/*! 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 22 The design of the rotating arm \... [FREE SOLUTION] | 91Ó°ÊÓ

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The design of the rotating arm \(O A\) of a control mechanism requires that it rotate about the vertical \(Z\) -axis at the constant rate \(\Omega=\dot{\beta}=\pi \mathrm{rad} / \mathrm{s}\). Simultaneously, \(O A\) oscillates according to \(\theta=\theta_{0} \sin 4 \Omega t\) where \(\theta_{0}=\pi / 6\) radians and \(t\) is in seconds measured from the time when \(\beta=0 .\) Determine the angular velocity \(\omega\) and the angular acceleration \(\alpha\) of \(O A\) for the instant \((a)\) when \(t=1 / 2\) s and \((b)\) when \(t=1 / 8\) s. The \(x-y\) reference axes rotate in the \(X\) -Y plane with the angular velocity \(\Omega\)

Short Answer

Expert verified
At \( t=1/2 \) s: \( \omega = \pi \hat{k} + \frac{2\pi^2}{3} \hat{i}, \alpha = 0 \). At \( t=1/8 \) s: \( \omega = \pi \hat{k}, \alpha = -\frac{8\pi^3}{3} \hat{i} \).

Step by step solution

01

Define the Problem Parameters

We are given that the arm rotates about the vertical Z-axis at a constant rate of \( \Omega = \dot{\beta} = \pi \) rad/s. Also, it oscillates according to \( \theta = \theta_0 \sin(4 \Omega t) \) with \( \theta_0 = \frac{\pi}{6} \), and \( \Omega = \pi \) rad/s. We need to find the angular velocity \( \omega \) and the angular acceleration \( \alpha \) for two instances of time \( t = \frac{1}{2} \) s and \( t = \frac{1}{8} \) s.
02

Calculate Angular Position

For \( t = \frac{1}{2} \) s:\[ \theta = \frac{\pi}{6} \sin(4 \pi \times \frac{1}{2}) = \frac{\pi}{6} \sin(2 \pi) = 0 \]For \( t = \frac{1}{8} \) s:\[ \theta = \frac{\pi}{6} \sin(4 \pi \times \frac{1}{8}) = \frac{\pi}{6} \sin(\pi / 2) = \frac{\pi}{6} \]
03

Determine Angular Velocity \( \omega \)

The angular velocity \( \omega \) is given by:\[ \omega = \dot{\beta} \hat{k} = \Omega \hat{k} + \dot{\theta} \hat{i} \]To find \( \dot{\theta} \), we differentiate \( \theta \) with respect to \( t \):\[ \dot{\theta} = \frac{d}{dt}\left(\frac{\pi}{6} \sin(4 \pi t)\right) = \frac{\pi}{6} \times 4 \pi \cos(4 \pi t) \]Calculate for \( t = \frac{1}{2} \) s:\[ \dot{\theta} = \frac{2 \pi^2}{3} \cos(2 \pi) = \frac{2 \pi^2}{3} \]And for \( t = \frac{1}{8} \) s:\[ \dot{\theta} = \frac{2 \pi^2}{3} \cos(\pi / 2) = 0 \]
04

Calculate Angular Acceleration \( \alpha \)

The angular acceleration \( \alpha \) is the derivative of \( \omega \), which can be expressed as:\[ \alpha = \ddot{\theta} \hat{i} + \ddot{\beta} \hat{k} = -\frac{\pi}{6} \times 16 \pi^2 \sin(4 \pi t) \hat{i} \]Calculate for \( t = \frac{1}{2} \) s:\[ \ddot{\theta} = -\frac{\pi}{6} \times 16 \pi^2 \sin(2 \pi) = 0 \]And for \( t = \frac{1}{8} \) s:\[ \ddot{\theta} = -\frac{\pi}{6} \times 16 \pi^2 \sin(\pi / 2) = -\frac{8 \pi^3}{3} \]
05

Compose Final Results

For \( t = \frac{1}{2} \) s, \( \omega = \pi \hat{k} + \frac{2 \pi^2}{3} \hat{i} \) and \( \alpha = 0 \).For \( t = \frac{1}{8} \) s, \( \omega = \pi \hat{k} \) and \( \alpha = -\frac{8 \pi^3}{3} \hat{i} \).

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Key Concepts

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

Angular Velocity
Angular velocity refers to the rate of change of angular position of an object. It is often denoted as \( \omega \). In the context of a rotating arm like the one in the exercise, angular velocity measures how fast the arm is spinning around the vertical Z-axis. There are two main aspects to consider:
  • Constant Angular Velocity: Here, \( \Omega = \pi \) rad/s represents a steady spinning, reflecting uniform circular motion.
  • Variable Component: Due to oscillation, there is an additional velocity component from the changing \( \theta \).
For rotating systems, angular velocity is a vector quantity, typically expressed in components. In this exercise, it's defined as \( \omega = \dot{\beta} \hat{k} + \dot{\theta} \hat{i} \), where \( \hat{k} \) and \( \hat{i} \) are unit vectors. Understanding these components helps students perceive motion in multi-directional dynamics.
When \( t = \frac{1}{2} \) s, \( \omega = \pi \hat{k} + \frac{2 \pi^2}{3} \hat{i} \), incorporating rotation due to \( \Omega \) and oscillation from \( \dot{\theta} \). At \( t = \frac{1}{8} \) s, we see only \( \omega = \pi \hat{k} \) because oscillatory angular velocity is zero.
Angular Acceleration
Angular acceleration is the rate at which angular velocity changes. It is denoted as \( \alpha \). In dynamics, especially for our rotating arm, it’s crucial for analyzing how quickly the arm's speed of rotation changes as it oscillates.
  • The angular acceleration from rotating about the Z-axis is \( \alpha = 0 \) as it rotates at constant speed \( \Omega \).
  • However, due to the arm’s oscillatory motion, \( \alpha \) also has a contribution from \( \ddot{\theta} \), the second derivative of \( \theta \).
Substituting, this acceleration along the x-axis, expressed as \( \alpha = \ddot{\theta} \hat{i} \), reflects the varying nature of motion. For our exercise:
At \( t = \frac{1}{2} \) s, the oscillatory component is zero because \( \sin(2 \pi) = 0 \). Contrarily, at \( t = \frac{1}{8} \) s, the acceleration is maximized with \( \alpha = -\frac{8 \pi^3}{3} \hat{i} \) as \( \sin(\pi/2) = 1 \). This demonstrates active deceleration or acceleration along the i-axis during oscillation.
Oscillatory Motion
Oscillatory motion describes a repeating back-and-forth movement around an equilibrium position. In our example, the arm not only spins about the Z-axis, but also oscillates according to the equation \( \theta = \theta_0 \sin(4 \Omega t) \). Here, the maximum amplitude \( \theta_0 = \frac{\pi}{6} \) dictates the farthest extent of motion from its central axis.
This motion mirrors the behavior of pendulums and other systems in harmonic motion. It is vital in many mechanical designs, especially those that combine rotational dynamics with oscillations.
  • At \( t = \frac{1}{2} \) s, the arm returns to its equilibrium point with \( \theta = 0 \) since \( \sin(2 \pi) = 0 \).
  • At \( t = \frac{1}{8} \) s, the arm is at one extreme of its oscillation, producing a maximum \( \theta = \frac{\pi}{6} \).
Understanding such motion helps predict how systems with similar motion patterns behave over time.
Rotational Dynamics
Rotational dynamics involves the study of the rotational motion of objects and the forces that cause such motion. It covers everything from simple to complex moving systems, much like linear dynamics but with rotation-specific considerations.
In the described exercise, rotational dynamics encompasses the constant rotation about the Z-axis and the oscillation of the arm, linking angular velocity and acceleration to the forces and torques applied.
  • Constant Rotational Motion: This is evoked by \( \Omega \), presenting uniform circular motion without further acceleration.
  • Non-Uniform Motion: The oscillation due to \( \theta \), introduces variable torque,

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Most popular questions from this chapter

A dynamics instructor demonstrates gyroscopic principles to his students. He suspends a rapidly spinning wheel with a string attached to one end of its horizontal axle. Describe the precession motion of the wheel.

The housing of the electric motor is freely pivoted about the horizontal \(x\) -axis, which passes through the mass center \(G\) of the rotor. If the motor is turning at the constant rate \(\dot{\phi}=p,\) determine the angular acceleration \(\ddot{\psi}\) which will result from the application of the moment \(M\) about the vertical shaft if \(\dot{\gamma}=\dot{\psi}=0 .\) The mass of the frame and housing is considered negligible compared with the mass \(m\) of the rotor. The radius of gyration of the rotor about the \(z\) -axis is \(k_{z}\) and that about the \(x\) -axis is \(k_{x}\)

The disk rotates with a spin velocity of 15 rad/s about its horizontal \(z\) -axis first in the direction \((a)\) and second in the direction (b). The assembly rotates with the angular velocity \(N=10 \mathrm{rad} / \mathrm{s}\) about the vertical axis. Construct the space and body cones for each case.

Each of the identical wheels has a mass of \(4 \mathrm{kg}\) and a radius of gyration \(k_{z}=120 \mathrm{mm}\) and is mounted on a horizontal shaft \(A B\) secured to the vertical shaft at \(O .\) In case \((a),\) the horizontal shaft is fixed to a collar at \(O\) which is free to rotate about the vertical \(y\) -axis. In case \((b),\) the shaft is secured by a yoke hinged about the \(x\) -axis to the collar. If the wheel has a large angular velocity \(p=3600\) rev/min about its \(z\) -axis in the position shown, determine any precession which occurs and the bending moment \(M_{A}\) in the shaft at \(A\) for each case. Neglect the small mass of the shaft and fitting at \(O\)

The 5 -kg disk and hub \(A\) have a radius of gyration of \(85 \mathrm{mm}\) about the \(z_{0}\) -axis and spin at the rate \(p=1250\) rev/min. Simultaneously, the assembly rotates about the vertical \(z\) -axis at the rate \(\Omega=400\) rev/min. Calculate the gyroscopic moment \(\mathbf{M}\) exerted on the shaft at \(C\) by the disk and the bending moment \(M_{O}\) in the shaft at \(O .\) Neglect the mass of the shaft but otherwise account for all forces acting on it.

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