The problem of velocity synthesis and decomposition, the problem of velocity synthesis and decomposi

I'm going to talk about it briefly, and I hope it will help you.

The speed problem related to rope is generally to break down the actual velocity into velocity along the rope and velocity perpendicular to the rope. The actual motion of the object drives the movement of the rope, and the rope has two motion effects: shortening or elongation (movement along the rope); The swing of the rope (the movement perpendicular to the rope).

For example, if a person stands on a cliff and pulls a boat in the water under the cliff through a rope, the speed of the person is the speed at which the rope is shortened, and after breaking down the speed of the thread into the speed along the rope and perpendicular to the rope, the speed of the boat can be found according to the speed of the rope shortening, and of course the speed perpendicular to the rope can also be found.

The best example of the second problem is gears and conveyor belts. The linear velocity on the common point of the two rotating organisms is the same (unless the two "slip"), but the angular velocity is not necessarily the same, because the linear velocity is the same, and the angular velocity of the random radius is small, and the angular velocity of the small radius is hit. The velocity of the conveyor belt is the same as the linear velocity of the wheels at both ends, the linear velocity of the contact parts of the two gears is the same, and the angular velocity is inversely proportional to its radius.

Can you figure it out, it's actually not difficult, as long as you figure it out once, you can do it.

The master is here!!

You must have seen a lot of this kind of picture, don't talk about it).

First of all, let's talk about the "actual speed", the so-called "actual speed" is the speed that causes the change effect, as shown in the figure of the wooden block from a to a' is the change effect, so the speed of the wooden block is the actual speed, so the partial velocity is not generated on the wooden block, so only the rope can be related to the partial velocity, what is the change of the rope? 1.The rope has become shorter by 2

The rope is angled. Generally, the decomposition speed is a right-angle decomposition, that is, try to make the two partial velocities perpendicular, so you can take the rope at the initial position as the axis and decompose the velocity along the rope and the vertical rope (note: the partial velocity does not exist, it is just a problem-solving technique to help you solve the problem).

Proof : V1 is the partial velocity of velocity vb in the direction of V1, and Va is the rate at which the rope of object A1 is partially shortened on the right side of the pulley.

The speed at which the rope is shortened.

The speed at which the rope turns (perpendicular to the shortened velocity) = vb(1) v1 + v2

vb(2) (both vector additive).

v2 is obviously not equal to the speed at which the rope turns, unless v1 is perpendicular to v2.

Therefore, it can be obtained from the above two equations that v1 is not equal to the speed of the rope shortening.

Break down all the velocity crosses, and then synthesize the velocities on the x-axis and y-axis using the parallelogram rule, and any vector can be combined and decomposed to do so.

I disagree with what you said about you: "Velocity can only be broken down into two vertical components, what's going on".

Orthogonal decomposition. Just one of the decomposition methods. Actually, as long as the sum of the two velocity vectors decomposed is equal to the original velocity, it is fine.

Of course, velocity synthesis is relatively easy, and vectors can be summed (satisfying the parallelogram rule).

The decomposition of speed, pay attention not to decompose blindly, if you want to decompose, you must decompose the actual speed, that is, objective (for some.

frame of reference) velocity. How to break it down. Generally, high orthogonal decomposition is OK, but it should be noted that orthogonal decomposition should also be selected in a good direction (for example, it should be decomposed along the inclined plane and perpendicular to the inclined plane). The decomposition of the speed is actually and.

Decomposition of forces. Almost, analogy is sufficient.

If the angle between the velocities of two balls is 90°, we can do a horizontal and vertical orthogonal decomposition of their velocities.

The horizontal velocity of the two velocities is v1v2 and the vertical velocity is equal (because the time is the same), so when would the angle be 90°?

When the velocity of the first ball is at an angle a with the horizontal plane and the velocity of the second ball is at the angle b of the horizontal plane, it meets the requirements.

That is, tana=cotb

It's v1 v=v v2

v=under the root number (v1v2).

Time is t = root (v1v2) g

Absolute velocity = traction velocity + relative velocity (vector sum) Absolute velocity usually means that the object of study has the ground (or other reference frame) as the speed of the reference frame;

Relative velocity is the velocity of the object of study relative to the current frame of reference;

The traction velocity is the velocity of the current frame of reference relative to the reference frame.

The coincident point is generally a moving point relative to an object and a moving point relative to another object.

In general, the point velocity on a relatively motionless object is determined to be absolute.

The motion of the relative object is considered to be implication motion (the motion of the whole object), while the motion of the object with the coincident point relative to the implicated motion is relative motion.

The analysis in the textbook is very detailed! Didn't you take a good class!!

If the angle between the velocities of the two balls is 90°.

We can do an orthogonal decomposition of their velocities horizontally and vertically.

The horizontal velocity of the two speeds is v1v2

The vertical velocity is equal (because the time is the same).

So under what circumstances will the angle be 90°?

When the velocity of the first ball is interconnected with the angle a of the dry surface of the horizontal key and the velocity of the second ball is the angle b of the horizontal plane, it meets the requirements.

That is, tana=cotb

It's v1 v=v v2

v=under the root number (v1v2).

Time is t = root (v1v2) g