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For a rigid body in translation, show that the system of the inertial terms consists of vectors \(\left(\Delta m_{i}\right) \overline{\mathbf{a}}\) attached to the various particles of the body, where \(\overline{\mathbf{a}}\) is the acceleration of the mass center \(G\) of the body. Further show, by computing their sum and the sum of their moments about \(G,\) that the inertial terms reduce to a single vector \(m \overline{\mathbf{a}}\) attached at \(G\)

Short Answer

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The inertial terms reduce to a single vector \( m \overline{\mathbf{a}} \) attached at the center of mass \( G \).

Step by step solution

01

Understanding Inertial Terms in Rigid Body Translation

In a rigid body in translation, every point of the body has the same acceleration due to the motion of the center of mass, denoted as \( \overline{\mathbf{a}} \). If a body is composed of discrete particles with masses \( \Delta m_i \), then for each particle, the inertial term can be represented as \( \Delta m_i \overline{\mathbf{a}} \). This indicates that the inertial forces acting on each particle are collinear with the acceleration of the center of mass.
02

Summing the Inertial Forces

To analyze the system of inertial forces, sum the force vectors \( \Delta m_i \overline{\mathbf{a}} \) for all particles in the body. Since each \( \overline{\mathbf{a}} \) is the same, the summation becomes: \[ \sum (\Delta m_i \overline{\mathbf{a}} ) = \overline{\mathbf{a}} \sum \Delta m_i. \]This simplifies to \( m \overline{\mathbf{a}} \) where \( m \) is the total mass of the body \( m = \sum \Delta m_i \). This shows that the inertial force system reduces to a single vector \( m \overline{\mathbf{a}} \) applied at the center of mass.
03

Summing the Moments About G

For the moments, consider the moment of each inertial force \( \Delta m_i \overline{\mathbf{a}} \) about the center of mass \( G \). The moment is given by the cross product \( \mathbf{r}_i \times (\Delta m_i \overline{\mathbf{a}}) \), where \( \mathbf{r}_i \) is the position vector from \( G \) to the particle. Since \( \overline{\mathbf{a}} \) is the same for each particle and collinear with the force, \[ \sum \mathbf{r}_i \times (\Delta m_i \overline{\mathbf{a}}) = \overline{\mathbf{a}} \times \sum (\mathbf{r}_i \Delta m_i) = 0, \]as \( \sum (\mathbf{r}_i \Delta m_i) = 0 \) because the sum of moments \( \mathbf{r}_i \Delta m_i \) about the center of mass is zero due to symmetry or definition of the center of mass.
04

Final Conclusion About the Inertial System

The summation of inertial forces reduces to a single vector \( m \overline{\mathbf{a}} \) at the center of mass, with no net moment about \( G \) under translation. This simplification shows that the system of inertial forces acts as if all the mass were concentrated at the center of mass, with the same linear acceleration as the whole body, effectively making it behave as a particle with mass \( m \).

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Key Concepts

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

Center of Mass
The center of mass (often denoted as \(G\)) of a rigid body is a crucial point to understand when analyzing motion. It is a point where we can consider the entire mass of an object to be concentrated for the sake of calculations. This simplifies the study of the body's movement, especially when it's in translation motion, similar to how we consider the motion of a simple particle. The center of mass helps us predict and analyze the motion of complex objects more easily.
  • All forces and motions can be referenced from this point.
  • It is essentially the average position of all mass elements in the body.
Understanding the center of mass is fundamental in solving dynamics problems because it connects the body's mass distribution to its motion. With the acceleration of the center of mass \( \overline{\mathbf{a}} \), we can analyze how the whole body moves, even if it is made up of many particles.
Inertial Forces
Inertial forces are those forces that appear to act on a body when its state of motion changes, a concept derived from Newton's second law of motion. When a rigid body translates, each particle of the body feels an inertial force when observed from a non-inertial frame of reference. These forces are due to the acceleration of the body's center of mass, \( \overline{\mathbf{a}} \).
  • For any small mass element \( \Delta m_i \) in the body, the inertial force experienced is \( \Delta m_i \overline{\mathbf{a}} \).
  • This inertial term is crucial for understanding how forces propagate through a body that is not stationary.
Effectively, all parts of the body experience this force, pulling them along with the center of mass. This simplifies down to the whole body experiencing a force \( m \overline{\mathbf{a}} \), represented by a single vector.
Translation Motion
Translation motion in rigid body dynamics refers to the movement where every point of the object moves the same distance, in the same direction, and at the same time. In this kind of motion, the orientation of the body does not change, simplifying the analysis of forces and moments.
  • All parts of the body experience the same acceleration as the center of mass.
  • Rotation about any axis does not occur; any two vectors representing the velocities or accelerations of different parts of the body are parallel.
Such motion allows the system of inertial forces to be reduced to a more straightforward analysis since we can treat the whole body as if a single particle (center of mass) is experiencing the net force. This makes solving dynamics problems like the original exercise clearer and more accessible, breaking down complex motion into manageable parts.
Acceleration Analysis
Acceleration analysis involves understanding how quickly an object changes its velocity, specifically focusing on the center of mass for rigid bodies in translation. This is a crucial part of dynamics as it allows us to predict how a body will move under various forces.
  • The object's response to forces and its resultant acceleration is found using \( \overline{\mathbf{a}} \), the acceleration of the center of mass.
  • This vector is used to deduce the motion of the entire body.
By knowing that \( \overline{\mathbf{a}} \) applies the same to every constituent particle, it becomes vastly easier to conduct a thorough analysis of overall movement. This gives insight into how the forces are balanced or create resultant accelerations that define the motion picture the rigid body will follow. Understanding and calculating a body's acceleration leads effectively to a comprehensive grasp of its dynamics.

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Most popular questions from this chapter

A force \(P\) with a magnitude of \(3 \mathrm{N}\) is applied to a tape wrapped around the body indicated. Knowing that the body rests on a frictionless horizontal surface, determine the acceleration of \((a)\) point \(A,(b)\) point \(B .\) A uniform disk of mass \(2.4 \mathrm{kg}\).

In order to determine the mass moment of inertia of a flywheel of radius \(600 \mathrm{mm}, \mathrm{a} 12\) -kg block is attached to a wire that is wrapped around the flywheel. The block is released and is observed to fall \(3 \mathrm{m}\) in \(4.6 \mathrm{s}\). To eliminate bearing friction from the computation, a second block of mass \(24 \mathrm{kg}\) is used and is observed to fall \(3 \mathrm{m}\) in \(3.1 \mathrm{s}\). Assuming that the moment of the couple due to friction remains constant, determine the mass moment of inertia of the flywheel.

Disk \(B\) is at rest when it is brought into contact with disk \(A,\) which has an initial angular velocity \(\omega_{0} .(a)\) Show that the final angular velocities of the disks are independent of the coefficient of friction \(\mu_{k}\) between the disks as long as \(\mu_{k} \neq 0 .(b)\) Express the final angular velocity of disk \(A\) in terms of \(\omega_{0}\) and the ratio of the masses of the two disks \(m_{A} / m_{B} .\)

For a rigid body in plane motion, show that the system of the inertial terms consists of vectors \(\left(\Delta m_{i}\right) \overline{\mathbf{a}},-\left(\Delta m_{i}\right) \omega^{2} \mathbf{r}_{i}^{\prime},\) and \(\left(\Delta m_{i}\right)\left(\mathbf{Q} \times \mathbf{r}_{i}\right)\) attached to the various particles \(P_{i}\) of the body, where \(\overline{\mathbf{a}}\) is the acceleration of the mass center \(G\) of the body, \(\boldsymbol{\omega}\) is the angular velocity of the body, \(\alpha\) is its angular acceleration, and \(r_{i}^{\prime}\) denotes the position vector of the particle \(P_{i}\), relative to \(G\). Further show, by computing their sum and the sum of their moments about \(G,\) that the inertial terms reduce to a vector \(m \bar{a}\) attached at \(G\) and a couple \(\bar{I} \alpha\).

End \(A\) of the 6 -kg uniform rod \(A B\) rests on the inclined surface, while end \(B\) is attached to a collar of negligible mass than slide along the vertical rod shown. When the rod is at rest, a vertical force \(P\) is applied at \(B,\) causing end \(B\) of the rod to start moving upward with an acceleration of \(4 \mathrm{m} / \mathrm{s}^{2}\). Knowing that \(\theta=35^{\circ},\) determine the force \(\mathbf{P}\).

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