Chapter 2 basic equations of particle dynamics 2

§ 11-2Differential Equations of Particle Motion

1.Vector-form motion differential equations

(1) Moving Differential Equations

The particle depends on several ForcesF1,F2 ,...,FNVector

Equations of the equation are

(10-3)

(2) Another Vector Form of the motion differential equation

(10-3a)

2.Projection of Differential Equations on Cartesian Axes

(1) projection of forces on Cartesian Axes

When calculating the actual problem, the application type (10-3a) is required.

Form of shadow.

Vector pathRThe projection on the Cartesian axis isX,Y,Z, ForceFThe projection on the Cartesian axis isFxi,Fyi,Fzi.

(2) Cartesian projection expression

Projection of formula (10-3a) on the Cartesian axis is

(10-4)

3.Projection of Differential Equations on natural Axes

(1) Full acceleration of points

Full acceleration of pointsAWithin the close surface of the tangent and the main normal, the projection of the point's acceleration on the subnormal is equal to zero, that is

NeutralizationNIt is the unit vector along the trajectory tangent and the main normal, as shown in 10-1.

Figure 10-1

(2) Projection on natural Shaft Systems

Type (10-3) the projection type on the natural shaft system is

(10-5)

FormulaFti,Fni,FbiIt is the projection of each force on the particle on the tangent, main normal, and subnormal, respectively, and is the curvature radius of the trajectory.

4.Two basic problems of particle dynamics

(1) first basic issues

The first basic problem is to know the motion of a particle and find the force acting on the particle.

For the first basic problem, you only need to find two derivatives of the known equations of motion of the particle to obtain the acceleration of the particle and import them into the operation of the particle.

The first type of basic problems can be solved by using dynamic differential equations.

(2) Basic Issues of the second category

The second basic problem is: known the force acting on the particle

The movement of a particle.

For the second type of basic problem, it is to solve the differential equation, that is

Use the forced function law to perform the integral, and determine the integral constant based on the specific motion conditions of the problem.

5.Example

Example10-1

The 10-2a crank linkage mechanism is shown. CrankOARotate at a uniform velocity,OA=R,AB= L, when compared to hourOIs the coordinate origin, SliderBThe equations of motion can be approximately written

If the slider quality isM, Ignoring friction and Connecting RodABQuality, when and when, connecting rodABStrength.

Figure 10-2

Solution:

This is the first basic problem of dynamics. Take the sliderBAs the research object, at that time, the stress.

SliderBEdgeXThe moving differential equation of the axis is

The slider specified by the questionBEquations of motion, which can be obtained through differential

At that time

And, available

ABRod tension.

At that time

However

Substitute, get

ABThe rod is under pressure.

Example10-2

Quality isMThe particle carries a chargeE, At speedV0 input strengthE=ACosKtIn the uniformly changing electric field, the initial velocity is perpendicular to the electric field intensity, as shown in 10-3.

The force action of the particle in the electric field. Known ConstantsA,KIgnore the gravity of the particle and try to find the trajectory of the quality point.

Figure 10-3

Solution:

This problem is a basic problem of the second category of dynamics. Initial particle fetch

Start positionOCreates a coordinate system for the origin.

The moving differential equations of the particle areXAxis andYProjection score on the Axis

Do not

()

According to the question, the initial condition of particle motion is: whenT= 0

Vx=V0,Vy= 0,X = y= 0

Therefore, the definite integral of the motion differential equation (a) is

Particle velocity is

(B)

(B) The fixed points are

The particle motion equation is

(C)

Remove time from particle motion equation (c)TTo obtain the trajectory equation.

The trajectory is a cosine curve ,.

Example10-3

A taper pendulum, as shown in Figure 10-4. QualityM= 1kg of the ball LengthL= M rope, the other end of the rope is tied to a fixed pointOAnd straight line with lead. For example, if the ball moves at a constant speed in the horizontal plane, the speed of the ball is obtained.VTension with ropesF.

Figure 10-4

Solution:

This problem is a hybrid problem between the first basic problem of dynamics and the second basic problem of dynamics. Small ball as the research particle. The force acting on the particle has gravityMGAnd rope TensionF.

The projection formula of the particle motion differential equation on the natural axis is

Because, you can understand

Rope Tension and tensionF.

This example shows that the projection to the natural axis can enable solving the two types of dynamics problems.

Example10-4

The Crusher drum radius isR, Rotate at a constant speed through the central horizontal axis, and the iron ball in the cylinder increases from the convex edges on the cylinder wall. In order to make the iron ball obtain the energy of crushing ore, the iron ball should fall down only when (10-5. Calculates the number of turns per minute on the drum.N.

Figure 10-5

Solution:

The iron ball is regarded as a particle. Iron Balls are subject to gravity in the rising process.MGLegal binding force on the wallFN. tangent binding forceF.

Projection of particle motion differential equations on the Primary Normal

The speed of the iron ball before leaving the cylinder wall is equal to the speed of the cylinder wall, that is

Therefore

At that time, the iron ball will fall.FN = 0, so

Obviously, the smaller the requirementNLarger. At that time, at this time, the iron ball will be close to the cylinder wall to turn the highest point without leaving the cylinder wall, can not afford Powder