Equations of plane motion.
The main theorem
The movement of a flat figure in its plane consists of two movements: translational together with an arbitrarily chosen point (pole), and rotational around this pole.
The position of a plane figure on a plane is determined by the position of the selected pole and the angle of rotation around this pole, therefore, plane motion is described by three equations:
The first two equations (Fig. 5) determine the movement that the figure would perform at φ = const, it is obvious that this movement will be translational, in which all points of the figure will move in the same way as the pole BUT.
The third equation determines the movement that the figure would perform if x A = const and y A = const, those. when the pole BUT will be motionless; this movement will be the rotation of the figure around the pole BUT.
In this case, the rotational movement does not depend on the choice of the pole, and the translational movement is characterized by the movement of the pole.
The relationship between the speeds of two points of a flat figure.
Consider two points A and B of a flat figure. Point position IN relative to the fixed coordinate system Oxy is determined by the radius vector r B (fig. 5):
r B = r A + ρ,
Where r A - radius vector of the point BUT, ρ = AB
the vector that defines the position of the point IN
with respect to moving axes Ah 1 y 1 moving translationally with the pole BUT parallel to fixed axes Ooh.
Then the speed of the point IN will be equal
.
In the obtained equality, the quantity is the velocity of the pole BUT.
The magnitude is equal to the speed that the point IN gets at = const, those. relative to the axes Ah 1 y 1 when the figure rotates around the pole BUT... Let us introduce the designation for this speed:
Hence,
|
The speed of the rotational movement of the point is directed perpendicular to the segment AB and is equal
The module and direction of the speed of point B is found by plotting the corresponding parallelogram(fig. 6).
Example 1. Find the speeds of points A, B and D of the rim of a wheel rolling on a straight rail without sliding if the speed of the center of the wheel C is equal to V C.
Decision. We choose point C, the speed of which is known for the pole. Then the speed of point A is
where and modulo.
The value of the angular velocity ω is found from the condition that the point R the wheels do not slide on the rail and, therefore, at the moment is equal to zero V P = 0.
At the moment the speed of the point R is equal to
Since at the point R speed and directed in one straight line opposite sides and V P = 0 then V PC = V C, whence we get that ω = V C. / R, hence, V AC = ω R = V C.
Point speed BUT is the diagonal of a square built on mutually perpendicular vectors and, whose moduli are equal, therefore
The speed of point D is determined in a similar way.The speed of point B is equal to
In this case, the velocities and are equal in magnitude and directed along one straight line, therefore V B = 2V C .
Kernel AB makes a plane motion, which can be represented as a fall without initial velocity under the action of gravity and rotation around the center of gravity FROM with constant angular velocity.
Determine the equations of motion for a point IN if at the initial moment the rod AB was horizontal and point IN was on the right. Acceleration of gravity q... Bar length 2l... Point start position FROM take the origin of coordinates, and direct the coordinate axes, as shown in the figure.
Based on relations (2) and (3), equations (1) will take the form:
Integration and noticing that at the initial moment t = 0, x B = l and y B = 0, we get the coordinates of the point IN in the following form.
Determination of 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 poles BUT, and from a rotational motion around this pole. Let us show that the speed of any point M figures 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 BUT , - vector defining the position of the point M with respect to the axesmoving with the pole BUT translational (the movement of the figure in relation to these axes is a rotation around the pole BUT). Then
In the obtained equality, the quantityis the pole speed BUT; the magnitude equal to speed which point M gets at, i.e. with respect to the axes, or, in other words, when the figure rotates around the pole BUT... Thus, it really follows from the previous equality that
Speed which point M gets when the figure rotates around the pole BUT :
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 BUT taken for the pole, and the speed that the point M gets when the figure rotates around this pole. Speed module 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
Determining the velocities of the 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 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 BUT and IN a flat figure (or body). Taking the point BUT for the pole (Fig. 5), we get... Hence, projecting both sides of the equality onto the axis directed along AB, and taking into account that the vectorperpendicular AB, we find
and the theorem is proved.
Determination of the velocities of points of a flat figure using the instantaneous center of velocities.
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 tthere is and, moreover, the only one. Let at the moment of time t points BUT and IN flat figures have speeds and not parallel to each other (Fig. 6). Then the point R lying at the intersection of perpendiculars Aa to vector and 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 at the same time perpendicular and AR(as) and BP(as), which is impossible. From the same theorem it is seen that no other point of the figure at this moment in time can have a speed equal to zero.
Fig. 6
If now at the moment of time we take the point R beyond the pole, then the speed of the point BUT 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 velocities, you only need to know the directions of the velocities and any two points BUT and IN a flat figure (or the trajectory of these points); the instantaneous center of velocities is at the point of intersection of the perpendiculars retrieved from the points BUT 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 BUT figures and the direction of the speed of its other point IN... Then, recovering from the points BUT and IN perpendiculars to and , 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 stationary surface (Fig. 7), at a given time, due to the absence of sliding, has a velocity equal to zero (), and, therefore, is the instantaneous center of velocities. An example is the rolling of a wheel on a rail.
b) If the velocities of the points BUT and IN planar figures are parallel to each other, and the line AB not perpendicular(Fig. 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 that i.e. ; a similar result is obtained for all other points. Consequently, in the case under consideration, the velocities of all points of the figure at a given moment of 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 instantly translational) Angular velocitythe body at this moment in time, as seen, is zero.
Fig. 7
Fig. 8
c) If the velocities of the points BUT and IN planar figures are parallel to each other and at the same time the line AB perpendicular, then the instant center of velocities 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 the 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 the 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. With the determination of these characteristics according to the given tasks, the solution should begin.
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 an 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 moment in 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= 2 cm (Fig. 9). .
Fig. 9
Decision.Define the velocity of the point A, B and FROM.
The instantaneous center of speeds is at the point where the coil touches the plane.
Pole speed FROM .
|
|
Point speeds BUT and IN directed perpendicular to the line segments connecting these points with the instantaneous center of velocities. The magnitude of the speeds:
Example 2.Radius wheel R= 0.6 m rolls without sliding along a straight section of the track (Figure 9.1); the velocity of its center C is constant and equal tov c
= 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= 0.5 m is rolling with sliding (slipping) along a straight section of the highway; the speed of its center 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 BUT and IN its vertical diameter.
Figure 9.2
Decision.Wheel angular speed
Find the speed of the points BUT 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, but). Given the angular velocity of the crank OA and sizes: ω OA = 2 s -1, OA =AB = 0.36 m, AS= 0.18 m.
but)
b)
Figure 9.3
Decision. Crank OAmakes a rotational movement, connecting rod AB- plane-parallel movement (Figure 9.3, b).
Find the speed of the point BUT link
OA
Point speed IN directed horizontally. Knowing the direction of the velocities of the points BUT 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 AB slides its ends along mutually perpendicular straight lines so that at an angle speed (fig. 10). Bar length AB = l... Determine the speed of the end BUT and the angular velocity of the rod.
Fig. 10
Decision.It is not difficult to determine the direction of the point velocity vector BUT sliding along a vertical line. Thenis at the intersection of perpendiculars and (fig. 10).
Angular velocity
Point speed BUT :
|
|
Speed plan.
Let the velocities of several points of a plane section of the body be known (Fig. 11). If these velocities are plotted to scale from some point ABOUT and connect their ends with straight lines, you get a picture, which is called a speed plan. (On the image) .
Fig. 11
Properties of the velocity plan.
|
|
Really, ... But on the plan of speeds
. Means moreover 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 is 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 the points of the body graphically.
Example 6.Figure 12 is a scaled representation of the mechanism. Known angular velocity link OA.
Fig. 12
Decision.To construct a plan of velocities, the velocity of one point must be known, even though the direction of the velocity vector of another. In our example, you can determine the speed of the point BUT : and direction of the vector.
Fig. 13
|
|
Point speed E equal to zero, therefore the point e on the plan of velocities 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-linkOABS drive crankOAcm rotates evenly around the axis ABOUT with angular velocityω = 4 s -1 and using a connecting rod AB= 20 cm drives the crank Sun around the axis FROM(Figure 13.1, but). Determine Point Velocities BUT and IN, as well as the angular speeds of the connecting rod AB and crank Sun.
but)
b)
Figure 13.1
Decision.Point speed BUT crank OA
Taking a point BUT 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 on 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 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 ABOUT xy (see Figure 30) is determined radius vector- the angle between the vectorand a segment MA(fig. 14).
Thus, the acceleration of any point M a flat figure is geometrically composed of the acceleration of some other point BUT taken for the pole, and the acceleration, which is the point M gets when the figure rotates around this pole. Acceleration module and direction, are found by constructing the corresponding parallelogram (Fig. 23).
However, the calculation and acceleration any point BUT 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":
“The projections of the acceleration vectors of two points of an absolutely rigid body, which performs a 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 known and - 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 using 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.
Arbitrary point speed M figures are defined as the sum of the velocities that the point receives at translational motion together with the pole and rotational motion around the pole.
Imagine the position of the point M like (Figure 1.6).
Differentiating this expression in time, we get:
since 
.
In this case, the speed v MA... which point M gets when the figure rotates around the pole BUT, will be determined from the expression
v MA=ω · MA,
Where ω is the angular velocity of a flat figure.
Any point speed M a flat figure is geometrically composed of the speed of a point BUT taken as a pole, and speed, point M when rotating the figure around the pole. The module and direction of the speed of this speed are found by plotting a parallelogram of speeds.
Problem 1
Determine point speed BUT, if the speed of the center of the roller is 5 m / s, the angular speed of the roller is 
... Roller radius r = 0.2m, angle. Roller to roll without slipping.
Since the body performs a plane-parallel motion, the speed of the point BUT will consist of the pole velocity (point FROM) and the speed obtained by the point BUT when rotating around the pole FROM.
,
Answer: 
Theorem on the projections of the velocities of two points of a body moving in a plane-parallel
Consider any two points BUT and IN flat figure. Taking the point BUT for the pole (Figure 1.7), we get
.
Hence, projecting both sides of the equality onto the axis directed along AB, and given that the vector is perpendicular AB, we find
v B· cosβ=v A· cosα + v V A· cos90 °.
since v In A· cos90 ° = 0 we get: the projections of the velocities of two points of the rigid body on the axis passing through these points are equal.
Problem 1
Kernel AB glides down smooth wall and smooth floor, point speed A V A = 5m / s, angle between floor and bar AB is equal to 30 0 . Determine point speed IN.
Determination of the velocities of points of a flat figure using the instantaneous center of velocities
When determining the speeds of points of a flat figure through the speed of the pole, the speed of the pole and the speed of rotational movement around the pole can be equal in magnitude and opposite in direction and there is such a point P, the speed of which at a given moment in time is equal to zero 
, call it the instantaneous center of velocities.
Instantaneous speed center is called a point associated with a flat figure, the speed of which at a given time is zero.
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 instantaneously rotational around the axis passing through the instantaneous center of velocities (Fig. 1.8).
v A=ω · PA; ().
Because v B=ω · PB; (), then w = v B/PB=v A/PA
The velocities of the points of a flat figure are proportional to the shortest distances from these points to the instantaneous center of velocities.
The results obtained lead to the following conclusions:
1) to determine the position of the instantaneous center of velocities, it is necessary to know the magnitude and direction of the velocity and the direction of the velocity of some two points BUT and IN flat figure; instant center of speeds P located at the point of intersection of perpendiculars derived from points BUT and IN to the speeds of these points;
2) angular velocity ω of a flat figure at a given time is equal to the ratio of the speed to the distance from it to the instantaneous center R speeds: ω =v A/PA;
3) The speed of the point in relation to the instantaneous center of the velocities P will indicate the direction of the angular velocity w.
4) The magnitude of the speed of the point is directly proportional to the shortest distance from the point IN to the instant center of speeds R v А = ω ВР
Problem 1
Crank OA the length 0.2m rotates uniformly with angular velocity ω = 8 rad / s... To the connecting rod AB at the point FROM pivotally attached connecting rod CD. For a given position of the mechanism, determine the speed of the point D slider if angle.
Point movement IN limited by horizontal guides, the slider can only make a translational movement along the horizontal guides. Point speed IN directed in the same direction as. Since the two points of the connecting rod have the same direction of speeds, the body makes an instantaneous translational motion, and the speeds of all points of the connecting rod have the same direction and value.
5) Translational movement. Examples.
Determination of the rotational motion of the body around a fixed axis.
Equation of rotational motion.
- such a movement in which all its points move in planes perpendicular to some fixed straight line, and describe circles with centers lying on this straight line, called the axis of rotation.
The motion is set by the law of variation of the dihedral angle φ (angle of rotation) formed by the fixed plane P passing through the axis of rotation and the plane Q rigidly connected to the body:
Angular velocity is a quantity that characterizes the rate of change in the angle of rotation.
Angular acceleration is a quantity that characterizes the rate of change in the angular velocity.
Determination of the speed of any point of a flat figure.
1 way to determine velocities is through vectors. The speed of any point of a flat figure is equal to the geometric sum of the speeds of the pole and the rotational speed of this point around the pole. Thus, the speed of point B is equal to the geometric sum of the speed of pole A and the rotational speed of point B around the pole:
2 way of determining speeds - through projections. (theorem about the projections of velocities) The projections of the velocities of the points of a plane figure on the axis passing through these points are equal.
3) Formulas for calculating the speed and acceleration of a point in the natural way of specifying its movement.
Velocity vector; - Projection of speed to tangent;
Acceleration vector components; -projections of acceleration on the t and n axes;
Thus, the total acceleration of a point is the vector sum of two accelerations:
tangent, directed tangentially to the trajectory in the direction of increasing the arc coordinate, if (otherwise, in the opposite direction) and
normal acceleration normal to the tangent towards the center of curvature (concavity of the trajectory): Total acceleration modulus:
4) Formulas for calculating the speed and acceleration of a point in the coordinate way of specifying its movement in Cartesian coordinates.
Velocity vector components: - Velocity projections on the coordinate axes:
- components of the acceleration vector; -projection of acceleration on the coordinate axis;
5) Translational movement. Examples.
(a slider, a pump piston, a pair of wheels of a steam locomotive moving along a straight path, an elevator car, a compartment door, a Ferris wheel cabin) is a movement in which any straight line rigidly connected to the body remains parallel to itself. Usually the translational movement is identified with the rectilinear movement of its points, but this is not the case. The points and the body itself (the center of mass of the body) can move along curved paths, see, for example, the movement of the Ferris wheel's cockpit. In other words, it is movement without turns.
Recall that the motion of a plane figure can be considered as a sum of translational motion together with the pole and rotational motion around the pole.
In accordance with this the speed of an arbitrary point M of a flat figure is geometrically summed up of the speed of some point A, taken as a pole, and the speed that point M receives when the figure rotates around this pole, i.e.
In this case, the speed V MA is defined as the speed of the point M when the body rotates around a fixed axis passing through the point BUT perpendicular to the plane of motion (see § 7.2), i.e.
Thus, if the speed of the pole is known V A and the angular velocity of the body ω, then
speed of any point M body is determined in accordance with equality (8.2), the diagonal of the parallelsgram built on the vectors V A and V MA, as on the sides (fig. 8.3), and the speed module V M calculated by the formula
where y is the angle between vectors V A and V MA
Task 8.1. The wheel rolls on a stationary surface without slipping (fig. 8.4, but). Find the speed of points TO and D wheels, if the speed is known V c center C wheel, radius R wheels, distance KS = b and angle a.
Decision. 1. The considered motion of the wheel is plane-parallel. Taking point C as a pole (since its velocity is known), in accordance with the general equality (8.2), for the point TO we can write
However, there is no way to determine the value V KC, since the angular velocity ω is unknown.
To determine ω, consider the speed of another point, namely the point R the wheel touches a stationary surface (fig. 8.4, b). For this point, we can write the equality
Feature point R is the fact that at a given moment in time V p - 0 because the wheel rolls without slipping. Then equality (b) takes the form
where do we get
Hence follows: 1) the velocity vectors V PC and V c should be directed in opposite directions; 2) from the equality of modules V PC - V c we get sRS = V c, from here we find ω = V c / PC = V c / R. According to the direction of the vector V PC we determine the direction of the arc arrow from and show it in the drawing (Fig. 8.4, b).
Now back to the definition V K by equality (a). We find
Vcs = about KS - V ^ b / R. Knowing the direction of the angular velocity ω, we represent the vector V KC perpendicular to line KS and construct a parallelogram on the vectors V c and V KC(fig. 8.4, in). Since in this case V c and V KC mutually perpendicular, we finally find
2. Point speed D on the wheel rim, we determine from the equality V D = V C + V DC. Since numerically V DC - with R - V c, then the parallelogram built on the vectors V c and V DC, will be a rhombus. Angle between V c and V DC is equal to 2a. Having defined V D as the length of the corresponding diagonal of the rhombus, we obtain
Theorem on the projections of the velocities of two points of a rigid body
According to equality (8.2) for two_ arbitrary points BUT and IN a rigid body, the equality V B = V A + V BA, in accordance with which we carry out the construction shown in Fig. 8.5. By projecting this equality onto the axis Az, aimed at A B, get Mind + V BAz. Considering that the vector V BA perpendicular to a straight line
A B, find
This result also expresses the theorem: the projections of the velocities of two points of a rigid body onto an axis passing through these points are equal to each other.
Note that equality (8.5) mathematically reflects the fact that the body is considered as absolutely rigid and the distance between the points BUT and IN does not change. therefore equality (8.5) holds not only for plane-parallel, but also for any motion of a rigid body.
Task 8.2. Sliders BUT and IN, connected by a rod with hinges at the ends, are mixed along mutually perpendicular guides in the plane of the drawing (Figure 8.6, but). Determine at a given angle a the speed of a point IN, if the speed is known V A.
Decision. Let's draw the x-axis through the points BUT and IN. Knowing the direction V A,
we find the projection of this vector on the line AB: V Ax - V A cos a (in Fig.8.6, b this will be a segment Aa). Further in the drawing from point IN put off Bb - Aa(since the segment Aa located on the x-axis to the right of the point BUT, then the segment Bb postpone from the point IN on the x-axis to the right). Rising to the point B perpendicular to line AB, find the end point of the vector V B.
According to the projection theorem V A cos a = K ^ cosp. Hence (taking into account that P = 90 ° - a) we finally obtain V B = V A cos a / cos (90 ° - a) or V B = = V A ctg a.
Determining Point Velocities Using Instant Velocity Center
To determine the velocities of the points of a flat figure, we will select any point as a pole R. Then, according to the formula
(8.2), the speed of an arbitrary point M is defined as the sum of two vectors:
If the speed of the pole R at a given time was equal to zero, then the right-hand side of this equality would be represented by one term MR and the speed of any point would be defined as the speed of the point M body when rotating it around a fixed pole R.
Therefore, if we choose the point R, whose speed at a given time is equal to zero, then the moduli of the velocities of all points of the figure will be proportional to their distances to the pole P, and the directions of the velocity vectors of all points will be perpendicular to the straight line connecting the point under consideration and the pole P. Naturally, the calculation by formulas (8.6) is much simpler than the calculation by general formula (8.2).
The point of a flat figure, the speed of which at a given time is equal to zero, is called the instantaneous center of velocities (IMC). It is easy to verify that if a figure moves non-translationally, then such a point exists at each moment of time and, moreover, is unique. Note that the instantaneous center of velocities can be located both on the figure itself and on its mental continuation.
Consider ways to determine the position of the instantaneous center of velocities.
1. Let at the moment of time tjum of a plane figure, its angular velocity ω and the velocity V A any of her points BUT(fig. 8.7, but). Then, choosing a point BUT as a pole, the _speed_ of our point R can be determined by the formula V p = V A + Vp A -
The task is ^ to find such a point R, in which V P= 0, hence, for her V A + Y RL= 0 and hence U RA = -U A. Therefore, for the point R speed Have RA which point R gets when the figure rotates around the pole BUT, and speed U A poles BUT equal in modulus (U RA = U A) or about zAR = U A and are opposite in direction. Also, the point R must lie perpendicular to the vector Have A. Determining the position of a point R is carried out by such a construction: from the point BUT(fig. 8.7, b) raise the perpendicular to the vector U A and postpone the distance on it AR = Y A / s to the other side of the point BUT, where will the vector "point" Have And, if you turn it 90 ° in the direction of the arc arrow from.
The instantaneous center of velocities is the only point of a flat figure whose velocity at a given moment in time is equal to zero.
At another moment in time, the instantaneous center of velocities may already be another point of the plane figure.
2. Let the directions of the velocities be known V A and In(fig. 8.8, but) two points BUT and IN a flat figure (and the velocity vectors of these points are not parallel), or the elementary displacements of these points are known. The instantaneous center of velocities will be located at the point of intersection of the perpendiculars reconstructed from points A and B to the velocities of these points (or to the elementary displacements of points). Such a construction is performed in Fig. 8.8, b. It is based on the fact that for any points A and B figures apply provisions (8.6):
It follows from these equalities that
Knowing the position of the MCS and the angular velocity of the body, applying formulas (8.6), it is easy to determine the velocity of any point of this body. For example ^ for point TO(see fig. 8.8, b) module speed V K = coKP, vector To directed perpendicular to a straight line KR in accordance with
the direction of the arc arrow y.
Hence, the velocities of the points of a flat figure are determined at a given moment in time as if this figure rotates around the instantaneous center of velocities.
3. If the speed of the points BUT and IN planar figures are parallel to each other, then three options are possible, which are shown in Fig. 8.9. For cases where straight AB perpendicular to vectors V A and V B(fig. 8.9, a, b), constructions are based on proportion (8.7).
If the speed of the points Li W are parallel, and the straight line AB_nt perpendicular VBUT(fig. 8.9, in), then the perpendiculars to U A and V B are parallel and the instantaneous center of velocities is at infinity (AP = oo); the angular velocity of rotation of the figure ω = VJAP = V A / cc = 0. In this case, the velocities of all points of the figure at a given moment of time are equal to each other, ie, the figure has a velocity distribution as in translational motion. This state of body movement is called instantly translational. Note that in this state the accelerations of all points of the body will not be the same.
4. If the plane movement of the body is carried out by rolling it without sliding on a fixed surface (Fig. 8.10), then the point of contact R will be the instantaneous center of velocities (see Problem 8.1).
Task 8.3. The flat mechanism consists of 7 rods, 2, 3, 4 and slider IN(fig. 8.11), connected to each other and with fixed supports 0 { and 0 2 hinges; point D is in the middle of the rod AB. The lengths of the rods: / 2 = 0.4 m, / 2 = 1.2 m, / 3 = 0.7 m, / 4 = 0.3 m. The angular velocity of the rod 7 in a given position of the mechanism with, = 2 s -1 and directed counterclockwise. Define V A, V B, V D, V E, oo 2, co 3, to 4 and point speed TO in the middle of the rod DE (DK = KE).
Decision. In the considered mechanism, rods 7, 4 rotate, slide IN- translational, and the rods 2, 3 -
plane-parallel movement.
Point speed BUT define as belonging to the rod 7, performing rotational motion:
Consider the movement of the rod 2. Point speed BUT is determined, and the direction of the point velocity IN due to the fact that it belongs to the rod at the same time 2 and half-
a zoon moving along the guides. Now, raising from points BUT and IN perpendiculars to U A and the direction of movement of the slide IN, find the position of point C 2 - MDS of the rod 2.
In the direction of the vector U A, considering that in the considered position of the mechanism, the rod 2 rotates around point C 2, we determine the direction of angular velocity from 2 rods 2 and find its numerical value (about 2 = V a / AC 2 = 0.8 / 1.04 = 0.77 s -1, where AC 2 - AB sin 60 ° = 1.04 m (when considering A AC ~, B).
Now we determine the numerical values and directions of the velocities of the points IN and D rod 2 (as ABDC 2 equilateral, then ВС 2 - DC 2 - - 0.6 m):
Consider the movement of the rod 3. Point speed D known. Since the point E belongs to the rod at the same time 4, rotating around an axis 0 4 , then Y e 10 4 E. Then, drawing through the points D and E straight, perpendicular to velocities V D wV E, find the position of point C 3 - MDS of the rod
3. In the direction of the vector V D, looking from a fixed point С 3, we determine the direction of the angular velocity с 3, and we find its numerical value (having previously determined from AZ) C 3? segment Z) C 3 = Desin 30 ° = 0.35 m): c 3 = V d / C 3 D = 1.32 s -1.
To determine the speed of a point TO let's draw a straight line COP 3 and considering that AR K C 3 equilateral ( COP 3 = 0.35 m), we calculate Y k = = 0.462 m / s, U to AKS 3.
Consider the movement of the rod_4 rotating around the axis 0 4 . Knowing the direction and numerical value V E, we find the direction and value of the angular velocity from 4: from 4 = V e / 0 4 E - 2.67 s.
Answer: V A= 0.8 m / s, V B = V D= 0.462 m / s, V E = 0.8 m / s, with 2 = 0.77 s "1, with 3 = 1.32 s -1, (about 4 = 2.67 s -1, the directions of these quantities are shown in Fig. 8.11.
Note.In a mechanism consisting of several bodies, each non-translationally moving body at a given time has its own instantaneous center of velocities and its own angular velocity.
Task 8.4. The flat mechanism consists of rods 1, 2, 3 and a roller rolling without sliding on a fixed plane (Fig.8.12, but). Connections of rods to each other and a rod 3 to the skating rink at the point D - articulated. Rod lengths: 1 { - 0.4 m, / 2 = 0.6 m, / 3 = 0.8 m.For given angles a = 60 °, B = 30 °, the values and directions of the angular velocity ω, = = 2 s and the speed of the center are known ABOUT skating rink V 0= 0.346 m / s, ZABD= 90 °. Determine point speed IN and angular velocity from 2.
Decision. The mechanism has two degrees of freedom (its position is determined by two angles a and p, independent of each other) and the speed of the point IN(common point of the bars 2 and 3) depends on the speeds of the points BUT and D.
Considering the movement of the rod /, n we find the direction and value of the speed of the point A: V A= coj / j = 0.8 m / s, V a AjO (A.
Consider the movement of the ice rink. Its instantaneous center of velocities is located at the point R; then V D find from the proportion
Since A DOP isosceles and acute angles in it are equal to 30 °, then DP- 2 OP cos 30 ° = ORL / 3. From equality (a) we find V D - 0.6 m / s. Vector V D directed perpendicularly DP.
Since the point IN belongs to the rods at the same time AB and BD, then by the theorem on the projections of velocities there should be: 1) the projection of the vector In on a straight line A B U A(section Aa in fig. 8.12, but), i.e. U A cos a = 0.4 m / s; 2) vector projection In on a straight line DB is equal to the projection on this line of the vector Have 0(section Dd in fig. 8.12, but), i.e. Have 0 cos y = 0.3 m / s (y = 60 °).
Next, we decide graphically. We postpone from the point IN segments in the corresponding directions Bb (= Aa and Bb 2 = Dd. Point speed IN is equal to the sum of vectors V B = Bb + Bbj. Recovering from the point B ( perpendicular to BL X, and from
points b 2 - perpendicular to Bb 2. The point of intersection of these perpendiculars determines the end of the desired vector V B.
Since the directions of the segments Bb and Bb 2 are mutually perpendicular, then
Determine with 2. In fig. 8.12, b the so-called speed plan is shown, which graphically depicts the vector equality
where vectors V A and V B defined (see fig. 8.12, but), and direction V BA perpendicular to the bar AB. From the drawing (fig.8.12, b) find
Now we determine with 2 = V ba / AB- 1.66 s -1 (direction from 2 - counterclockwise).
Answer: V B - 0.5 m / s, s 2 = 1.66 s -1.
