/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 A smooth uniform rod of length \... [FREE SOLUTION] | 91Ó°ÊÓ

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A smooth uniform rod of length \(L\) and mass \(M\) has two identical beads ( 1 and 2) of negligible size, each of mass \(m\) which can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating with angular velocity \(\omega_{0}\) about an axis perpendicular to the rod and is passing through its midpoint. There are no external forces when the beads reach the ends of the rod, the angular velocity of the system is (1) \(\frac{M \omega_{b}}{M+6 m}\) (2) \(\frac{M \omega_{0}}{m}\) (3) \(\frac{M \omega_{0}}{M+12 m}\) (4) \(\omega_{0}\)

Short Answer

Expert verified
The angular velocity of the system when the beads reach the ends of the rod is \(\frac{M \omega_0}{M+6m}\).

Step by step solution

01

Understanding the Initial Condition

Initially, the rod of length \(L\) and mass \(M\) has two beads, each of mass \(m\), at its center. The system rotates with angular velocity \(\omega_0\) about the midpoint of the rod.
02

Calculating the Initial Moment of Inertia

The initial moment of inertia \(I_1\) is comprised of the rod and the two beads located at the center. The moment of inertia for the rod about its center is \(\frac{1}{12}ML^2\), and for the beads is \(2 \times m \times 0^2 = 0\). Therefore, \(I_1 = \frac{1}{12}ML^2\).
03

Applying Conservation of Angular Momentum

Since there are no external forces, the angular momentum is conserved. Thus, we have \( I_1 \omega_0 = I_2 \omega_f \), where \(\omega_f\) is the final angular velocity and \(I_2\) is the final moment of inertia when the beads are at the ends of the rod.
04

Calculating the Final Moment of Inertia

When the beads move to the ends of the rod, the distance from the center is \(\frac{L}{2}\). Thus, for the beads, the moment of inertia is \( 2 \times m \times (\frac{L}{2})^2 = \frac{1}{2}mL^2 \). The final moment of inertia \( I_2 \) is the sum of this and the rod's inertia: \( I_2 = \frac{1}{12}ML^2 + \frac{1}{2}mL^2 \).
05

Solving for the Final Angular Velocity

Substituting \(I_1\) and \(I_2\) back into the conservation equation: \[ \left(\frac{1}{12}ML^2\right) \omega_0 = \left(\frac{1}{12}ML^2 + \frac{1}{2}mL^2\right) \omega_f \]. Simplifying, \[ \omega_f = \frac{M\omega_0}{M+6m} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
The moment of inertia is a fundamental concept in rotational dynamics that describes how mass is distributed in relation to an axis of rotation. It's like the rotational equivalent of mass in linear motion. For any object, its moment of inertia determines how difficult it is to change its rotation. In the exercise, we started by calculating the initial moment of inertia, \(I_1\), of the rod and the beads system when they are positioned at the center. This is expressed as
  • Rod: \( \frac{1}{12}ML^2 \)
  • Beads: \( 2 \times m \times 0^2 = 0 \)
Thus, the initial moment of inertia is \( \frac{1}{12}ML^2 \).
This value increases once the beads move to the ends of the rod, demonstrating that distance from the axis affects rotational inertia. For two beads located at the ends of a rod, each at \( \frac{L}{2} \), the formula becomes:
  • Beads: \( 2 \times m \times (\frac{L}{2})^2 = \frac{1}{2}mL^2 \)
This is summed with the rod's inertia to get the final moment of inertia \(I_2\):
  • \( I_2 = \frac{1}{12}ML^2 + \frac{1}{2}mL^2 \).
These calculations reveal how mass positioning transforms reluctance to changes in rotational motion.
Rotational Dynamics
Rotational dynamics involves the study of objects in motion about an axis, and key to understanding this is the concept of torques and angular momentum. These concepts are integral in systems like the rod and beads, as explored in the exercise.
When analyzing such a system, it's essential to remember that without external forces, the angular momentum remains conserved. The angular momentum \(L\) of a system is given by the product of its moment of inertia \(I\) and its angular velocity \(\omega\), formulated as \(L = I \omega\).
In our scenario, the rod-bead system initially had an angular momentum of \( I_1 \omega_0 \). As the beads moved, energy conservation principles dictated that the final angular momentum \( I_2 \omega_f \) must equal the initial, meaning:
  • \( I_1 \omega_0 = I_2 \omega_f \)
This shift shows how rotational dynamics rely heavily on the conservation laws, helping solve for unknowns like final angular velocity given initial conditions.
Angular Velocity
Angular velocity is a vector quantity expressing how fast an object rotates or revolves around an axis. In the exercise, this concept is central to determining how the system's rotation changes as the beads move.
We initially know the system rotates with an angular velocity \(\omega_0\). Applying conservation of angular momentum as described above, when the beads reach the rod's ends and the moment of inertia changes to \(I_2\), the angular velocity becomes \(\omega_f\).
Solving the equation \( \left(\frac{1}{12}ML^2\right) \omega_0 = \left(\frac{1}{12}ML^2 + \frac{1}{2}mL^2\right) \omega_f \) showcases:
  • \( \omega_f = \frac{M\omega_0}{M+6m} \)
This tells us how angular velocity adjusts due to changes in the system's mass distribution, essential for understanding rotational motion in practical applications. It's a powerful tool in predicting how systems behave dynamically, akin to speed in linear kinematics.

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Most popular questions from this chapter

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