# Find the angular velocity of this linkage

• Noob of the Maths
In summary: Draw a free body diagram of the system2) Label the axes3) Connect the dots to show the forces4) Calculate the velocitiesIn summary, the author was discussing how to solve a problem using trigonometry and instant centers. He stated that his advance can be seen in the next image. The instant centers procediment its (1) up and trigonometry procediment its (2) down. He knows that the result its 4rad/s, but in his answer its not make the sense; (and he doesn't know why the angular velocity of "D" its 4rad/s). He needs to resolve it with 2 procedures: trigonometry and instant
Noob of the Maths
Homework Statement
If the angular velocity of link AB is w(AB)=4 rad/s at the instant shown, determine the velocity of the sliding block E at that instant. Also, identify the type of motion of each of the four links.
Relevant Equations
V = w(r)

Hi! everyone! ;)

I have a problem with the development of this problem.

I need to resolve it with 2 procedures: trigonometry and instant centers. My advance can be see in the next image:

The instant centers procediment its (1) up and trigonometry procediment its (2) down.

I know that the result its 4 rads/s. But in my answer its not make the sense ;( and i don't know why the angular velocity of "D" its 4 rads/s.

PD: I don't can use cross product for resolve it... only trigonometry and instant centers (separately).

I would introduce xy coordinate as sketched.
From these linear equations for sin##\phi## and cos##\phi## we can delete ##\phi## and get x, the horizontal coordinate of E, as function of ##\theta## which is function of time.

[EDIT] The formula are written as
$$(x/2-\cos\phi-\beta)^2+(1-\sin\phi)^2=1/4$$
$$(x/2-\cos\phi-\cos\theta)^2+(1-\sin\phi-\sin \theta)^2=1/4$$
where
$$\beta=\frac{\sqrt{3}-1}{2}=\frac{1}{\sqrt{3}+1}$$
Solving this equation for sin##\phi## and cos##\phi## and using the relation
$$\sin^2\phi+\cos^2\phi=1$$
The relation of x and ##\theta## is given as
$$(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2=(a_1b_2-a_2b_1)^2...(1)$$
where
$$a_1=1$$
$$a_2=1-\sin\theta$$
$$b_1=\frac{x}{2}+\frac{\sqrt{3}-1}{2}$$
$$b_2=\frac{x}{2}-\cos\theta$$
$$c_1=\frac{3}{8}+\frac{a_1^2+b_1^2}{2}$$
$$c_2=\frac{3}{8}+\frac{a_2^2+b_2^2}{2}$$
Differentiating (1) by t for t=0 we would know ##\dot{x}(t=0)##. After calculation I got
$$\dot{x}=-3.017 r\dot{\theta}$$ for initial time t=0, where r is 1 length unit for rods. I do not have trust of my calculation. I should appreciate it if someone would challenge it too.

#### Attachments

• 2021-09-15 21.16.39.jpg
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Last edited:
anuttarasammyak said:
I would introduce xy coordinate as sketched.
From these linear equations for sin##\phi## and cos##\phi## we can delete ##\phi## and get x, the horizontal coordinate of E, as function of ##\theta## which is function of time.
yep, i can use a free body diagram, but in this pariticulary case i only can use the procedure as I put it. The most important are the traslation and rotation diagram and the use of angles in triangles.

Noob of the Maths said:
yep, i can use a free body diagram, but in this pariticulary case i only can use the procedure as I put it. The most important are the traslation and rotation diagram and the use of angles in triangles.
Hi, I think you should put this into the engineering forum. Does your method allow for sketching velocity diagrams?

Master1022 said:
Hi, I think you should put this into the engineering forum. Does your method allow for sketching velocity diagrams?
Yes, velocity diagrams its allowed. Hi :)

Noob of the Maths said:
Yes, velocity diagrams its allowed. Hi :)
Cool, so the steps I would think about when going down the velocity diagram route are:
1. A and C are fixed
2. E is constrained to pure horizontal movement, as is D (in this case) as CD is vertical
3. We know ##\omega_{AB}## and thus know where point B is
4. To get from point B to point D, we then know there must be a rotation between the two
5. Then to get from D to E, given that E is constrained to horizontal movement, it suggests to me that ##\omega_{DE} = 0## at this instant. However, you said that the answer is ##\omega_{DE} = 4 ## rad/s, or did I misunderstand what you said?

I hope this rough outline makes sense. I can try to make a sketch a bit later

Master1022 said:
Cool, so the steps I would think about when going down the velocity diagram route are:
1. A and C are fixed
2. E is constrained to pure horizontal movement, as is D (in this case) as CD is vertical
3. We know ##\omega_{AB}## and thus know where point B is
4. To get from point B to point D, we then know there must be a rotation between the two
5. Then to get from D to E, given that E is constrained to horizontal movement, it suggests to me that ##\omega_{DE} = 0## at this instant. However, you said that the answer is ##\omega_{DE} = 4 ## rad/s, or did I misunderstand what you said?

I hope this rough outline makes sense. I can try to make a sketch a bit later
This very helpful!, i know that my mistakes its in the diagram and their angles. wDE= 4 rad/s based on assumptions,

Consider that:
1) E can only have horizontal movement.
2) Only the horizontal component of the instantaneous velocity of D has influence on E.
3) Because both are connected by a horizontal link, the horizontal components of the instantaneous velocity of D and B must be equal.
4) Since you are given the angular velocity of the 2 feet long link AB, you can calculate the tangential velocity of B and then, its instantaneous horizontal component.

anuttarasammyak said:
I should appreciate it if someone would challenge it too.
You are way overthinking it.
It is really quite simple following @Lnewqban's scheme.

Lnewqban and anuttarasammyak

## 1. What is angular velocity?

Angular velocity is a measure of the rate of change of an object's angular position with respect to time. It is typically measured in radians per second or degrees per second.

## 2. How is angular velocity calculated?

Angular velocity is calculated by dividing the change in angular position by the change in time. It can also be calculated by multiplying the angular speed (in radians or degrees per second) by the radius of the circular motion.

## 3. What is the difference between angular velocity and linear velocity?

Angular velocity is a measure of how fast an object is rotating, while linear velocity is a measure of how fast an object is moving in a straight line. Angular velocity is measured in radians or degrees per second, while linear velocity is measured in meters or feet per second.

## 4. How does the linkage affect the angular velocity?

The linkage refers to the interconnected system of rotating objects that are being studied. The arrangement and size of the linkage can affect the angular velocity, as well as the forces and torques acting on the system.

## 5. What are some real-world applications of finding angular velocity of a linkage?

Angular velocity of a linkage is used in various fields such as robotics, mechanical engineering, and physics. It is used to design and analyze rotating machinery, such as engines and turbines, and to understand the motion of objects in circular motion, such as wheels and gears.

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