Center of gravity of a circular arc
The arc has an axis of symmetry. The center of gravity lies on this axis, i.e. y C = 0 .
dl– arc element, dl = Rdφ, R– radius of the circle, x = Rcosφ, L= 2αR,
Hence:
x C = R(sinα/α).
Center of gravity of a circular sector
Radius sector R with central angle 2 α has an axis of symmetry Ox, where the center of gravity is located.
We divide the sector into elementary sectors, which can be considered triangles. The centers of gravity of elementary sectors are located on a circular arc of radius (2/3) R.
The center of gravity of the sector coincides with the center of gravity of the arc AB:
Semicircle:
37. Kinematics. Kinematics of a point. Methods for specifying the movement of a point.
Kinematics– a branch of mechanics in which the movement of material bodies is studied from a geometric point of view, without taking into account mass and the forces acting on them. Methods for specifying the movement of a point: 1) natural, 2) coordinate, 3) vector.
Kinematics of a point- a branch of kinematics that studies the mathematical description of the movement of material points. The main task of kinematics is to describe movement using a mathematical apparatus without identifying the reasons causing this movement.
Natural sp. the trajectory of the point, the law of its movement along this trajectory, the beginning and direction of the arc coordinate are indicated: s=f(t) – the law of the point’s movement. For linear motion: x=f(t).
Coordinate sp. the position of a point in space is determined by three coordinates, changes in which determine the law of motion of the point: x=f 1 (t), y=f 2 (t), z=f 3 (t).
If the motion is in a plane, then there are two equations of motion. The equations of motion describe the trajectory equation in parametric form. By excluding the parameter t from the equations, we obtain the trajectory equation in the usual form: f(x,y)=0 (for a plane).
Vector sp. the position of a point is determined by its radius vector drawn from some center. A curve that is drawn by the end of a vector is called. hodograph this vector. Those. trajectory – radius vector hodograph.
38. Relationship between coordinate and vector, coordinate and natural methods of specifying the movement of a point.
RELATIONSHIP OF THE VECTOR METHOD WITH THE COORDINATE AND NATURAL METHOD expressed by the ratios:
where is the unit unit of the tangent to the trajectory at a given point, directed towards the distance reference, and is the unit unit of the normal to the trajectory at a given point, directed towards the center of curvature (see Fig. 3).
CONNECTION OF THE COORDINATE METHOD WITH THE NATURAL. Trajectory equation f(x, y)=z; f 1 (x, z)=y is obtained from the equations of motion in coordinate form by eliminating the time t. Additional analysis of the values that the coordinates of a point can take determines that section of the curve that is a trajectory. For example, if the motion of a point is given by the equations: x=sin t; y=sin 2 t=x 2 , then the trajectory of the point is that section of the parabola y=x 2 for which -1≤x≤+1, 0≤x≤1. The beginning and direction of distance counting are chosen arbitrarily, this further determines the sign of the speed and the magnitude and sign of the initial distance s 0 .
The law of motion is determined by the dependence:
the sign + or - is determined depending on the accepted direction of distance measurement.
Point speed is a kinematic measure of its motion, equal to the time derivative of the radius vector of this point in the reference system under consideration. The velocity vector is directed tangent to the trajectory of the point in the direction of movement
Velocity vector (v) is the distance that a body travels in a certain direction per unit of time. Please note that the definition velocity vector is very similar to the definition of speed, except for one important difference: the speed of a body does not indicate the direction of movement, but the velocity vector of a body indicates both the speed and direction of movement. Therefore, two variables are needed that describe the velocity vector of the body: speed and direction. Physical quantities that have a value and a direction are called vector quantities.
Speed vector body may change from time to time. If either its speed or direction changes, the speed of the body also changes. A constant velocity vector implies a constant speed and a constant direction, whereas the term constant speed only implies a constant value without taking direction into account. The term "velocity vector" is often used interchangeably with the term "velocity". They both express the distance a body travels per unit time
Point acceleration is a measure of the change in its speed, equal to the derivative with respect to time of the speed of this point or the second derivative of the radius vector of the point with respect to time. Acceleration characterizes the change in the velocity vector in magnitude and direction and is directed towards the concavity of the trajectory.
Acceleration vector
This is the ratio of the change in speed to the period of time during which this change occurred. The average acceleration can be determined by the formula:
Where - acceleration vector.
The direction of the acceleration vector coincides with the direction of change in speed Δ = - 0 (here 0 is the initial speed, that is, the speed at which the body began to accelerate).
At time t1 (see Fig. 1.8) the body has a speed of 0. At time t2 the body has speed . According to the rule of vector subtraction, we find the vector of speed change Δ = - 0. Then you can determine the acceleration like this:
Center of gravity of the triangle. Let's use the partitioning method and divide the triangle ABC into elementary strips by drawing lines parallel to the side AC triangle. Each such strip can be taken as a rectangle; the centers of gravity of these rectangles are in their middles, i.e. at the median BD triangle. Therefore, the center of gravity of the triangle must lie on the same median BD.
Now dividing the triangle into elementary strips with lines parallel to the side AB, we conclude that the center of gravity of the triangle must be located on the median EU.
Hence, the center of gravity of a triangle is at the point of intersection of its medians
. This point, as is known, divides each of the medians into segments in the ratio, i.e. 
.
Center of gravity of the trapezoid. Similar to the previous one, let’s split the trapezoid ABCD into elementary strips parallel to the bases Sun And AD. The centers of gravity of the strips will be located on a straight line KL connecting the midpoints of the bases of the trapezoid. Consequently, the center of gravity of the trapezoid lies on this straight line. In order to find its distance from the lower base, we divide the trapezoid into triangles ABC And ACD. For these triangles, we have, respectively, , , , .
Using formula (8.20), we obtain
.
Center of gravity of a circular arc. Consider the arc ADV circles of radius with a central angle. Place the origin of coordinates at the center of the circle and direct the axis perpendicular to the chord AB.
Since, due to the symmetry of the figure relative to the axis, the center of gravity will lie on this axis, i.e. , then all that remains is to find the abscissa of the center of gravity; for this we use formula (8.18).
According to Fig. we have , , and, therefore,
, (8.22) where is half of the central angle in radians.
In particular, for a semicircle arc we will have
Center of gravity of a circular sector. To determine the position of the center of gravity of a circular sector, we divide it into elementary sectors, as shown in Fig. Each elementary sector can be taken as an isosceles triangle with a height equal to . But the altitude in an isosceles triangle is also its median; therefore, the center of gravity of each elementary triangle lies at a distance from the origin ABOUT. Accordingly, the geometric locus of the centers of gravity of all elementary triangles is an arc of a circle with radius .
This means that the center of gravity of the area of a circular sector can be sought as the center of gravity of the material line along which the weight of this sector is continuously and uniformly distributed. Applying formula (8.22), we obtain the coordinate of the center of gravity of the sector area
, (8.23) where is half of the central angle in radians. In particular, for a sector in the form of a semicircle we obtain
Problem 8.3. The plate is obtained from a square whose side is equal to , after cutting out of it a part constituting a quarter of a circle of radius centered at the vertex A square. Determine the center of gravity of the plate.
or, substituting the appropriate values,
.
Let us present without derivation the formulas that determine the positions of the centers of gravity of some of the simplest homogeneous bodies.
Based on the general formulas obtained above, it is possible to indicate specific methods for determining the coordinates of the centers of gravity of bodies.
1. Symmetry. If a homogeneous body has a plane, axis or center of symmetry (Fig. 7), then its center of gravity lies, respectively, in the plane of symmetry, axis of symmetry or in the center of symmetry.
Fig.7
2. Splitting. The body is divided into a finite number of parts (Fig. 8), for each of which the position of the center of gravity and area are known.
Fig.8
3.Negative area method. A special case of the partitioning method (Fig. 9). It applies to bodies that have cutouts if the centers of gravity of the body without the cutout and the cutout part are known. A body in the form of a plate with a cutout is represented by a combination of a solid plate (without a cutout) with an area S 1 and an area of the cut out part S 2 .
Fig.9
4.Grouping method. It is a good complement to the last two methods. After dividing a figure into its component elements, it is convenient to combine some of them again in order to then simplify the solution by taking into account the symmetry of this group.
Centers of gravity of some homogeneous bodies.
1) Center of gravity of a circular arc. Consider the arc AB radius R with a central angle. Due to symmetry, the center of gravity of this arc lies on the axis Ox(Fig. 10).
Fig.10
Let's find the coordinate using the formula. To do this, select on the arc AB element MM' length, the position of which is determined by the angle. Coordinate X element MM' will . Substituting these values X and d l and keeping in mind that the integral must be extended over the entire length of the arc, we obtain:
Where L- arc length AB, equal to .
From here we finally find that the center of gravity of a circular arc lies on its axis of symmetry at a distance from the center ABOUT, equal
where the angle is measured in radians.
2) Center of gravity of the triangle's area. Consider a triangle lying in the plane Oxy, the coordinates of the vertices of which are known: A i(x i,y i), (i= 1,2,3). Breaking the triangle into narrow strips parallel to the side A 1 A 2, we come to the conclusion that the center of gravity of the triangle must belong to the median A 3 M 3 (Fig. 11).
Fig.11
Breaking a triangle into strips parallel to the side A 2 A 3, we can verify that it must lie on the median A 1 M 1 . Thus, the center of gravity of a triangle lies at the point of intersection of its medians, which, as is known, separates a third part from each median, counting from the corresponding side.
In particular, for the median A 1 M 1 we obtain, taking into account that the coordinates of the point M 1 is the arithmetic mean of the coordinates of the vertices A 2 and A 3:
x c = x 1 + (2/3)∙(x M 1 - x 1) = x 1 + (2/3)∙[(x 2 + x 3)/2-x 1 ] = (x 1 +x 2 +x 3)/3.
Thus, the coordinates of the triangle’s center of gravity are the arithmetic mean of the coordinates of its vertices:
x c =(1/3)Σ x i ; y c =(1/3)Σ y i.
3) Center of gravity of the area of a circular sector. Consider a sector of a circle with radius R with a central angle of 2α, located symmetrically relative to the axis Ox(Fig. 12) .
It's obvious that y c = 0, and the distance from the center of the circle from which this sector is cut to its center of gravity can be determined by the formula:
Fig.12
The easiest way to calculate this integral is by dividing the integration domain into elementary sectors with an angle dφ. Accurate to infinitesimals of the first order, such a sector can be replaced by a triangle with a base equal to R× dφ and height R. The area of such a triangle dF=(1/2)R 2 ∙dφ, and its center of gravity is at a distance of 2/3 R from the vertex, therefore in (5) we put x = (2/3)R∙cosφ. Substituting in (5) F= α R 2, we get:
Using the last formula, we calculate, in particular, the distance to the center of gravity semicircle.
Substituting α = π/2 into (2), we obtain: x c = (4R)/(3π) ≅ 0.4 R .
Example 1. Let us determine the center of gravity of the homogeneous body shown in Fig. 13.
Fig.13
The body is homogeneous, consisting of two parts with a symmetrical shape. Coordinates of their centers of gravity:
Their volumes:
Therefore, the coordinates of the center of gravity of the body
Example 2. Let us find the center of gravity of a plate bent at a right angle. Dimensions are in the drawing (Fig. 14).
Fig.14
Coordinates of the centers of gravity:
Areas:
|
Fig.15
In this problem, it is more convenient to divide the body into two parts: a large square and a square hole. Only the area of the hole should be considered negative. Then the coordinates of the center of gravity of the sheet with the hole:
coordinate since the body has an axis of symmetry (diagonal).
Example 4. The wire bracket (Fig. 16) consists of three sections of equal length l.
Fig.16
Coordinates of the centers of gravity of the sections:
Therefore, the coordinates of the center of gravity of the entire bracket are:
Example 5. Determine the position of the center of gravity of the truss, all the rods of which have the same linear density (Fig. 17).
Let us recall that in physics the density of a body ρ and its specific gravity g are related by the relation: γ= ρ g, Where g- acceleration of gravity. To find the mass of such a homogeneous body, you need to multiply the density by its volume.
Fig.17
The term “linear” or “linear” density means that to determine the mass of a truss rod, the linear density must be multiplied by the length of this rod.
To solve the problem, you can use the partitioning method. Representing a given truss as a sum of 6 individual rods, we obtain:
Where L i length i th truss rod, and x i, y i- coordinates of its center of gravity.
The solution to this problem can be simplified by grouping the last 5 bars of the truss. It is easy to see that they form a figure with a center of symmetry located in the middle of the fourth rod, where the center of gravity of this group of rods is located.
Thus, a given truss can be represented by a combination of only two groups of rods.
The first group consists of the first rod, for it L 1 = 4 m, x 1 = 0 m, y 1 = 2 m. The second group of rods consists of five rods, for it L 2 = 20 m, x 2 = 3 m, y 2 = 2 m.
The coordinates of the center of gravity of the truss are found using the formula:
x c = (L 1 ∙x 1 +L 2 ∙x 2)/(L 1 + L 2) = (4∙0 + 20∙3)/24 = 5/2 m;
y c = (L 1 ∙y 1 +L 2 ∙y 2)/(L 1 + L 2) = (4∙2 + 20∙2)/24 = 2 m.
Note that the center WITH lies on the straight line connecting WITH 1 and WITH 2 and divides the segment WITH 1 WITH 2 regarding: WITH 1 WITH/SS 2 = (x c - x 1)/(x 2 - x c ) = L 2 /L 1 = 2,5/0,5.
Self-test questions
What is the center of parallel forces called?
How are the coordinates of the center of parallel forces determined?
How to determine the center of parallel forces whose resultant is zero?
What properties does the center of parallel forces have?
What formulas are used to calculate the coordinates of the center of parallel forces?
What is the center of gravity of a body?
Why can the gravitational forces of the Earth acting on a point on a body be taken as a system of parallel forces?
Write down the formula for determining the position of the center of gravity of inhomogeneous and homogeneous bodies, the formula for determining the position of the center of gravity of flat sections?
Write down the formula for determining the position of the center of gravity of simple geometric shapes: rectangle, triangle, trapezoid and half circle?
What is the static moment of area?
Give an example of a body whose center of gravity is located outside the body.
How are the properties of symmetry used in determining the centers of gravity of bodies?
What is the essence of the negative weights method?
Where is the center of gravity of a circular arc?
What graphical construction can be used to find the center of gravity of a triangle?
Write down the formula that determines the center of gravity of a circular sector.
Using formulas that determine the centers of gravity of a triangle and a circular sector, derive a similar formula for a circular segment.
What formulas are used to calculate the coordinates of the centers of gravity of homogeneous bodies, flat figures and lines?
What is called the static moment of the area of a plane figure relative to the axis, how is it calculated and what dimension does it have?
How to determine the position of the center of gravity of an area if the position of the centers of gravity of its individual parts is known?
What auxiliary theorems are used to determine the position of the center of gravity?
The result of the calculations depends not only on the cross-sectional area, therefore, when solving problems on strength of materials, one cannot do without determining geometric characteristics of figures: static, axial, polar and centrifugal moments of inertia. It is imperative to be able to determine the position of the center of gravity of the section (the listed geometric characteristics depend on the position of the center of gravity). In addition to geometric characteristics of simple figures: rectangle, square, isosceles and right triangles, circle, semicircle. The center of gravity and the position of the main central axes are indicated, and the geometric characteristics relative to them are determined, provided that the beam material is homogeneous.
Geometric characteristics of rectangle and square
Axial moments of inertia of a rectangle (square)
Geometric characteristics of a right triangle
Axial moments of inertia of a right triangle
Geometric characteristics of an isosceles triangle
Axial moments of inertia of an isosceles triangle
The mathematical technique of calculating the center of mass belongs to the field of mathematics courses; there, similar problems serve as good examples in integral calculus. But even if you know how to integrate, it is useful to know some tricks for calculating the position of the center of mass. One such trick is based on the use of the so-called Pappus theorem, which works as follows. If we take some closed figure and form a rigid body, rotating this figure in space so that each point moves perpendicular to the plane of the figure, then the volume of the resulting body is equal to the product of the area of the figure and the distance traveled by its center of gravity! Of course, this theorem is also true in the case when a flat figure moves in a straight line perpendicular to its area, but if we move it along a circle or some other
curve, then this results in a much more interesting body. When moving along a curved path, the inner part of the figure moves less than the outer part and these effects compensate each other. So if we want to define; the center of mass of a flat figure with uniform density, then you need to remember that the volume formed by its rotation about the axis is equal to the distance traveled by the center of mass multiplied by the area of the figure.
For example, if we need to find the center of mass of a right triangle with base D and height H (Fig. 19.2), then this is done as follows. Imagine an axis along H and rotate the triangle 360° around this axis. This gives us a cone. The distance traveled by the x-coordinate of the center of mass is 2πx, and the area of the region that moved, i.e. the area of the triangle, is l/2 HD. The product of the distance traveled by the center of mass and the area of the triangle is equal to the volume of the cone, i.e. 1/3 πD 2 H. Thus, (2πх) (1/2HD) = 1/3D 2 H, or x = D/З. Quite analogously, by rotating around the second leg or simply for reasons of symmetry, we find that y = H/3. In general, the center of mass of any homogeneous triangle is located at the intersection point of its three medians (lines connecting the vertex of the triangle with the middle of the opposite side), which is located at a distance from the base equal to 1/3 of the length of each median.
How to see it? Cut the triangle with lines parallel to the base into many strips. Notice now that the median divides each strip in half, therefore the center of mass must lie on the median.
Let us now take a more complex figure. Suppose that we need to find the position of the center of mass of a homogeneous semicircle, that is, a circle cut in half. Where will the center of mass be in this case? For a full circle, the center of mass is located at the geometric center, but for a semicircle it is more difficult to find its position. Let r be the radius of the circle, and x be the distance of the center of mass from the straight boundary of the semicircle. By rotating it around this edge as if around an axis, we get a ball. In this case, the center of mass travels a distance of 2πx, and the area of the semicircle is equal to 1/2πr 2 (half the area of the circle). Since the volume of the ball is, of course, 4πg 3 /3, then from here we find
or
There is another theorem of Pappus, which is actually a special case of the theorem formulated above, and therefore is also valid. Suppose that instead of a solid semicircle we take a semicircle, for example a piece of wire in the form of a semicircle with uniform density, and we want to find its center of mass. It turns out that the area that is “swept out” by a flat curve during its movement, similar to that described above, is equal to the distance traveled by the center of mass multiplied by the length of this curve. (The curve can be viewed as a very narrow strip and the previous theorem applied to it.)