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Show that the system of momenta for a rigid body in plane motion reduces to a single vector, and express the distance from the mass center \(G\) to the line of action of this vector in terms of the centroidal radius of gyration \(\bar{k}\) of the body, the magnitude \(\bar{v}\) of the velocity of \(G,\) and the angular velocity \(\omega .\)

Short Answer

Expert verified
The distance is given by \(\bar{r} = \frac{\bar{k}^2\omega}{\bar{v}}\), reducing the momentum to a single vector.

Step by step solution

01

Understand the Problem Statement

We need to demonstrate that the system of momenta for a rigid body in plane motion reduces to a single vector. Then, we need to find the expression for the distance from the mass center \(G\) to the line of action of this vector in terms of the centroidal radius of gyration \(\bar{k}\), the magnitude \(\bar{v}\) of the velocity of \(G\), and the angular velocity \(\omega\).
02

Conceptualize the Moment of Momentum

The moment of momentum (also known as angular momentum) of a rigid body in plane motion can be expressed as \(H = I_{G}\omega\), where \(I_{G}\) is the moment of inertia about the mass center \(G\), and \(\omega\) is the angular velocity. The inertia itself is expressed as \(I_{G} = m\bar{k}^2\), with \(m\) being the mass and \(\bar{k}\) the radius of gyration.
03

Express the Linear Momentum

The linear momentum of the rigid body moving in a plane at the center of mass \(G\) is given by \(p = m\bar{v}\), where \(\bar{v}\) is the velocity of the center of mass.
04

Combine Linear and Angular Momentum

Combine the linear and angular momentum by representing the system of momenta as a vector \(M\) such that \(M = p \times r + H_G\), where \(r\) is the position vector from the mass center \(G\) to the line of action of the resultant vector.
05

Simplify to Single Vector Form

Substituting the expressions for linear and angular momentum yields \(p \times ar{r} = m\bar{v} \bar{r} = I_{G}\omega = m\bar{k}^2\omega\). Solving for \(\bar{r}\), we get \(\bar{r} = \frac{\bar{k}^2\omega}{\bar{v}}\).
06

Conclusion and Expression of the Distance

The system of momenta has consequently been reduced to a single vector whose line of action is at a perpendicular distance \(\bar{r} = \frac{\bar{k}^2\omega}{\bar{v}}\) from the mass center \(G\). This distance depends on the centroidal radius of gyration, the velocity of the mass center, and the angular velocity.

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Key Concepts

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

Rigid Body Motion
Rigid body motion refers to the movement of a solid object where all points within the body move in such a way that the relative positions of the points remain constant throughout the motion. This means the shape and size of the object do not change.
In plane motion, which is a type of rigid body motion, every point on the body moves along a path that lies in a single plane. This can include both translation (moving from one place to another) and rotation (spinning around an axis). Together, they form what is known as compound motion.
Understanding rigid body motion is essential as it forms the basis of analyzing both linear and angular momentum in objects. It’s particularly useful in engineering and physics to predict the motion of vehicles, machinery, and even celestial bodies.
  • Translation: All points of the body move in the same direction and at the same speed.
  • Rotation: The body spins about an axis, which can be internal or external.
The combination of these motions determines the overall trajectory and behavior of the rigid body.
Angular Momentum
Angular momentum is a measure of the quantity of rotation of a body, taking account of its mass distribution and speed of rotation. It is conserved in an isolated system, meaning the total angular momentum remains constant, unless an external torque acts on the system.
Mathematically, for a rigid body in plane motion, angular momentum about the mass center can be expressed as:\[H = I_{G}\omega\]Where:
  • \(H\) is the angular momentum.
  • \(I_{G}\) is the moment of inertia about the mass center \(G\).
  • \(\omega\) is the angular velocity.
The moment of inertia \(I_{G}\) depends on the mass and how that mass is distributed relative to the axis of rotation. It represents how hard it is to change the rotational motion of the body.
In the context of the problem, angular momentum interplays with linear momentum, ultimately simplifying the system of momenta to a single vector that aligns all active forces acting on the body.
Linear Momentum
Linear momentum is the product of the mass of an object and its velocity. It is a vector quantity, meaning it has both magnitude and direction, and it represents the quantity of motion possessed by an object moving in a straight path or a linear trajectory.
For a rigid body in planar motion, the expression of linear momentum at the mass center \(G\) is:\[p = m\bar{v}\]Where:
  • \(p\) is the linear momentum.
  • \(m\) is the mass of the rigid body.
  • \(\bar{v}\) is the velocity of the mass center \(G\).
Linear momentum is vital in analyzing the motion since changes to this momentum result from external forces applied to the body. It helps in understanding how the motion distributes across the body during translational movement.
The linear momentum, combined with angular momentum, is crucial for obtaining the system of momenta which emphasizes its role in evaluating dynamic systems.
Centroidal Radius of Gyration
The centroidal radius of gyration is a parameter that simplifies the calculation of the rotational characteristics of a body. It essentially represents the average radius at which the entire mass of the body could be concentrated to exhibit the same moment of inertia about an axis.
Mathematically, the radius of gyration \(\bar{k}\) is expressed in relation to the moment of inertia \(I_{G}\) and mass \(m\):\[I_{G} = m\bar{k}^2\]Where \(\bar{k}\) is the centroidal radius of gyration.
This concept is crucial because it helps in determining how the mass distribution affects rotational inertia and provides an intuitive measure for how mass is spread across the body. It factors into calculations for angular momentum, affecting how a body rotates around a particular axis.
In practical terms, the centroidal radius of gyration helps engineers design objects with specific rotational characteristics, ensuring stability and efficiency in movement.

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Most popular questions from this chapter

A bullet weighing 0.08 lb is fired with a horizontal velocity of \(1800 \mathrm{ft}\) s into the lower end of a slender \(15-\mathrm{lb}\) bar of length \(L=30 \mathrm{in}\). Knowing that \(h=12 \mathrm{in}\), and that the bar is initially at rest, determine \((a)\) the angular velocity of the bar immediately after the bullet becomes embedded, (b) the impulsive reaction at \(C,\) assuming that the bullet becomes embedded in \(0.001 \mathrm{s}\).

At what height h above its center G should a billiard ball of radius r be struck horizontally by a cue if the ball is to start rolling without sliding?

Show that the sum \(\mathbf{H}_{A}\) of the moments about a point \(A\) of the momenta of the particles of a rigid body in plane motion is equal to \(I_{A} \omega,\) where \(\omega\) is the angular velocity of the body at the instant considered and \(I_{A}\) the moment of inertia of the body about \(A,\) if and only if one of the following conditions is satisfied: \((a) A\) is the mass center of the body, (b) \(A\) is the instantaneous center of rotation, (c) the velocity of \(A\) is directed along a line joining point \(A\) and the mass center \(G .\)

The circular platform A is fitted with a rim of 200-mm inner radius and can rotate freely about the vertical shaft. It is known that the platform-rim unit has a mass of 5 kg and a radius of gyration of 175 mm with respect to the shaft. At a time when the platform is rotating with an angular velocity of 50 rpm, a 3-kg disk B of radius 80 mm is placed on the platform with no velocity. Knowing that disk B then slides until it comes to rest relative to the platform against the rim, determine the final angular velocity of the platform.

A \(5-m-\) -long ladder has a mass of \(15 \mathrm{kg}\) and is placed against a house at an angle \(\theta=20^{\circ} .\) Knowing that the ladder is released from rest, determine the angular velocity of the ladder and the velocity of end \(A\) when \(\theta=45^{\circ}\) Assume the ladder can slide freely on the horizontal ground and on the vertical wall.

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