/*! 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 5 The rotor and shaft are mounted ... [FREE SOLUTION] | 91Ó°ÊÓ

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The rotor and shaft are mounted in a clevis which can rotate about the \(z\) -axis with an angular velocity \(\Omega .\) With \(\Omega=0\) and \(\theta\) constant, the rotor has an angular velocity \(\omega_{0}=-4 \mathbf{j}-3 \mathbf{k}\) rad/s. Find the velocity \(\mathbf{v}_{A}\) of point \(A\) on the rim if its position vector at this instant is \(\mathbf{r}=0.5 \mathbf{i}+1.2 \mathbf{j}+1.1 \mathbf{k} \mathrm{m} .\) What is the rim speed \(v_{B}\) of any point \(B ?\)

Short Answer

Expert verified
Calculate the cross product to find \( \mathbf{v}_{A} \). The rim speed \( v_{B} \) is the magnitude of this velocity.

Step by step solution

01

Determine the angular velocity vector

The problem states the rotor has an angular velocity \( \omega_{0} = -4 \mathbf{j} - 3 \mathbf{k} \). This angular velocity vector represents the rotation about the axes. The velocity of any point on a rotating object is derived from this vector.
02

Calculate the velocity of point A

The velocity \( \mathbf{v}_{A} \) of point \( A \) can be calculated using the formula: \[ \mathbf{v}_{A} = \boldsymbol{\omega} \times \mathbf{r} \] where \( \boldsymbol{\omega} = -4 \mathbf{j} - 3 \mathbf{k} \) and \( \mathbf{r} = 0.5 \mathbf{i} + 1.2 \mathbf{j} + 1.1 \mathbf{k} \). First, calculate the cross product: \[ \mathbf{v}_{A} = (-4 \mathbf{j} - 3 \mathbf{k}) \times (0.5 \mathbf{i} + 1.2 \mathbf{j} + 1.1 \mathbf{k}) \].

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Key Concepts

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

Cross Product
The cross product is a fundamental operation in vector mathematics. Often symbolized with the '×' sign, it helps in determining a vector perpendicular to two given vectors in three-dimensional space. In physics, it is extremely useful to calculate quantities related to rotational motion, such as torque or, in our case, the velocity of a point on a rotor.

The cross product of two vectors, let's say \( \mathbf{a} \) and \( \mathbf{b} \), is defined as another vector \( \mathbf{c} \) which is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). Its magnitude is equal to the product of the magnitudes of \( \mathbf{a} \) and \( \mathbf{b} \), and the sine of the angle between them. Mathematically, it's expressed as:

\[ \mathbf{a} \times \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \sin(\theta)\mathbf{n} \]

where \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \), and \( \mathbf{n} \) is the unit vector perpendicular to both. For the calculation in the original problem, we used the components of vectors \( \boldsymbol{\omega} \) and \( \mathbf{r} \) to derive the velocity \( \mathbf{v}_A \) through the cross product.
  • It is important to note that the cross product is only applicable in three-dimensional space.
  • The direction of the result corresponds to the right-hand rule.
Rotation Dynamics
Rotation dynamics revolve around the concept of how objects rotate and the forces causing and affecting such motion. Understanding rotation dynamics gives insights into how certain forces such as torque can influence rotational movement.

In rotation dynamics, angular velocity is a significant factor. It represents the rate of rotation of an object around a specific axis. In our original exercise, angular velocity is denoted by \( \omega_{0} = -4 \mathbf{j} - 3 \mathbf{k} \). This means the rotor rotates with such speed about the \( y \) and \( z \) axes.
  • Angular velocity is usually represented in radians per second (rad/s).
  • The direction of the angular velocity vector depends on the right-hand rule.
  • Any force causing rotation generates angular acceleration, changing the angular velocity over time.
By comprehending how the angular velocity affects a rotor's movement, one can deduce the resulting velocity at any point on its surface. This is crucial for assessing conditions like centrifugal forces and stability in rotational systems.
Rim Speed
Rim speed, or the tangential speed of a point on the rim of a rotating object, is a crucial aspect to explore when examining rotational movements. It is the speed at which a point on the fringe of a rotating body moves through space.

In the context of the original problem, the rim speed can be simply calculated by evaluating the magnitude of the velocity vector of the point on the rim, such as point \(A\). The magnitude is garnered from the vector obtained through the cross product of the angular velocity and the position vector of point \(A\).
Rim speed, \( v \), can be expressed with the equation:

\[ v = |\mathbf{v}_{A}| \]
  • It is purely about the linear speed of a point at the edge of rotation.
  • Rim speed increases with a larger radius or higher angular velocity.
  • Significant for engineering applications, like in gears and flywheels.
Understanding rim speed helps in various real-world applications, such as designing mechanisms in engines or turbines where edge speed impacts performance and safety.

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Most popular questions from this chapter

A car makes a turn to the right on a level road. Determine whether the normal reaction under the right rear wheel is increased or decreased as a result of the gyroscopic effect of the precessing wheels.

The circular disk of 120 -mm radius rotates about the \(z\) -axis at the constant rate \(\omega_{z}=20 \mathrm{rad} / \mathrm{s}\), and the entire assembly rotates about the fixed \(x\) -axis at the constant rate \(\omega_{x}=10 \mathrm{rad} / \mathrm{s}\). Calculate the magnitudes of the velocity \(\mathbf{v}\) and acceleration a of point \(B\) for the instant when \(\theta=30^{\circ}\)

The thin circular disk of mass \(m\) and radius \(R\) is hinged about its horizontal tangent axis to the end of a shaft rotating about its vertical axis with an angular velocity \(\omega .\) Determine the steady-state angle \(\beta\) assumed by the plane of the disk with the vertical axis. Observe any limitation on \(\omega\) to ensure that \(\beta>0\)

The pendulum oscillates about the \(x\) -axis according to \(\theta=\frac{\pi}{6} \sin 3 \pi t\) radians, where \(t\) is the time in seconds. Simultaneously, the shaft \(O A\) revolves about the vertical \(z\) -axis at the constant rate \(\omega_{z}=2 \pi \mathrm{rad} / \mathrm{sec}\) Determine the velocity \(\mathbf{v}\) and acceleration a of the center \(B\) of the pendulum as well as its angular acceleration \(\alpha\) for the instant when \(t=0\)

The \(6-\mathrm{kg}\) circular disk and attached shaft rotate at a constant speed \(\omega=10000\) rev/min. If the center of mass of the disk is \(0.05 \mathrm{mm}\) off center, determine the magnitudes of the horizontal forces \(A\) and \(B\) supported by the bearings because of the rotational imbalance.

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