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Chapter 7: Problem 42

The collar and clevis \(A\) are given a constant upward velocity of 8 in./sec for an interval of motion and cause the ball end of the bar to slide in the radial slot in the rotating disk. Determine the angular acceleration of the bar when the bar passes the position for which \(z=3\) in. The disk turns at the constant rate of 2 rad/sec.

Short Answer

Expert verified
The angular acceleration of the bar is -10.67 rad/sec虏.

Step by step solution

01

Identify Variables and Relationships

First, identify the given conditions and variables: the initial velocity of the collar is 8 in./sec upward, the bar slides in a rotating disk that spins at a constant rate of 2 rad/sec, and we need to find the angular acceleration of the bar when the position is such that \( z = 3 \) in.
02

Relation between Linear and Angular Quantities

The upward velocity \( v_A = 8 \text{ in./sec} \) refers to the linear motion. We convert this into angular velocity and acceleration of the bar. Remember the disk has a constant angular velocity \( \omega = 2 \text{ rad/sec} \). Let \( z \) be the radial distance from the axis, which changes as the collar moves upward.
03

Finding Angular Velocity of the Bar's Motion

The motion in the slot has a radial component based on the disk's angular velocity and the collar's vertical velocity. Use the relationship \( v = r \cdot \omega \). Here, \( r = z = 3 \) in at the given instant, and \( \omega_{disk} = 2 \text{ rad/sec} \). This affects the bar's motion due to the disk's rotation.
04

Calculate Angular Acceleration

The bar's angular acceleration can be found using kinematic equations for rotational motion. Consider \( a_t = r \cdot \alpha_{B} + 2 \cdot v_r \cdot \omega \), where \( v_r \) is the velocity component due to collar movement, and \( a_t \) is tangential acceleration given as zero due to constant \( \omega \). Solve for \( \alpha_{B} \).
05

Calculate Specific Angular Values

Substitute specific values: \( v_r = 8 \text{ in./sec} \), \( r = 3 \text{ in} \), and \( \omega = 2 \text{ rad/sec} \). We have the equation \( 0 = 3 \cdot \alpha_{B} + 2 \cdot 8 \cdot 2\). Simplify to find \( \alpha_{B} \).
06

Solution for Angular Acceleration

From previously, \( 0 = 3\alpha_{B} + 32\). Solving \( 3\alpha_{B} = -32 \), gives \( \alpha_{B} = -\frac{32}{3} \text{ rad/sec}^2\). Thus, the angular acceleration of the bar is -10.67 rad/sec虏.

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Key Concepts

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

Rotational Motion
Rotational motion involves an object rotating around an internal axis. Imagine a spinning disk where each part of the disk follows a circular path around the center. This is rotational motion, a concept often compared to objects moving in a straight line, or linear motion.

Key aspects to note:
  • All points of a rotating object have the same angular velocity and acceleration.
  • Angular velocity (\(\omega\)) measures how fast an object rotates. It is the rotational equivalent of linear velocity.
  • Angular acceleration (\(\alpha\)) measures the change in angular velocity over time, similar to how linear acceleration for a straight-moving object measures changes in speed.
The exercise provided involves calculating the angular acceleration, determining how fast the disk's rotation speed changes due to the movement of a sliding bar inside it.
Linear to Angular Conversion
Linear to angular conversion translates straight line motion into rotational motion. Here鈥檚 how it works: consider a point on the rotating disk that moves in a straight line as the disk spins. This linear motion is due to its rotational movement.

To convert linear velocity (\(v\)) into an angular counterpart, the relationship is:
  • \(v = r \cdot \omega\)
Where:
  • \(v\) is the linear velocity of the point,\(r\) is the radial distance (distance from the rotation axis), and\(\omega\) is the angular velocity.
In our exercise, the collar's upward movement translates into an angular change of the bar, influenced by the disk's rotation. This conversion helps calculate how quickly the rotation changes, using linear movement data.
Radial Slot Mechanism
The radial slot mechanism in a rotating disk adds complexity by allowing movement along a radial line while the disk spins. This setup means a sliding bar can change its position relating to the rotation center.

Here's what happens:
  • The collar moves upward through the radial slot, changing its radial distance from the disk's axis.
  • As the position (\(z\) in the exercise) changes, the relationship between the linear movement (the collar) and rotational movement (the sliding bar and disk) creates a dynamic system.
  • This interaction affects how the bar rotates, as the radial movement directly influences rotational dynamics.
Understanding this mechanism is crucial when solving for rotational accelerations influenced by non-fixed radial movements.
Kinematic Equations for Rotation
Kinematic equations for rotation link angular displacement, velocity, and acceleration. They鈥檙e similar to linear kinematic equations but adapted for rotation.

A typical equation for rotational motion is:
  • \(\omega_f = \omega_0 + \alpha \cdot t\)
Where:
  • \(\omega_f\) is final angular velocity,\(\omega_0\) is initial angular velocity,\(\alpha\) is angular acceleration, and\(t\) is time elapsed.
In the given exercise, the tangential acceleration comes into play because of constant angular velocity from the rotating disk and affects the rotational equation used. This form of kinematic analysis helps calculate changes in motion, which is essential for determining the angular acceleration required. It melds together the concepts of angular velocity change due to both radial motion and the disk's inherent rotation.

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Most popular questions from this chapter

For the instant represented collar \(B\) is moving along the fixed shaft in the \(X\) -direction with a constant velocity \(v_{B}=4 \mathrm{m} / \mathrm{s}\). Also at this instant \(X=0.3 \mathrm{m}\) and \(Y=0.2 \mathrm{m} .\) Calculate the velocity of collar \(A,\) which moves along the fixed shaft parallel to the \(Y\) -axis. Solve, first, by differentiating the relation \(X^{2}+Y^{2}+\) \(Z^{2}=L^{2}\) with respect to time and, second, by using the first of Eqs. \(7 / 4\) with translating axes attached to \(B\). Each clevis is free to rotate about the axis of the rod.

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 8 -lb rotor with radius of gyration of 3 in. rotates on ball bearings at a speed of 3000 rev/min about its shaft \(O G\). The shaft is free to pivot about the \(X\) -axis, as well as to rotate about the \(Z\) -axis. Calculate the vector \(\Omega\) for precession about the \(Z\) -axis. Neglect the mass of shaft \(O G\) and compute the gyroscopic couple M exerted by the shaft on the rotor at \(G\)

The blades and hub of the helicopter rotor weigh 140 lb and have a radius of gyration of \(10 \mathrm{ft}\) about the \(z\) -axis of rotation. With the rotor turning at 500 rev/min during a short interval following vertical liftoff, the helicopter tilts forward at the rate \(\dot{\theta}=10\) deg/sec in order to acquire forward velocity. Determine the gyroscopic moment \(M\) transmitted to the body of the helicopter by its rotor and indicate whether the helicopter tends to deflect clockwise or counterclockwise, as viewed by a passenger facing forward.

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 \(\Omega\)
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