momentum \( L_i = \) the final angular momentum \( L_f \) when the rings are at the ends.\[ I_i \cdot \omega_i = I_f \cdot \omega_f \] At the ends of the rod, each ring is at a distance of \(0.200\, \text{m}\) from the axis of rotation. Calculate \( I_f \):\[ I_f = I_{rod} + 2(m_{ring} r^2) = 4.00 \times 10^{-4} + 2(0.0200)(0.200^2) \].Simplify:\[ I_f = 4.00 \times 10^{-4} + 2(0.0200 \cdot 0.0400) = 2.00 \times 10^{-3}\, \text{kg}\cdot\text{m}^2 \].Solving for \( \omega_f \):\[ \omega_f = \frac{I_i\cdot\omega_i}{ I_f} = \frac{5.00 \times 10^{-4} \cdot \pi}{ 2.00 \times 10^{-3}} = \frac{\pi}{4} \approx 0.785\, \text{rad/s} \].

Step 4: Calculate angular speed after rings leave

When the rings leave the rod, only the rod remains. The moment of inertia is now \( I_{rod} \), which remains as \( 4.00 \times 10^{-4}\, \text{kg}\cdot\text{m}^2 \).Applying conservation of angular momentum again:\[ I_f \cdot \omega_f = I_{rod} \cdot \omega_{final} \].Solving for \( \omega_{final} \):\[ \omega_{final} = \frac{I_f \cdot \omega_f}{ I_{rod}} = \frac{2.00 \times 10^{-3} \cdot \frac{\pi}{4}}{ 4.00 \times 10^{-4}} = \frac{\pi}{2} \approx 1.57\, \text{rad/s} \].

Key Concepts

Angular Momentum

Angular momentum is a measure of the rotational equivalent of linear momentum. It occurs in any object that is rotating or moving along a circular path. Angular momentum depends largely on two factors:

• the object's moment of inertia
• its angular velocity

The formula for angular momentum (\( L \)) is:\[ L = I \cdot \omega \]where\( I \) is the moment of inertia and\( \omega \) is the angular velocity.
When an object is rotating, such as the rod in our example, it possesses angular momentum. This quantity tells us how much motion the object has as it rotates. It is a pivotal concept in rotational motion because it introduces the idea of how motion is conserved or altered when conditions change, such as the rings moving along the rod in our exercise.
Understanding angular momentum is essential when analyzing or predicting the effects of forces on rotating objects, as it lays the foundational groundwork for more complex systems.

Conservation of Angular Momentum

The principle of the conservation of angular momentum states that the total angular momentum of a closed system remains constant if no external torques act on it. In simpler terms, if nothing outside is trying to stop or change the rotation of the system, the amount of spin or rotation energy stays the same.
In our exercise, releasing the rings from the catches along the rod exemplifies this principle. Before the rings are released, we calculate the total initial angular momentum by considering both the rods and the rings’ moments of inertia and their shared angular velocity: \[L_i = I_i \cdot \omega_i\] Once the catches are released, the rings move outward to the ends of the rod, changing the distribution of mass. However, because no external torque is applied, the system's angular momentum remains constant: \[L_i = L_f = I_f \cdot \omega_f\] This conservation allows us to determine how the angular speed changes as the rings reach the ends. Even after the rings leave, the rod itself continues to follow this principle, but now considering only the rod's own moment of inertia. It helps us understand phenomena where, despite changes internally, the rotational behavior follows an intuitive conservation rule.

Rotational Motion

Rotational motion refers to an object spinning around an axis. Unlike linear motion, where movement is in a straight line, rotational motion entails a circular path, a common occurrence in mechanical systems.
Understanding the basic concepts of rotational motion includes grasping terms like:

• **Moment of Inertia**: This is a measure of how much an object resists changes to its rotational state. It's akin to mass in linear motion. For the rod, its moment of inertia is calculated using the given length and mass, which explains how mass distribution affects rotation.
• **Angular Velocity**: This indicates how fast an object rotates. Initially given in revolutions per minute (rpm), it needs to be converted to radians per second for precise calculations.

In our practical exercise, the movement of the rod and rings exemplifies rotational motion principles. When the rings slide to the rod's ends, their increased radius significantly affects the rotational speed due to the increased moment of inertia. Observing these phenomena in action makes rotational dynamics clearer and explains how and why objects behave the way they do when spinning.