Velocity and acceleration in spherical coordinates. Determination of speed using the coordinate method Determination of kinematic quantities

motion tasks

Let's use equation (4) and take its derivative with respect to time

In (8) for unit vectors there are projections of the velocity vector onto the coordinate axes

Projections of velocity onto coordinate axes are defined as the first time derivatives of the corresponding coordinates.

Knowing the projections, one can find vector module and its direction

, (10)

Determining speed using the natural method

motion tasks

Let the trajectory of a material point and the law of change of the curvilinear coordinate be given. Suppose, at t 1 point had
and the coordinate s 1 , and at t 2 – coordinate s 2. During
the coordinate has been incremented
, Then average speed points

.

To find the speed at a given time, let's go to the limit

,

. (12)

The velocity vector of a point in the natural way of specifying motion is defined as the first derivative with respect to time of the curvilinear coordinate.

Point acceleration

Under the acceleration of a material point understand a vector quantity that characterizes the rate of change in the velocity vector of a point in magnitude and direction over time.

Acceleration of a point using the vector method of specifying motion

Consider a point at two points in time t 1 (
) And t 2 (
), Then
- time increment,
- speed increment.

Vector
always lies in the plane of motion and is directed towards the concavity of the trajectory.

P od average acceleration of a point during  t understand the magnitude

. (13)

To find the acceleration at a given time, let's go to the limit

,

. (14)

The acceleration of a point at a given time is defined as the second derivative with respect to time of the radius vector of the point or the first derivative of the velocity vector with respect to time.

The acceleration vector is located in the contacting plane and is directed towards the concavity of the trajectory.

Acceleration of a point with the coordinate method of specifying motion

Let us use the equation for the connection between the vector and coordinate methods of specifying movement

And let's take the second derivative from it

,

. (15)

In equation (15) for unit vectors there are projections of the acceleration vector onto the coordinate axes

. (16)

Acceleration projections onto coordinate axes are defined as the first derivatives with respect to time from the velocity projections or as the second derivatives of the corresponding coordinates with respect to time.

The magnitude and direction of the acceleration vector can be found using the following expressions

, (17)

,
,
. (18)

Acceleration of a point using the natural method of specifying motion

P
Let the point move along a curved path. Let us consider its two positions at moments of time t (s, M, v) And t 1 (s 1, M 1, v 1).

In this case, the acceleration is determined through its projections on the axes of the natural coordinate system moving together with the point M. The axes are directed as follows:

M  - tangent, directed along the tangent to the trajectory, towards the positive distance reference,

M n– main normal, directed along the normal lying in the contacting plane, and directed towards the concavity of the trajectory,

M b– binormal, perpendicular to plane M  n and forms a right-hand triple with the first axes.

Since the acceleration vector lies in the touching plane, then a b = 0. Let's find the projections of acceleration onto other axes.

. (19)

Let's project (19) onto the coordinate axes

, (20)

. (21)

Let's draw the 1st axis through the point M parallel to the axes at point M and find the velocity projections:

Where  - the so-called angle of adjacency.

Substitute (22) into (20)

.

At  t 0  0, cos 1 then

. (23)

The tangential acceleration of a point is determined by the first time derivative of the velocity or the second time derivative of the curvilinear coordinate.

Tangential acceleration characterizes the change in the velocity vector in magnitude.

Let's substitute (22) into (21)

.

Multiply the numerator and denominator by s to get known limits

Where
(the first wonderful limit),

,
,

, Where  - radius of curvature of the trajectory.

Substituting the calculated limits into (24), we obtain

. (25)

The normal acceleration of a point is determined by the ratio of the square of the velocity to the radius of curvature of the trajectory at a given point.

Normal acceleration characterizes the change in the velocity vector in direction and is always directed towards the concavity of the trajectory.

Finally, we obtain the projections of the acceleration of the material point on the axis of the natural coordinate system and the magnitude of the vector

, (26)

. (27)

The movement of a point in space can be considered given if the laws of change of its three Cartesian coordinates x, y, z as a function of time are known. However, in some cases of spatial motion of material points (for example, in areas limited by surfaces of various shapes), the use of equations of motion in Cartesian coordinates is inconvenient, since they become too cumbersome. In such cases, you can choose other three independent scalar parameters $q_1,(\q)_2,\\q_3$, called curvilinear or generalized coordinates, which also uniquely determine the position of the point in space.

The speed of point M, when specifying its movement in curvilinear coordinates, will be determined in the form of a vector sum of velocity components parallel to the coordinate axes:

\[\overrightarrow(v)=\frac(d\overrightarrow(r))(dt)=\frac(\partial \overrightarrow(r))(\partial q_1)\dot(q_1)+\frac(\partial \ overrightarrow(r))(\partial q_2)\dot(q_2)+\frac(\partial \overrightarrow(r))(\partial q_3)\dot(q_3)=v_(q_1)\overline(e_1)+v_( q_2)\overline(e_2)\ +v_(q_3)\overline(e_3)\]

The projections of the velocity vector onto the corresponding coordinate axes are equal to: $v_(q_i)=\overline(v\ )\cdot \overline(e_i)=H_i\dot(q_i)\ \ ,\ \ i=\overline(1,3)$

Here $H_i=\left|(\left(\frac(\partial \overrightarrow(r))(\partial q_i)\right))_M\right|$ is a parameter called i-th coefficient Lame and is equal to the modulus of the partial derivative of the radius vector of the point along the i-th curvilinear coordinate calculated at a given point M. Each of the vectors $\overline(e_i)$ has a direction corresponding to the direction point movement the end of the radius vector $r_i$ as the i-th generalized coordinate increases. The velocity module in an orthogonal curvilinear coordinate system can be calculated from the dependence:

In the above formulas, the values ​​of derivatives and Lamé coefficients are calculated for the current position of point M in space.

The coordinates of a point in a spherical coordinate system are the scalar parameters r, $(\mathbf \varphi ),\ (\mathbf \theta )$, measured as shown in Fig. 1.

Figure 1. Velocity vector in a spherical coordinate system

System of equations of motion of a point in in this case has the form:

\[\left\( \begin(array)(c) r=r(t) \\ \varphi =\varphi (t \\ \theta =\theta (t \end(array) \right.\]

In Fig. Figure 1 shows the radius vector r drawn from the origin, angles $(\mathbf \varphi )$ and $(\mathbf \theta )$, as well as coordinate lines and axes of the system under consideration at an arbitrary point M of the trajectory. It can be seen that the coordinate lines $((\mathbf \varphi ))$ and $((\mathbf \theta ))$ lie on the surface of a sphere of radius r. This curvilinear coordinate system is also orthogonal. Cartesian coordinates can be expressed in terms of spherical coordinates like this:

Then the Lame coefficients: $H_r=1;\ \ H_(\varphi )=rsin\varphi ;\ \ H_0=r$ ; projections of the point's velocity on the axis of the spherical coordinate system $v_r=\dot(r\ \ );$ $v_(\theta )=r\dot(\theta )$; $\ v_(\varphi )=r\dot(\varphi )sin\theta $, and the magnitude of the velocity vector

Acceleration of a point in a spherical coordinate system

\[\overrightarrow(a)=a_r(\overrightarrow(e))_r+a_(\varphi )(\overrightarrow(e))_(\varphi )+a_(\theta )(\overrightarrow(e))_( \theta),\]

projections of the acceleration of a point on the axis of a spherical coordinate system

\ \

Acceleration module $a=\sqrt(a^2_r+a^2_(\varphi )+a^2_(\theta ))$

Problem 1

The point moves along the line of intersection of the sphere and the cylinder according to the equations: r = R, $\varphi $ = kt/2, $\theta $ = kt/2 , (r, $\varphi $, $\theta $ --- spherical coordinates ). Find the modulus and projections of the velocity of the point on the axis of the spherical coordinate system.

Let's find the projections of the velocity vector on the spherical coordinate axes:

Velocity modulus $v=\sqrt(v^2_r+v^2_(\varphi )+v^2_(\theta ))=R\frac(k)(2)\sqrt((sin)^2\frac(kt )(2)+1)$

Problem 2

Using the condition of problem 1, determine the acceleration modulus of the point.

Let's find the projections of the acceleration vector on the spherical coordinate axes:

\ \ \

Acceleration module $a=\sqrt(a^2_r+a^2_(\varphi )+a^2_(\theta ))=R\frac(k^2)(4)\sqrt(4+(sin)^2 \frac(kt)(2))$

Formulas for calculating the speed of a point, acceleration, radius of curvature of a trajectory, tangent, normal and binormal from given coordinates versus time. An example of solving a problem in which, using given equations of motion, it is necessary to determine the speed and acceleration of a point. The radius of curvature of the trajectory, tangent, normal and binormal are also determined.

Introduction

The conclusions of the formulas below and the presentation of the theory are given on the page “Kinematics of a material point”. Here we will apply the main results of this theory to the coordinate method of specifying the motion of a material point.

Let us have a fixed rectangular coordinate system with a center at a fixed point. In this case, the position of point M is uniquely determined by its coordinates (x, y, z). Coordinate method of specifying the movement of a point

- this is a method in which the dependence of coordinates on time is specified. That is, three functions of time are specified (for three-dimensional motion):

Determination of kinematic quantities
,
Knowing the dependence of coordinates on time, we automatically determine the radius vector of the material point M using the formula:

where are unit vectors (orts) in the direction of the x, y, z axes.
;
;
Differentiating with respect to time, we find the projections of velocity and acceleration on the coordinate axes:
;
.

.

Speed ​​and acceleration modules:
.
Tangential (tangential) acceleration is the projection of the total acceleration onto the direction of velocity:

Tangential (tangential) acceleration vector:
.
; .
Normal acceleration:
.

Unit vector in the direction of the main normal of the trajectory:
.
Radius of curvature of the trajectory:
.

.

Center of curvature of the trajectory:

Example of problem solution

Using the given equations of motion of a point, establish the type of its trajectory and, for a moment in time, find the position of the point on the trajectory, its speed, total, tangential and normal accelerations, as well as the radius of curvature of the trajectory.

Equations of motion of a point:
, cm;
, cm.

Solution

Determining the type of trajectory

We exclude time from the equations of motion. To do this, we rewrite them in the form:
; .
Let's apply the formula:
.
;
;
;
.

So, we got the trajectory equation:
.
This is the equation of a parabola with a vertex at a point and an axis of symmetry.

Because the
, That
;
.
or
;
;

In a similar way we obtain a constraint for the coordinate:
,
Thus, the trajectory of the point’s movement is the arc of a parabola
located at

And .

12
;
.

We determine the position of the point at the moment of time.

Determining the speed of a point
.
Differentiating the coordinates and with respect to time, we find the velocity components.
To differentiate, it is convenient to apply the trigonometry formula:
;
.

.
;
.
Then
.

We calculate the values ​​of the velocity components at the moment of time:

Speed ​​Module:
;
.

Determining the acceleration of a point
;
.
Differentiating the components of velocity and time, we find the components of the acceleration of the point.
.

We calculate the values ​​of the acceleration components at the moment of time:
.
Acceleration module:

Tangential (tangential) acceleration vector:
.
Tangential acceleration is the projection of the total acceleration onto the direction of velocity:

Unit vector in the direction of the main normal of the trajectory:
.

Since, the tangential acceleration vector is directed opposite to the speed.
; .
The vector and is directed towards the center of curvature of the trajectory.
The trajectory of a point is the arc of a parabola
Point speed: .

Point acceleration: ;

;
.
; ;
Radius of curvature of the trajectory: .
; ;
Determination of other quantities
; ;
When solving the problem we found:

vector and speed module:

vector and module of total acceleration:
.
tangential and normal acceleration:

.
radius of curvature of the trajectory: .

.
Let's determine the remaining quantities.
.
Unit vector in the tangent direction to the path:

.

Tangential acceleration vector:
; .
Normal acceleration vector:


.