Problem 59 - Fundamentals of Physics

Question

An overhead view of a thin uniform rod of length \(0.800 \mathrm{~m}\) and mass \(M\) rotating horizontally at angular speed \(20.0 \mathrm{rad} / \mathrm{s}\) about an axis through its center. A particle of mass \(M / 3,00\) initially attached to one end is ejected from the rod and travels along a path that is perpendicular to the rod at the instant of ejection. If the particle's speed \(v_{p}\) is \(6.00 \mathrm{~m} / \mathrm{s}\) greater than the speed of the rod end just after ejection, what is the value of \(v_{p} ?\)

Step-by-Step Solution

Answer

The speed of the ejected particle, \(v_p\), is 14.0 m/s.

Step 1: Determine the Speed of the Rod End

The end of the rod moves at a tangential speed due to the rotation. The speed of the rod end, \(v_{r}\), is calculated using the formula \(v_{r} = \omega \cdot r\), where \(\omega\) is the angular speed and \(r\) is the radius of the rod. Since the rod is rotating about its center, \(r = \frac{0.800}{2} = 0.400 \: \text{m}\). Thus, \(v_{r} = 20.0 \: \text{rad/s} \times 0.400 \: \text{m} = 8.0 \: \text{m/s}\).

Step 2: Apply the Condition for Particle Speed

According to the problem, the particle's speed \(v_p\) is \(6.00 \: \text{m/s}\) greater than the speed of the rod end just after ejection. Therefore, we can write the equation: \(v_p = v_r + 6.0 \: \text{m/s}\). Substituting the value calculated earlier, \(v_p = 8.0 \: \text{m/s} + 6.0 \: \text{m/s}\).

Step 3: Solve for Particle Speed

Perform the addition from the condition. From step 2, \(v_p = 8.0 + 6.0 = 14.0 \: \text{m/s}\). Therefore, the speed of the ejected particle is \(14.0 \: \text{m/s}\).

Key Concepts

Angular Speed

Angular speed is a concept that helps describe how fast an object rotates. It's denoted by the Greek letter \( \omega \) (omega) and is measured in radians per second (rad/s). In simple terms, angular speed tells us how many radians an object passes through per second as it spins around an axis.
For any rotating object, such as the rod in our exercise, knowing the angular speed is crucial because it directly influences the tangential speed of any point on the object. To calculate it, you need to know the number of turns in a given time or, more formally, the angular displacement per unit time. In our problem, the rod has an angular speed of \( 20.0 \text{ rad/s} \), which helps us find the tangential speed of the rod's ends.

Tangential Speed

Tangential speed is how fast a point on a rotating object is moving in a straight line (tangential to the circle of rotation). It's measured in meters per second (m/s) and is important because it affects how an object behaves when released, as happens with the particle in our exercise.
The formula for tangential speed is \( v = \omega \cdot r \), where \( \omega \) is the angular speed and \( r \) is the radius of rotation. For the rod, the radius \( r \) is half its length since it's rotating about its center. By substituting the values:

• \( r = 0.400 \text{ m} \)
• \( \omega = 20.0 \text{ rad/s} \)

We get a tangential speed of \( v = 8.0 \text{ m/s} \). This tells us how quickly the rod's end moves, which is essential for calculating the ejected particle's speed.

Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause this motion. It includes concepts like velocity, speed, distance, and acceleration.
In this exercise, kinematics helps us understand the relationship between the speeds of different parts of the rotating system, such as the rod and the ejected particle. The given information states that the particle's speed \( v_p \) is \( 6.00 \text{ m/s} \) greater than the speed of the rod's end immediately after ejection. Kinematics lets us calculate the final speed of the particle by adding this velocity difference to the previously determined tangential speed of the rod's end. Thus, we have:

• \( v_p = v_r + 6.0 \text{ m/s} \)
• \( v_p = 8.0 \text{ m/s} + 6.0 \text{ m/s} \)
• \( v_p = 14.0 \text{ m/s} \)

This direct calculation demonstrates how kinematic concepts make difficult problems accessible.

Conservation of Angular Momentum

The conservation of angular momentum is a fundamental idea in rotational dynamics. It states that if no external torque acts on a system, the total angular momentum remains constant. Angular momentum is the rotational equivalent of linear momentum and plays a significant role in problems involving rotating objects.
In our exercise, although the specific values didn't require calculation of angular momentum, understanding the concept is critical. When the particle is ejected, the system's angular momentum (i.e., rod plus particle) could be considered unchanged if we imagined a situation without external influences.
Applying these principles illuminates how systems can change in interesting ways, like a figure skater pulling in arms to spin faster. Similarly, if the particle interaction didn't involve external forces, the velocity changes would be guided by angular momentum conservation.