The center of gravity of a flat body of irregular shape. Determining the coordinates of the center of gravity of plane figures
Physics lesson notes, grade 7
Topic: Determining the center of gravity
Physics teacher, Argayash Secondary School No. 2
Khidiyatulina Z.A.
Laboratory work:
"Determination of the center of gravity of a flat plate"
Target : finding the center of gravity of a flat plate.
Theoretical part:
All bodies have a center of gravity. The center of gravity of a body is the point relative to which the total moment of gravity acting on the body is zero. For example, if you hang an object by its center of gravity, it will remain at rest. That is, its position in space will not change (it will not turn upside down or on its side). Why do some bodies tip over while others don't? If you draw a line perpendicular to the floor from the center of gravity of the body, then if the line goes beyond the boundaries of the body’s support, the body will fall. The larger the area of support, the closer the center of gravity of the body is to the central point of the area of support and the center line of the center of gravity, the more stable the position of the body will be. For example, the center of gravity of the famous Leaning Tower of Pisa is located just two meters from the middle of its support. And the fall will happen only when this deviation is about 14 meters. The center of gravity of the human body is approximately 20.23 centimeters below the navel. An imaginary line drawn vertically from the center of gravity passes exactly between the feet. For a tumbler doll, the secret also lies in the center of gravity of the body. Its stability is explained by the fact that the center of gravity of the tumbler is at the very bottom; it actually stands on it. The condition for maintaining the balance of a body is the passage of the vertical axis of its common center of gravity within the area of the body’s support. If the vertical center of gravity of the body leaves the support area, the body loses balance and falls. Therefore, the larger the area of support, the closer the center of gravity of the body is located to the central point of the area of support and the central line of the center of gravity, the more stable the position of the body will be. The area of support when a person is in a vertical position is limited by the space that is under the soles and between the feet. The center point of the vertical line of the center of gravity on the foot is 5 cm in front of the heel tubercle. The sagittal size of the support area always prevails over the frontal one, therefore the displacement of the vertical line of the center of gravity occurs more easily to the right and left than backward, and is especially difficult forward. In this regard, stability during turns during fast running is significantly less than in the sagittal direction (forward or backward). A foot in shoes, especially with a wide heel and a hard sole, is more stable than without shoes, as it acquires a larger area of support.
Practical part:
Purpose of the work: Using the proposed equipment, experimentally find the position of the center of gravity of two figures made of cardboard and a triangle.
Equipment:Tripod, thick cardboard, triangle from a school kit, ruler, tape, thread, pencil...
Task 1: Determine the position of the center of gravity of a flat figure of arbitrary shape
Using scissors, cut out a random shape from cardboard. Attach the thread to it at point A with tape. Hang the figure by the thread to the tripod leg. Using a ruler and pencil, mark the vertical line AB on the cardboard.
Move the thread attachment point to position C. Repeat the above steps.
Point O of the intersection of lines AB andCDgives the desired position of the center of gravity of the figure.
Task 2: Using only a ruler and pencil, find the position of the center of gravity of a flat figure
Using a pencil and ruler, divide the shape into two rectangles. By construction, find the positions O1 and O2 of their centers of gravity. It is obvious that the center of gravity of the entire figure is on the O1O2 line
Divide the figure into two rectangles in another way. By construction, find the positions of the centers of gravity O3 and O4 of each of them. Connect points O3 and O4 with a line. The intersection point of lines O1O2 and O3O4 determines the position of the figure’s center of gravity
Task 2: Determine the position of the center of gravity of the triangle
Using tape, secure one end of the thread at the top of the triangle and hang it from the tripod leg. Using a ruler, mark the direction AB of the gravity line (make a mark on the opposite side of the triangle)
Repeat the same procedure, hanging the triangle from vertex C. On the opposite vertex C side of the triangle, make a markD.
Using tape, attach pieces of thread AB andCD. Point O of their intersection determines the position of the center of gravity of the triangle. In this case, the center of gravity of the figure is outside the body itself.
III . Solving quality problems
1.For what purpose do circus performers hold heavy poles in their hands when walking on a tightrope?
2. Why does a person carrying a heavy load on his back lean forward?
3. Why can’t you get up from a chair unless you tilt your body forward?
4.Why does the crane not tip towards the load being lifted? Why, without a load, does the crane not tip towards the counterweight?
5. Why do cars and bicycles, etc. Is it better to put brakes on the rear wheels rather than the front wheels?
6. Why does a truck loaded with hay overturn more easily than the same truck loaded with snow?
Draw a diagram of the system and mark the center of gravity on it. If the found center of gravity is outside the object system, you received an incorrect answer. You may have measured distances from different reference points. Repeat the measurements.
- For example, if children are sitting on a swing, the center of gravity will be somewhere between the children, and not to the right or left of the swing. Also, the center of gravity will never coincide with the point where the child is sitting.
- These arguments are valid in two-dimensional space. Draw a square that will contain all the objects of the system. The center of gravity should be inside this square.
Check your math if you get a small result. If the reference point is at one end of the system, a small result places the center of gravity near the end of the system. This may be the correct answer, but in the vast majority of cases this result indicates an error. When you calculated the moments, did you multiply the corresponding weights and distances? If instead of multiplying you added the weights and distances, you would get a much smaller result.
Correct the error if you found multiple centers of gravity. Each system has only one center of gravity. If you found multiple centers of gravity, you most likely did not add up all the moments. The center of gravity is equal to the ratio of the “total” moment to the “total” weight. There is no need to divide “every” moment by “every” weight: this way you will find the position of each object.
Check the reference point if the answer differs by some integer value. In our example, the answer is 3.4 m. Let's say you got the answer 0.4 m or 1.4 m, or another number ending in ".4". This is because you did not choose the left end of the board as your starting point, but a point that is located a whole amount to the right. In fact, your answer is correct no matter what reference point you choose! Just remember: the reference point is always at position x = 0. Here's an example:
- In our example, the reference point was at the left end of the board and we found that the center of gravity was 3.4 m from this reference point.
- If you choose as a reference point a point that is located 1 m to the right from the left end of the board, you will get the answer 2.4 m. That is, the center of gravity is 2.4 m from the new reference point, which, in turn, is located 1 m from the left end of the board. Thus, the center of gravity is at a distance of 2.4 + 1 = 3.4 m from the left end of the board. It turned out to be an old answer!
- Note: when measuring distances, remember that the distances to the “left” reference point are negative, and to the “right” reference point are positive.
Measure distances in straight lines. Suppose there are two children on a swing, but one child is much taller than the other, or one child is hanging under the board rather than sitting on it. Ignore this difference and measure the distances along the straight line of the board. Measuring distances at angles will give close but not entirely accurate results.
- For the see-saw board problem, remember that the center of gravity is between the right and left ends of the board. Later, you will learn to calculate the center of gravity of more complex two-dimensional systems.
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 bearing 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:
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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- free fall acceleration. 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 a 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?
Before finding the center of gravity of simple figures, such as those that have a rectangular, round, spherical or cylindrical, as well as square shape, you need to know at what point the center of symmetry of a particular figure is located. Because in these cases, the center of gravity will coincide with the center of symmetry.
The center of gravity of a homogeneous rod is located at its geometric center. If you need to determine the center of gravity of a round disk of a homogeneous structure, then first find the point of intersection of the diameters of the circle. It will be the center of gravity of this body. Considering such figures as a ball, a hoop and a uniform rectangular parallelepiped, we can say with confidence that the center of gravity of the hoop will be in the center of the figure, but outside its points, the center of gravity of the ball is the geometric center of the sphere, and in the latter case, the center of gravity is considered to be the intersection diagonals of a rectangular parallelepiped.
Center of gravity of inhomogeneous bodies
To find the coordinates of the center of gravity, as well as the center of gravity of an inhomogeneous body, it is necessary to figure out on which segment of a given body the point is located at which all the gravity forces intersect, acting on the figure if it is turned over. In practice, to find such a point, the body is suspended on a thread, gradually changing the points of attachment of the thread to the body. In the case when the body is in equilibrium, the center of gravity of the body will lie on a line that coincides with the line of the thread. Otherwise, gravity causes the body to move.
Take a pencil and a ruler, draw vertical straight lines that will visually coincide with the thread directions (threads attached to various points of the body). If the body shape is quite complex, then draw several lines that will intersect at one point. It will become the center of gravity for the body on which you performed the experiment.
Center of gravity of the triangle
To find the center of gravity of a triangle, you need to draw a triangle - a figure consisting of three segments connected to each other at three points. Before finding the center of gravity of the figure, you need to use a ruler to measure the length of one side of the triangle. Place a mark in the middle of the side, then connect the opposite vertex and the middle of the segment with a line called the median. Repeat the same algorithm with the second side of the triangle, and then with the third. The result of your work will be three medians that intersect at one point, which will be the center of gravity of the triangle.
If you are faced with a task regarding how to find the center of gravity of a body in the shape of an equilateral triangle, then you need to draw a height from each vertex using a rectangular ruler. The center of gravity in an equilateral triangle will be at the intersection of altitudes, medians and bisectors, since the same segments are simultaneously altitudes, medians and bisectors.
Coordinates of the center of gravity of the triangle
Before finding the center of gravity of the triangle and its coordinates, let’s take a closer look at the figure itself. This is a homogeneous triangular plate, with vertices A, B, C and, accordingly, coordinates: for vertex A - x1 and y1; for vertex B - x2 and y2; for vertex C - x3 and y3. When finding the coordinates of the center of gravity, we will not take into account the thickness of the triangular plate. The figure clearly shows that the center of gravity of the triangle is indicated by the letter E - to find it, we drew three medians, at the intersection of which we placed point E. It has its own coordinates: xE and yE.
One end of the median drawn from vertex A to segment B has coordinates x 1 , y 1 (this is point A), and the second coordinates of the median are obtained based on the fact that point D (the second end of the median) is in the middle of segment BC. The ends of this segment have coordinates known to us: B(x 2, y 2) and C(x 3, y 3). The coordinates of point D are denoted by xD and yD. Based on the following formulas:
x=(X1+X2)/2; y=(U1+U2)/2
Determine the coordinates of the middle of the segment. We get the following result:
xd=(X2+X3)/2; уd=(У2+У3)/2;
D *((X2+X3)/2, (U2+U3)/2).
We know what coordinates are typical for the ends of the segment AD. We also know the coordinates of point E, that is, the center of gravity of the triangular plate. We also know that the center of gravity is located in the middle of the segment AD. Now, using formulas and data known to us, we can find the coordinates of the center of gravity.
Thus, we can find the coordinates of the center of gravity of the triangle, or rather, the coordinates of the center of gravity of the triangular plate, given that its thickness is unknown to us. They are equal to the arithmetic mean of the homogeneous coordinates of the vertices of the triangular plate.