Find the calculation method of the center of gravity of the sector and the ring

Let the angle a of the density p radius r of the fan-shaped object, and the position of the symmetrical centroid is on the angular bisector, and the polar axis or is established on the bisector, and the origin is at the center of the circle, which is defined by the centroid.

rm=∫rdm/∫dm=∫rpds/∫pds=∫rrdrdθ/∫rdrdθ=(1/3)ar^3/（1/2）ar^2=(2/3)r

The ring is at the center of the circle.

The center of gravity is the point at which the resultant force of all the constituent fulcrums of the gravity force passes through the object in any direction in the gravitational field. The center of gravity of a regular and uniform object is its geometric center. The center of gravity of an irregular object can be determined by the suspension method.

The center of gravity of an object, not necessarily on the object. Also, the center of gravity can refer to the center or main part of the matter.

Find the center of gravity method to collapse this paragraph.

Here are some ways to find the center of gravity of an object with an irregular shape or uneven mass.

a.Suspension method.

Apply only to thin sheets (not necessarily uniform). First find a thin rope, find a point on the object, hang it with a rope, draw the gravity line after the object is stationary, and find a little bit of suspension in the same way, the intersection of the two gravity lines is the center of gravity of the object.

Support method. Apply only to fine sticks (not necessarily uniform). Supporting the object with a fulcrum, constantly changing position, the more stable the position, the closer to the center of gravity.

One possible workaround is to use two fulcrums to support and then apply a smaller force to bring the two fulcrums closer together, because the fulcrum closer to the center of gravity will have more friction, so the object will move with it, bringing the other fulcrum closer to the center of gravity, so that the approximate position of the center of gravity can be found.

c.The pintop method is also only suitable for thin plates. Use a thin needle against the underside of the board, and when the board is able to maintain balance, the top of the needle is positioned close to the center of gravity.

In the same way as the support method, you can use the method of approaching three thin needles to each other to find the range of the center of gravity, but this is not as convenient as the workaround of the support method.

d.Use the plumb line to find the center of gravity (any shape, uniform texture).

Hang it with a rope at one end of it, and then hang it on this end with a plumb line (trace it down). Then use the same method to make another line. The intersection of two lines is its center of gravity.

The sector chord length is c arc length l radius r

Then its center of gravity is 2rc 3l away from the vertex and this point is on the axis of symmetry! And so it is determined.

If the radius of the major circle is set to r and the minor circle is set to r, then half of the circumferential center of gravity is separated from the center of the circle.

d=[4 (3*pi)]*r 2+r 2+r*r] [r+r], especially when r=0, it is the center of gravity of the semicircle 4*r (3*pi), and when r=r it is the center of gravity of the semicircle curve.

2*r/pi.

Let the radius of this sector r angle a

First of all, the center of gravity must be bisector of the fan angle. The length of the angular bisector is r, and one of the segments is x, starting from the center of the circle. Then there is a sector with an area of r 2 a 360° then the sum of the areas of the two symmetrical triangles formed by x should be half of the total area.

That is, x= ra 720sin(a 2) xr=1 2 r 2 a 360° is the center of gravity of the point according to the center of the circle ra 720sin(a 2) on the bisector of the sector angle.

Wow, wow, wow... I've forgotten about this, I'm sorry!

Are you talking about the part that is cut out with a circle and a straight line?

This shape should be called a bow.

The distance from the center of the circle to the chord is c, and the distance from the center of gravity is d. An object can be seen as the sum of straight lines that are continuously distributed from left to right (blue line in the diagram). The gravitational force of the straight line is proportional to the length, and the force arm is proportional to the distance to the green line.

Let the distance from the blue line to the center of the circle be x. The force arm is d-x, the green line is negative to the right force arm, and the sum of the left and right moments is zero.

There is c r arm * area = 0

c r (d-x)[ r 2-x 2)]dx=0 and then it depends on how well you are calculus.

c and r are known, and d can be found after integration.

Sector area = r n 360 ( is the pi, r is the radius of the sector, and n is the degree of the central angle. ）

The sector has 1 axis of symmetry.

Annular area = (r r ).

Fan ring area = (r r) n 360