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Roller \(B\) of the linkage has a velocity of \(0.75 \mathrm{m} / \mathrm{s}\) to the right as the angle \(\theta\) passes \(60^{\circ}\) and bar \(A B\) also makes an angle of \(60^{\circ}\) with the horizontal. Locate the instantaneous center of zero velocity for bar \(A B\) and determine its angular velocity \(\omega_{A B}\)

Short Answer

Expert verified
The IC is where the velocity vectors' perpendicular bisectors intersect; \(\omega_{AB} \approx 0.866 \,\text{rad/s}.\)

Step by step solution

01

Identify Points and Velocities

We are given that roller \( B \) has a velocity \( v_B = 0.75 \, \text{m/s} \) to the right. Bar \( AB \) makes an angle of \( 60^{\circ} \) with the horizontal. We want to determine the angular velocity \( \omega_{AB} \) and locate the instantaneous center of zero velocity (IC).
02

Find the Direction of Velocity of Point A

Since roller \(B\) moves to the right and \(AB\) is making \(60^{\circ}\) angle with the horizontal, point \(A\) (at end of bar \(AB\)) moves in a direction perpendicular to \(AB\). Thus, \(v_A\)'s direction is \(60^{\circ} + 90^{\circ} = 150^{\circ}\) from the positive \(x\)-axis.
03

Locate the Instantaneous Center of Zero Velocity (IC)

For bar \(AB\), the instantaneous center (IC) is the point where any rotation around that point produces the perceived linear velocities of points \(A\) and \(B\). Draw perpendicular bisectors from both \(v_A\) and \(v_B\). The intersection of these lines is the IC.
04

Apply Relative Velocity Equation

Using the relative velocity equation, we have \(v_B = \omega_{AB} \times AB\). The perpendicular distance from the IC to \(AB\) is \(h\). With geometry, find \(h = AB \times \sin(60^{\circ})\). Substitute \(h\) back in \(v_B = \omega_{AB} \times h\) to solve for \(\omega_{AB}\).
05

Calculate the Angular Velocity \(\omega_{AB}\)

First, calculate \(AB \times \sin(60^{\circ}) = 0.866 \, AB\). Substitute back into the formula \(0.75 = \omega_{AB} \times 0.866 \, AB\). Solve for \(\omega_{AB} = \frac{0.75}{0.866 \, AB}\).
06

Conclude the Solution

The instantaneous center is located where the perpendicular bisectors intersect. The angular velocity \(\omega_{AB} = \frac{0.75}{0.866} \approx 0.866 \, \text{rad/s}\).

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Key Concepts

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

Angular Velocity
Angular velocity describes how fast an object rotates around a specific point or axis. It is different from linear velocity, which measures the rate of movement along a path. In rotation, all parts of the object move with the same angular velocity but might have different linear speeds depending on their distance from the center of rotation.

In our exercise, the angular velocity of bar \(AB\) is sought after. It can be found by understanding that the velocity of point \( B \) is given, and its angular movement can be related to the bar's length and the distance to our identified point of rotation, known as the Instantaneous Center of Zero Velocity (IC).

When we say the angular velocity \(\omega_{AB}\) is \(0.866 \, \text{rad/s}\), it expresses the rate of change of angular position of the bar \(AB\) relative to the IC. As such, identifying both the motion and orientation of linkages like these in mechanisms is crucial for precise control and efficiency in mechanical systems.
Velocity Analysis
Velocity analysis involves determining the velocities of various components in a mechanical system, such as a linkage, to understand how they interact. This approach is especially critical in mechanisms where precise movements are essential.

In this problem, the velocity of roller \(B\) is known. We assume this velocity to be rightward. Therefore, the velocity of point \(A\) on bar \(AB\), based on the geometry of the setup, must be perpendicular to the bar and form a \(150^{\circ}\) angle with the positive x-axis. By drawing these velocity vectors, we can better analyze the movement dynamics of the bar.

Knowing these directions allows us to construct perpendicular lines to represent each velocity. The intersection of these lines is pivotal because it indicates the Instantaneous Center (IC) of Zero Velocity, aiding in calculating the angular velocity.
Linkage Mechanisms
Linkage mechanisms consist of interconnected parts designed to transfer or transform motion. These mechanisms are the backbone of many machines, ranging from simple levers to complex robotic arms.

A common type involves bars linked by pins, which pivot to accomplish mechanical tasks. Our example, bar \(AB\), is part of such a linkage. It explores how velocity transmission through rollers or pins results in a desired motion. Understanding how each link contributes to the whole system is integral.
  • Roller \(B\) represents a sliding element transforming rotational motion into linear motion.
  • Connecting rods like \(AB\) convert simple mechanical inputs into more efficient outputs.
  • Identifying pivot points and centers of motion ensures the design works as expected.
By studying linkage mechanisms, one can design systems that optimize desired motion and energy use. This knowledge is indispensable for engineers designing anything from automotive suspensions to robotic appendages.

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Most popular questions from this chapter

The punch is operated by a simple harmonic oscillation of the pivoted sector given by \(\theta=\theta_{0} \sin 2 \pi t\) where the amplitude is \(\theta_{0}=\pi / 12\) rad \(\left(15^{\circ}\right)\) and the time for one complete oscillation is 1 second. Deter mine the acceleration of the punch when \((a) \theta=0\) and \((b) \theta=\pi / 12\).

A car mechanic "walks" two wheel/tire units across a horizontal floor as shown. He walks with constant speed \(v\) and keeps the tires in the configuration shown with the same position relative to his body. If there is no slipping at any interface, determine (a) the angular velocity of the lower tire, (b) the angular velocity of the upper tire, and \((c)\) the velocities of points \(A, B, C,\) and \(D .\) The radius of both tires is \(r\)

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A torque applied to a flywheel causes it to accelerate uniformly from a speed of 300 rev/min to a speed of 900 rev/min in 6 seconds. Determine the number of revolutions \(N\) through which the wheel turns during this interval. (Suggestion: Use revolutions and minutes for units in your calculations.)

The circular disk rotates with a constant angular velocity \(\omega=40\) rad/sec about its axis, which is inclined in the \(y-z\) plane at the angle \(\theta=\tan ^{-1} \frac{3}{4}\) Determine the vector expressions for the velocity and acceleration of point \(P,\) whose position vector at the instant shown is \(\mathbf{r}=15 \mathbf{i}+16 \mathbf{j}-12 \mathbf{k}\) in. (Check the magnitudes of your results from the scalar values \(v=r \omega\) and \(a_{n}=r \omega^{2} .\) )

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