In the direction of speed. Determination of the acceleration of points of a plane figure using the mtsu Methods for determining the acceleration of points of a plane
Considering the plane motion of a plane figure as the sum of translational motion, in which all points of the figure move with acceleration a A of the pole A, and rotational
motion around this pole, we obtain a formula for determining the acceleration of any point B of a flat figure in the form
a B \u003d |
a A + |
a BA \u003d |
a A + a BAв + |
a BAc. |
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Here a |
acceleration |
poles A; a |
Acceleration |
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rotational motion of point B around pole A, which, as in the case of rotation of a body around a fixed axis, is vector
is the sum of the rotational acceleration a BA in and the center
rapid acceleration a BA c ... The modules of these accelerations are determined by the formulas
angular acceleration module. Rotational acceleration a BA in is directed perpendicularly to segment AB in the direction of the arc arrow ε, and centripetal acceleration a BA c is directed along line AB from point B to pole A (Fig. 12). The absolute acceleration modulus a BA of point B relative to pole A due to the condition a BA in a BA q is calculated by the formula
Fig 12. Determination of acceleration point B
using pole A.
To find the acceleration a B by the formula (2.18)
it is recommended to use analytical way... In this method, a rectangular Cartesian coordinate system is introduced (the Bxy system in Fig. 12) and the projections a Bx, a By
the required acceleration as algebraic sums of the projections of the accelerations included in the right-hand side of equality (2.18):
(a in |
(a c |
a cosα |
c; |
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(a in |
(a c |
sinα |
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where α is the angle between the vector a A |
and the Bx axis. By found |
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The described method for determining the accelerations of points of a plane figure is applicable to solving problems in which the motion of the pole A and the angle of rotation of the figure are specified
equations (2.14). If the dependence of the angle of rotation on time is unknown, then for a given position of the figure it is necessary to determine the instantaneous angular velocity and instantaneous angular acceleration. Methods for their determination are discussed further in the examples of the task 2.
Note also that when determining the accelerations of points of a plane figure, one can use instant acceleration center- a point whose acceleration at a given moment in time is equal to zero. However, the use of the instantaneous center of acceleration is associated with rather laborious methods of finding its position; therefore, it is recommended to determine the accelerations of the points of a flat figure using the formula
2.4 Task 2. Determination of the speeds and accelerations of points of a flat mechanism
Mechanisms (see p. 5) are called flat if all its points move in one or in parallel planes, otherwise the mechanisms are called space
nym.
IN task 2.1 deals withplanetary gears,
in task 2.2 - crank-posture mechanisms, and in task
2.3 in addition to the two types named, the movement of mechanisms of other types is studied. Most of the mechanisms considered are mechanisms with one degree of freedom,
in which, to determine the motion of all links, you need to set the law of motion of one link.
Assignment 2.1
In the planetary mechanism (Fig. 13), crank 1 with a length of OA \u003d 0.8 (m) rotates around a fixed axis O, perpendicular to the plane of the figure, according to the law
ϕ OA (t) \u003d 6t - 2t 2 (rad). At point A, the crank is pivotally connected
with the center of the disk 2 of radius r \u003d 0.5 (m), which is in internal engagement with the fixed wheel 3, coaxial with
crank OA. Point B is set on disk 2 at the time t 1 \u003d 1 (s), the position of which is determined by the distance AB \u003d 0.5 (m) and the angle α \u003d 135 °. (At a given moment in time, the angle α is measured from the Ax axis in the counterclockwise direction for α\u003e 0 or in the opposite direction for
α < 0).
Fig 13. Planetary mechanism and method of specifying the position of point B.
Determine at time t 1
1) the speed of point B in two ways: using the instantaneous center of velocities (IMC) of disk 2 and using the pole A;
2) Acceleration of point B using pole A.
1) Determination of the speed of point B.
First you need to perform a graphic image
mechanism in the selected scale (for example, in 1 cm of the figure - 0.1 m of the segment OA and radius r) and show the given position of point B (Fig. 14).
Fig 14. Determination of the speed of point B using the instantaneous center of velocities P and pole A.
According to the given law of rotation of the crank OA, we find the speed of the center A of the disc 2. We determine the angular speed of the crank at a given time t 1 \u003d 1 (c):
ω OA \u003d ϕ! OA \u003d (6 t - |
6 - 4 t; |
ω OA (t 1) \u003d 2 (rad / s). |
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The obtained value ω OA (t 1) is positive, therefore we direct the arc arrow ω OA counterclockwise, that is, in the positive direction of the angle ϕ.
Calculate the speed module
v A \u003d ω OA (t 1) OA \u003d 2 0.8 \u003d 1.6 (m / s)
and construct the velocity vector v A perpendicular to ОА towards the arc arrow ω OA.
the arc arrow ω OA and the vector v A are drawn in the opposite direction, and the modulus is used to calculate v A
ω OA (t 1).
The instantaneous center of speeds (point P) of disk 2 is located at the point of its contact with wheel 3 (see item 5 on p. 34). Let us determine the instantaneous angular velocity ω of the disk from the found value of the velocity v A:
ω \u003d v A / AP \u003d v A / r \u003d 1.6 / 0.5 \u003d 3.2 (rad / s)
and depict its arc arrow in the figure (Fig. 14).
To determine the speed of point B using the MCS, we find the distance BP according to the cosine theorem from the ABP triangle:
BP \u003d AB2 + AP2 - 2 AB AP cos135 "\u003d
0.5 2 + 0.52 - 2 0.52 (- 2/2) ≈ 0.924 (m).
The speed v B is equal in absolute value
v B \u003d ω PB \u003d 3.2 0.924 ≈ 2.956 (m / s)
and is directed perpendicular to the segment PB in the direction of the arc arrow ω.
The same vector v B can be found using pole A according to the formula (2.15): v B \u003d v A + v BA. We transfer the vector v A to point B and construct a vector v BA, perpendicular to the segment AB and directed towards the arc arrow ω. Module
that the angle between the vectors v A and v BA is 45 °. Then, by formula (2.16), we find
vB \u003d vA 2 + vBA 2 + 2 vA vBA cos 45 "\u003d
1.6 2 + 1.62 + 2 1.62 (2/2) ≈ 2.956 (m / s).
In the figure, the vector v B must coincide with the diagonal of the parallelogram, the sides of which are vectors v A and v BA. This is achieved by constructing vectors v A, v B and v BA in the selected
standard scale (for example, 1 cm in the figure corresponds to 0.5 m / s). Note that the scales shown in the considered example can be changed and assigned independently.
2). Determination of point B acceleration.
The acceleration of point B is determined by the formula (2.18) using the pole A, the acceleration of which is the sum of the vector from the tangential and normal accelerations:
a B \u003d a A + a BA в + a BA c \u003d a τ A + a A n + a BA в + a BA c.
According to the given law of rotation of the OA crank, we find its angular acceleration:
ε OA \u003d ω! OA \u003d (6 - 4t!) \u003d - 4 (rad / s 2).
The obtained value ε OA is negative, therefore, we direct the arc arrow ε OA clockwise, then
is in the negative direction, and in the further calculation we will take this value modulo.
The moduli of the tangential and normal accelerations of the pole A at a given time t 1 are found by the formulas (2.11):
a τ A \u003d ε OA OA \u003d 4 0.8 \u003d 3.2 (m / s 2); a n A \u003d ω OA 2 OA \u003d 22 0.8 \u003d 3.2 (m / s 2).
The tangential acceleration a τ A is directed perpendicularly to the crank OA towards the arc arrow ε OA, and the normal acceleration a A n is directed from longing A to point O at any direction of the angular velocity of the crank (Fig. 15). The total acceleration a A does not need to be determined.
Fig 15. Determining the acceleration of point B using pole A.
ω \u003d v A / r \u003d ω OA (OA / r). |
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by definition angular |
acceleration |
disk (at |
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OA / r \u003d const) equals |
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ε = ω ! = |
ω! OA (OA / r) \u003d ε OA (OA / r) \u003d - |
4 (0.8 / 0.5) = |
- 6.4 (rad / s 2). |
the angular arrow ε is directed in the opposite direction to the arc arrow ω.
Let us calculate the modules of rotational and centripetal accelerations of point B relative to pole A using the formulas
a BAв |
AB \u003d |
6.4 0.5 \u003d 3.2 (m / s 2); |
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a BAц |
2 AB \u003d |
3.22 0.5 \u003d 5.12 (m / s 2). |
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The vector a BA in is directed perpendicularly to the segment AB towards
arc arrow ε, and vector a BA c - from point B to pole A
We find the acceleration of point B by its projections on the axis of the coordinate system Axy:
a Bx \u003d (a τ A) x + |
(a An) x + (a BAc) x + (a BAc) x \u003d |
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0 - a n A - |
a BA at cos 45 "+ |
a BAц |
cos 45 "\u003d |
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3.2 − |
/ 2 + 5.12 |
2 / 2 ≈ |
- 1.84 (m / s 2); |
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a By \u003d (a τ A) y + |
(a An) y + (a BAc) y + (a BAc) y \u003d |
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a τ A + |
0 − |
a BAв |
cos45 " |
- a BA c cos 45 "\u003d |
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3.2 − |
/ 2 − 5.12 |
2 / 2 ≈ |
- 9.08 (m / s 2). |
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Module a B \u003d |
a Bx2 |
a By2 |
≈ 9.27 (m / s 2). |
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acceleration |
a τ A, |
a A n, |
a BA c, a BA c is required |
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to represent in the selected scale and to construct in the same scale the vector a B according to the found projections (Fig. 15).
The initial data for self-fulfillment of task 2.1 are given in the table on p. 44.
Rigid body kinematics |
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ϕ OA (t), rad |
α, deg |
t 1, s |
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t2 + 3t |
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8t - 3t2 |
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t2 - 4t |
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3t - 2t2 |
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2t2 - t |
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4t - t2 |
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2t2 - 6t |
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2t - 3t2 |
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3t2 - 4t |
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8t - 2t2 |
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4t2 - 6t |
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3t - 4t2 |
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4t2 - 2t |
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6t - t2 |
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2t2 - 4t |
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4t - 3t2 |
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2t2 + t |
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4t - 2t2 |
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3t2 - 10t |
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t - 2t2 |
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3t2 + 2t |
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6t - 3t2 |
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3t2 - 8t |
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2t - 4t2 |
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Determining the velocities of points of a flat figure
It was noted that the motion of a flat figure can be considered as a component of translational motion, in which all points of the figure move with a speedpoles AND , and from a rotational motion around this pole. Let us show that the speed of any point Mfigures are added geometrically from the speeds that the point receives in each of these movements.
Indeed, the position of any point M shapes are defined in relation to the axes Ooh radius vector(Fig. 3), where is the radius vector of the pole AND , - vector defining the position of the point Mrelative to the axesmoving with the pole ANDtranslational (the movement of the figure in relation to these axes is a rotation around the pole AND). Then
In the obtained equality, the quantityis the pole speed AND ; the magnitudeequal to speed which point M gets at, i.e. relative to the axes, or, in other words, when the figure rotates around the pole AND... Thus, it really follows from the previous equality that
Speed which point Mgets when the figure rotates around the pole AND :
where ω is the angular velocity of the figure.
Thus, the speed of any point M a flat figure is geometrically composed of the speed of some other point AND taken for the pole, and the speed that the point M gets when the figure rotates around this pole. Speed \u200b\u200bmodule and directionare found by constructing the corresponding parallelogram (Fig. 4).
Fig. 3 Fig. 4
Theorem on the projections of the velocities of two points of a body
Determination of the velocities of points of a plane figure (or a body moving in a plane-parallel way) is usually associated with rather complicated calculations. However, you can get a number of other, practically more convenient and simple methods for determining the velocities of the points of a figure (or body).
Fig. 5
One of such methods is given by the theorem: the projections of the velocities of two points of a rigid body on an axis passing through these points are equal to each other. Consider any two points AND and IN flat figure (or body). Taking the point AND for the pole (Fig. 5), we obtain... Hence, projecting both sides of the equality onto the axis directed along AB, and considering that the vectorperpendicular AB, we find
and the theorem is proved.
Determination of the speeds of points of a flat figure using the instantaneous center of speeds.
Another simple and intuitive method for determining the velocities of points of a plane figure (or a body in plane motion) is based on the concept of an instantaneous center of velocities.
Instantaneous speed center is called a point of a flat figure, the speed of which at a given time is equal to zero.
It is easy to make sure that if the figure is moving implicitly, then such a point at each moment of time t there is and, moreover, the only one. Let the moment in time t points AND and IN flat figures have speedsand not parallel to each other (Fig. 6). Then the point Rlying at the intersection of perpendiculars Aa to vectorand IN b to vector , and will be the instantaneous center of velocities since... Indeed, if we assume that, then by the velocity projection theorem the vectormust be simultaneously perpendicular and AR (as) and BP (as), which is impossible. From the same theorem it can be seen that no other point of the figure at this moment in time can have a speed equal to zero.
Fig. 6
If we now take the point R per pole, then the speed of the point AND will be
as ... A similar result is obtained for any other point in the shape. Consequently, the velocities of the points of a flat figure are determined at a given moment in time as if the movement of the figure were rotation around the instantaneous center of velocities. Wherein
It also follows from the equalities thatpoints of a flat figure are proportional to their distances from the MDC.
The results obtained lead to the following conclusions.
1. To determine the instantaneous center of speeds, you only need to know the directions of the speedsand any two points AND and IN a flat figure (or the trajectory of these points); the instantaneous center of velocities is at the point of intersection of perpendiculars retrieved from points AND and IN to the velocities of these points (or to the tangents to the trajectories).
2. To determine the speed of any point of a flat figure, you need to know the module and direction of the speed of any one point AND figures and the direction of speed of another point IN... Then, recovering from the points AND and IN perpendiculars toand , construct the instantaneous center of velocities R and towardsdetermine the direction of rotation of the figure. After that, knowing, find the speedany point M flat figure. Directed vectorperpendicular RM towards the rotation of the figure.
3. Angular velocityof a flat figure is equal at any given moment in time to the ratio of the speed of some point of the figure to its distance from the instantaneous center of speeds R :
Let us consider some special cases of determining the instantaneous center of velocities.
a) If the plane-parallel movement is carried out by rolling without sliding one cylindrical body on the surface of another stationary body, then the point R of a rolling body, touching a fixed surface (Fig. 7), has at a given time, due to the absence of sliding, a velocity equal to zero (), and, therefore, is the instantaneous center of speeds. An example is the rolling of a wheel on a rail.
b) If the speeds of the points AND and IN planar figures are parallel to each other, and the line AB not perpendicular(Figure 8, a), then the instantaneous center of velocities lies at infinity and the velocities of all points are parallel... Moreover, it follows from the theorem on the projections of velocities thati.e. ; the same result is obtained for all other points. Consequently, in the case under consideration, the velocities of all points of the figure at a given time are equal to each other both in magnitude and in direction, i.e. the figure has an instantaneous translational distribution of velocities (this state of motion of the body is also called instantaneous translational). Angular velocitythe body at this moment in time, as seen, is zero.
Fig. 7
Fig. 8
c) If the speeds of the points AND and IN planar figures are parallel to each other and the line ABperpendicular, then the instant center of speeds R is determined by the construction shown in Fig. 8, b. The fairness of the constructions follows from the proportion. In this case, unlike the previous ones, to find the center R apart from directions, you also need to know speed modules.
d) If the velocity vector is knownany point IN figures and its angular velocity, then the position of the instantaneous center of velocities R lying on the perpendicular to(fig. 8, b), can be found as.
Solving problems to determine speed.
To determine the desired kinematic characteristics (angular velocity of a body or the velocities of its points), it is necessary to know the modulus and direction of the velocity of any one point and the direction of the velocity of another point of the section of this body. The solution should begin with the definition of these characteristics according to the given tasks.
The mechanism, the movement of which is being investigated, must be depicted in the drawing in the position for which it is required to determine the corresponding characteristics. When calculating, it should be remembered that the concept of the instantaneous center of velocities takes place for a given rigid body. In a mechanism consisting of several bodies, each non-translational moving body at a given time has its own instantaneous center of velocities R and its angular velocity.
Example 1.The body, which has the shape of a coil, rolls with its middle cylinder on a fixed plane so that(cm). Radii of cylinders:R= 4 media r\u003d 2 cm (Fig. 9). .
Fig. 9
Decision. We define the velocity of the point A, Band FROM.
The instantaneous center of speeds is at the point where the coil touches the plane.
Pole speed FROM .
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Point speeds AND and INdirected perpendicular to the line segments connecting these points with the instantaneous center of velocities. The magnitude of the speeds:
Example 2. Radius wheel R \u003d 0.6 m rolls without sliding along a straight section of the track (Figure 9.1); the speed of its center C is constant and equal tov c
\u003d 12 m / s. Find the angular speed of the wheel and the speeds of the ends M 1 , M 2 , M 3 , M 4 vertical and horizontal wheel diameters.
Figure 9.1
Decision. The wheel makes a plane-parallel movement. The instantaneous center of speeds of the wheel is at the point M1 of contact with the horizontal plane, i.e.
Wheel angular speed
Find the speeds of points M2, M3 and M4
Example3 . Car driving wheel radius R \u003d 0.5 m is rolling with sliding (with slipping) along a straight section of the highway; its center speed FROM constant and equalv c
= 4 m / s. The instantaneous center of speed of the wheel is at the point R on distance h = 0.3 m from the rolling plane. Find the angular speed of the wheel and the speed of the points AND and IN its vertical diameter.
Figure 9.2
Decision. Wheel angular speed
Find the speed of points AND and IN
Example 4.Find the angular velocity of the connecting rod AB and speed of points IN and From the crank mechanism (Fig. 9.3, and). Given the angular velocity of the crank OA and sizes: ω OA \u003d 2 s -1, OA = AB \u003d 0.36 m, AS\u003d 0.18 m.
and)
b)
Figure 9.3
Decision. Crank OA makes a rotational movement, connecting rod AB - plane-parallel movement (Figure 9.3, b).
Find the speed of the point AND link
OA
Point speed IN directed horizontally. Knowing the direction of the speeds of points AND and IN connecting rod AB, determine the position of its instantaneous center of velocities - point R AB.
Link angular velocity AB and speed of points IN and C:
Example 5. Kernel ABslides its ends along mutually perpendicular straight lines so that at an anglespeed (fig. 10). Bar lengthAB \u003d l... Determine the speed of the end AND and the angular velocity of the rod.
Fig. 10
Decision. It is easy to determine the direction of the point velocity vector AND sliding along a vertical line. Thenis at the intersection of perpendicularsand (fig. 10).
Angular velocity
Point speed AND :
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Speed \u200b\u200bplan.
Let the velocities of several points of a plane section of the body be known (Fig. 11). If these speeds are plotted to scale from some point ABOUT and connect them with straight ends, you get a picture, which is called a speed plan. (On the picture) .
Fig. 11
Properties of the velocity plan.
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Really, ... But on the plan of speeds
. Meansmoreover perpendicular AB, therefore.Likewise, and.
b) The sides of the velocity plan are proportional to the corresponding line segments on the plane of the body.
As
, then it follows that the sides of the velocity plan are proportional to the line segments on the plane of the body.Combining these properties, we can conclude that the plan of velocities is similar to the corresponding figure and rotated relative to it by 90˚ in the direction of rotation. These properties of the plan of velocities allow you to determine the velocities of points of the body graphically.
Example 6. Figure 12 is a scaled illustration of the mechanism. Known angular velocitylink OA.
Fig. 12
Decision.To construct a plan of velocities, the velocity of one point must be known, although the direction of the velocity vector of another. In our example, you can determine the speed of the point AND : and direction of the vector.
Fig. 13
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Point speed E equal to zero, therefore the point e on the plan of speeds coincides with the point ABOUT.
Next, there should be
and ... We draw these lines, find their intersection pointd.Section about d will determine the velocity vector.Example 7.In articulated four-link OABS drive crankOA cm rotates evenly around the axis ABOUT angular velocityω \u003d 4 s -1 and using a connecting rod AB \u003d 20 cm drives the rotary crank Sun around the axis FROM (Figure 13.1, and). Determine Point Velocities AND and IN, as well as the angular speeds of the connecting rod ABand crank Sun.
and)
b)
Figure 13.1
Decision.Point speed AND crank OA
Taking a point AND for the pole, compose the vector equation
where
A graphical solution to this equation is given in Figure 13.1. , b (speed plan).
Using the speed plan, we get
Angular speed of the connecting rod AB
Point speed IN can be found using the theorem on the projections of the velocities of two points of the body onto the line connecting them
B and the angular velocity of the crank SV
Determining the acceleration of points of a plane shape
Let us show that the acceleration of any point M of a plane figure (as well as speed) is the sum of the accelerations that the point receives during the translational and rotational movements of this figure. Point position M in relation to the axes ABOUT xy (see Figure 30) is determined radius vectoris the angle between the vectorand a segment MA (fig. 14).
Thus, the acceleration of any point Ma flat figure is geometrically composed of the acceleration of some other point AND taken for the pole, and the acceleration, which is the point Mgets when the figure rotates around this pole. Module and direction of acceleration, are found by constructing the corresponding parallelogram (Fig. 23).
However, the calculation and acceleration any point AND this figure at the moment; 2) the trajectory of some other point IN figures. In some cases, instead of the trajectory of the second point of the figure, it is sufficient to know the position of the instantaneous center of velocities.
When solving problems, the body (or mechanism) must be depicted in the position for which it is required to determine the acceleration of the corresponding point. The calculation begins with the determination of the point taken as the pole according to the data of the problem.
Solution plan (if the speed and acceleration of one point of the plane figure and the directions of the speed and acceleration of another point of the figure are specified):
1) Find the instantaneous center of velocities by restoring perpendiculars to the velocities of two points of a flat figure.
2) Determine the instantaneous angular velocity of the figure.
3) Determine the centripetal acceleration of a point around the pole, equating to zero the sum of the projections of all acceleration terms on the axis perpendicular to the known direction of acceleration.
4) Find the modulus of rotational acceleration by equating to zero the sum of the projections of all acceleration terms on the axis perpendicular to the known direction of acceleration.
5) Determine the instantaneous angular acceleration of a flat figure from the found rotational acceleration.
6) Find the acceleration of a point of a flat figure using the formula for the distribution of accelerations.
When solving problems, you can apply the "theorem on the projections of the acceleration vectors of two points of an absolutely rigid body":
“Projections of the acceleration vectors of two points of an absolutely rigid body, which performs plane-parallel motion, onto a straight line rotated relative to a straight line passing through these two points, in the plane of motion of this body at an anglein the direction of angular acceleration are equal. "
It is convenient to apply this theorem if the accelerations of only two points of an absolutely rigid body are known both in absolute value and in direction, only the directions of the acceleration vectors of other points of this body are known (the geometric dimensions of the body are not known), are not knownand - respectively, the projection of the angular velocity and angular acceleration vectors of this body onto the axis perpendicular to the plane of motion, the velocities of the points of this body are not known.
There are 3 more methods for determining the acceleration of points of a flat figure:
1) The method is based on differentiating twice in time the laws of plane-parallel motion of an absolutely rigid body.
2) The method is based on the use of the instantaneous center of acceleration of an absolutely rigid body (the instantaneous center of acceleration of an absolutely rigid body will be discussed below).
3) The method is based on the use of an absolutely rigid body acceleration plan.
Lecture 3. Plane-parallel motion of a rigid body. Determination of speeds and accelerations.
This lecture addresses the following issues:
1. Plane-parallel motion of a rigid body.
2. Equations of plane-parallel motion.
3. Decomposition of motion into translational and rotational.
4. Determination of the speeds of points of a flat figure.
5. Theorem on the projections of the velocities of two points of the body.
6. Determination of the speeds of points of a flat figure using the instantaneous center of speeds.
7. Solving problems to determine the speed.
8. Speed \u200b\u200bplan.
9. Determination of the acceleration of points of a flat figure.
10. Solving problems for acceleration.
11. Instant acceleration center.
The study of these issues is necessary in the future for the dynamics of the plane motion of a rigid body, the dynamics of the relative motion of a material point, for solving problems in the disciplines "Theory of machines and mechanisms" and "Machine parts".
Plane-parallel motion of a rigid body. Equations of plane-parallel motion.
Decomposition of motion into translational and rotational
Plane-parallel (or flat) is a motion of a rigid body, at which all its points move parallel to some fixed plane P (fig. 28). Plane motion is performed by many parts of mechanisms and machines, for example, a rolling wheel on a straight track, a connecting rod in a crank-slider mechanism, etc. A particular case of plane-parallel motion is the rotational motion of a rigid body around a fixed axis.
Fig. 28 Fig. 29
Consider the section S body of some plane Oxyparallel to the plane P (fig. 29). In a plane-parallel motion, all points of the body lying on a straight line MM’Perpendicular to the flow S, i.e., the plane P, move identically.
Hence, we conclude that to study the movement of the whole body, it is enough to study how it moves in the plane Oohsection Sof this body or some flat figure S... Therefore, in what follows, instead of the plane motion of the body, we will consider the motion of a plane figure S in its plane, i.e. in plane Ooh.
Figure position S in plane Oohis determined by the position of some segment drawn on this figure AB (fig. 28). In turn, the position of the segment AB can be determined by knowing the coordinates x A and y A points AND and the angle that the segment AB forms with the axis x... Point ANDselected to define the position of the figure S, hereinafter referred to as a pole.
When the figure moves, the values x A and y A and will change. To know the law of motion, that is, the position of the figure in the plane Ooh at any time, you need to know the dependencies
The equations that determine the law of the ongoing motion are called the equations of motion of a flat figure in its plane. They are also equations of plane-parallel motion of a rigid body.
The first two of the equations of motion determine the motion that the figure would perform at \u003d const; this, obviously, will be a translational movement in which all points of the figure move in the same way as the pole AND... The third equation determines the movement that the figure would perform at and, i.e. when the pole ANDmotionless; this will rotate the figure around the pole AND... Hence, we can conclude that in the general case, the motion of a flat figure in its plane can be considered as a sum of translational motion, in which all points of the figure move in the same way as the pole AND, and from a rotational motion around this pole.
The main kinematic characteristics of the considered motion are the speed and acceleration of the translational motion equal to the speed and acceleration of the pole, as well as the angular velocity and angular acceleration of the rotational motion around the pole.
Determining the velocities of points of a flat figure
It was noted that the motion of a plane figure can be considered as a component of translational motion, in which all points of the figure move with the speed of the pole AND, and from a rotational motion around this pole. Let us show that the speed of any point Mfigures are formed geometrically from the speeds that the point receives in each of these movements.
Indeed, the position of any point M shapes are defined in relation to the axes Ooh radius vector (Fig. 30), where is the radius vector of the pole AND, is the vector defining the position of the point M relative to axes moving with the pole ANDtranslationally (the movement of the figure in relation to these axes is a rotation around the pole AND). Then
Let us show that the acceleration of any point M a plane figure (as well as speed) is the sum of the accelerations that a point receives during the translational and rotational movements of this figure. Point position M in relation to the axes Oxy(see Figure 30) is determined by the radius vector where. Then
On the right-hand side of this equality, the first term is the pole acceleration AND, and the second term determines the acceleration that point m receives when the figure rotates around the pole A... hence,
The value, as the acceleration of a point of a rotating rigid body, is defined as
where and are the angular velocity and angular acceleration of the figure, and is the angle between the vector and the segment MA (fig. 41).
Thus, the acceleration of any point Ma flat figure is geometrically composed of the acceleration of some other point ANDtaken for the pole, and the acceleration, which is the point Mgets when the figure rotates around this pole. The modulus and direction of acceleration are found by plotting the corresponding parallelogram (Fig. 23).
However, the calculation using the parallelogram shown in Fig. 23 complicates the calculation, since it will first be necessary to find the value of the angle, and then the angle between the vectors and, Therefore, when solving problems, it is more convenient to replace the vector with its tangent and normal components and represent it in the form
In this case, the vector is directed perpendicular to AM in the direction of rotation, if it is accelerated, and against rotation, if it is slower; the vector is always directed from the point M to the pole AND(fig. 42). Numerically
If the pole ANDdoes not move in a straight line, then its acceleration can also be represented as the sum of the tangent and normal components, then
Fig. 41 Fig. 42
Finally, when the point Mmoves curvilinearly and its trajectory is known, then it can be replaced by a sum.
Self-test questions
What motion of a rigid body is called flat? Give examples of links of mechanisms that make plane motion.
What are the simple motions that make up the plane motion of a rigid body?
How is the speed of an arbitrary point of a body determined in plane motion?
What motion of a rigid body is called plane-parallel?
Complex point movement
This lecture addresses the following issues:
1. Complex movement of a point.
2. Relative, figurative and absolute motion.
3. Speed \u200b\u200baddition theorem.
4. The theorem of addition of accelerations. Coriolis acceleration.
5. Complex motion of a rigid body.
6. Cylindrical gear drives.
7. Addition of translational and rotational movements.
8. Screw movement.
The study of these issues is necessary in the future for the dynamics of the plane motion of a rigid body, the dynamics of the relative motion of a material point, for solving problems in the disciplines "Theory of machines and mechanisms" and "Machine parts".
Instant Center of Speeds.
Instant speed center - in a plane-parallel motion, a point with the following properties: a) its speed at a given time is equal to zero; b) the body rotates relative to it at a given time.
In order to determine the position of the instantaneous center of velocities, it is necessary to know the directions of the velocities of any two different points of the body, the velocities of which not are parallel. Then, to determine the position of the instantaneous center of velocities, it is necessary to draw perpendiculars to straight lines parallel to the linear velocities of the selected points of the body. At the point of intersection of these perpendiculars, the instantaneous center of velocities will be located.
In the event that the vectors of linear velocities of two different points of the body are parallel to each other, and the segment connecting these points is not perpendicular to the vectors of these velocities, then the perpendiculars to these vectors are also parallel. In this case, they say that the instantaneous center of velocities is at infinity, and the body moves instantly translationally.
If the velocities of two points are known, and these velocities are parallel to each other, and, in addition, the indicated points lie on a straight line perpendicular to the velocities, then the position of the instantaneous center of velocities is determined as shown in Fig. 2.
The position of the instantaneous center of velocities in the general case not coincides with the position of the instantaneous center of accelerations. However, in some cases, for example, with a purely rotational movement, the positions of these two points may coincide.
21. Determination of the accelerations of body points. The method of the pole. The concept of the instantaneous center of accelerations.
Let us show that the acceleration of any point M a plane figure (as well as speed) is the sum of the accelerations that a point receives during the translational and rotational movements of this figure. Point position M in relation to the axes Oxy(see Figure 30) is determined by the radius vector where. Then
On the right-hand side of this equality, the first term is the pole acceleration AND, and the second term determines the acceleration that point m receives when the figure rotates around the pole A... hence,
The value, as the acceleration of a point of a rotating rigid body, is defined as
where and are the angular velocity and angular acceleration of the figure, and is the angle between the vector and the segment MA (fig. 41).
Thus, the acceleration of any point Ma flat figure is geometrically composed of the acceleration of some other point ANDtaken for the pole, and the acceleration, which is the point Mgets when the figure rotates around this pole. The modulus and direction of acceleration are found by plotting the corresponding parallelogram (Fig. 23).
However, the calculation using the parallelogram shown in Fig. 23 complicates the calculation, since it will first be necessary to find the value of the angle, and then the angle between the vectors and, Therefore, when solving problems, it is more convenient to replace the vector with its tangent and normal components and represent it in the form
In this case, the vector is directed perpendicular to AM in the direction of rotation, if it is accelerated, and against rotation, if it is slower; the vector is always directed from the point M to the pole AND(fig. 42). Numerically
If the pole ANDdoes not move in a straight line, then its acceleration can also be represented as the sum of the tangent and normal components, then
Fig. 41 Fig. 42
Finally, when the point Mmoves curvilinearly and its trajectory is known, then it can be replaced by a sum.
