The projections of the velocities of two points of a rigid body onto an axis passing through these points are equal to each other.
v A cos α = v B cos β.
Proof
Let us choose a rectangular fixed coordinate system Oxyz. Let's take two arbitrary points of a rigid body A and B. Let (x A , y A , z A ) And (x B , y B , z B )- coordinates of these points. When a rigid body moves, they are functions of time t. Differentiating with respect to time, we obtain projections of the velocities of the points.
,
.
Let us take advantage of the fact that when a rigid body moves, the distance | AB| between points remains constant, that is, does not depend on time t. Also constant is the square of the distance
.
Let's differentiate this equation with respect to time t, applying the rule for differentiating a complex function.
Let's shorten it by 2
.
(1)
Let's introduce the vector
.
Then the equation (1)
can be represented as a scalar product of vectors.
(2)
We carry out transformations.
;
(3)
.
By the scalar product property
,
.
Substitute in (3)
and reduce by | AB|.
;
Q.E.D.
Relative speed
Consider the movement of point B relative to point A. Let us introduce the relative speed of point B relative to A.
Then the equation (2)
can be rewritten in the form
.
That is, the relative speed is perpendicular to the vector drawn from point A to point B. Since point B is taken arbitrarily, the relative speed of any point on a rigid body is perpendicular to the radius vector drawn from point A. That is, relative to point A, the body undergoes rotational motion. The relative speed of body points is determined by the formula for rotational motion
.
Point A, relative to which motion is considered, is often called pole.
The absolute speed of point B relative to a fixed coordinate system can be written in the following form:
.
It is equal to the sum of the speed of translational motion of an arbitrary point A (pole) and the speed of rotational motion relative to pole A.
Example of problem solution
The task
Wheels 1 and 2 with radii R 1 = 0.15 m and R 2 = 0.3 m, respectively, are connected by hinges to a rod of 3 lengths | AB| = 0.5 m. Wheel 1 rotates with angular velocity ω 1 = 1 rad/s. For the position of the mechanism shown in the figure, determine the angular velocity ω 2 wheels 2. Take L = 0.3 m.
The solution of the problem
Point A moves in a circle radius R 1
around the center of rotation O 1
. The speed of point A is determined by the formula
V A = ω 1 R 1.
The vector is directed vertically (perpendicular to O 1 A).
Point B moves in a circle radius R 2
around the center of rotation O 2
. The speed of point B is determined by the formula
V B = ω 2 R 2.
From here
.
The vector is directed horizontally (perpendicular to O 2 B).
We are building right triangle ABC. We apply the Pythagorean theorem.
(m)
.
The cosine of the angle between the velocity vector and the straight line AB, in the direction of the vector, is equal to
.
By velocity projection theorem two points of a rigid body on a straight line we have:
V A cos α = V B cos β.
From here
.
Finding the angular velocity of wheel 2.
rad/s .
Uniform movement– this is movement at a constant speed, that is, when the speed does not change (v = const) and acceleration or deceleration does not occur (a = 0).
Straight-line movement- this is movement in a straight line, that is, the trajectory of rectilinear movement is a straight line.
Uniform linear movement- this is a movement in which a body makes equal movements at any equal intervals of time. For example, if we divide a certain time interval into one-second intervals, then with uniform motion the body will move the same distance for each of these time intervals.
The speed of uniform rectilinear motion does not depend on time and at each point of the trajectory is directed in the same way as the movement of the body. That is, the displacement vector coincides in direction with the velocity vector. In this case, the average speed for any period of time is equal to the instantaneous speed: v cp = v Speed of uniform rectilinear motion is a physical vector quantity equal to the ratio of the movement of a body over any period of time to the value of this interval t:
Thus, the speed of uniform rectilinear motion shows how much movement a material point makes per unit time.
Moving with uniform linear motion is determined by the formula:
Distance traveled in linear motion is equal to the displacement module. If the positive direction of the OX axis coincides with the direction of movement, then the projection of the velocity onto the OX axis is equal to the magnitude of the velocity and is positive:
V x = v, that is, v > 0 The projection of displacement onto the OX axis is equal to: s = vt = x – x 0 where x 0 is the initial coordinate of the body, x is the final coordinate of the body (or the coordinate of the body at any time)
Equation of motion, that is, the dependence of the body coordinates on time x = x(t), takes the form:
X = x 0 + vt If the positive direction of the OX axis is opposite to the direction of motion of the body, then the projection of the body’s velocity onto the OX axis is negative, the speed is less than zero (v x = x 0 - vt
Dependence of speed, coordinates and path on time
The dependence of the projection of the body velocity on time is shown in Fig. 1.11. Since the speed is constant (v = const), the speed graph is a straight line parallel to the time axis Ot.
Rice. 1.11. Dependence of the projection of body velocity on time for uniform rectilinear motion.
The projection of movement onto the coordinate axis is numerically equal to the area of the rectangle OABC (Fig. 1.12), since the magnitude of the movement vector is equal to the product of the velocity vector and the time during which the movement was made.
Rice. 1.12. Dependence of the projection of body displacement on time for uniform rectilinear motion.
A graph of displacement versus time is shown in Fig. 1.13. The graph shows that the projection of the velocity is equal to
V = s 1 / t 1 = tan α where α is the angle of inclination of the graph to the time axis. The larger the angle α, the faster the body moves, that is, the greater its speed (the longer the distance the body travels in less time). The tangent of the tangent to the graph of the coordinate versus time is equal to the speed: tg α = v
Rice. 1.13. Dependence of the projection of body displacement on time for uniform rectilinear motion.
The dependence of the coordinate on time is shown in Fig. 1.14. From the figure it is clear that
Tg α 1 > tan α 2 therefore, the speed of body 1 is higher than the speed of body 2 (v 1 > v 2). tg α 3 = v 3 If the body is at rest, then the coordinate graph is a straight line parallel to the time axis, that is, x = x 0
Rice. 1.14. Dependence of body coordinates on time for uniform rectilinear motion.
Speed is one of the main characteristics. It expresses the very essence of the movement, i.e. determines the difference that exists between a stationary body and a moving body.
The SI unit of speed is m/s.
It is important to remember that speed is a vector quantity. The direction of the velocity vector is determined by the movement. The velocity vector is always directed tangentially to the trajectory at the point through which the moving body passes (Fig. 1).
For example, consider the wheel of a moving car. The wheel rotates and all points of the wheel move in circles. The splashes flying from the wheel will fly along tangents to these circles, indicating the directions of the velocity vectors of individual points of the wheel.
Thus, speed characterizes the direction of movement of a body (direction of the velocity vector) and the speed of its movement (modulus of the velocity vector).
Negative speed
Can the speed of a body be negative? Yes maybe. If the speed of a body is negative, this means that the body is moving in the direction opposite to the direction of the coordinate axis in the chosen reference system. Figure 2 shows the movement of a bus and a car. The speed of the car is negative and the speed of the bus is positive. It should be remembered that when we talk about the sign of velocity, we mean the projection of the velocity vector onto the coordinate axis.
Uniform and uneven movement
In general, speed depends on time. According to the nature of the dependence of speed on time, movement can be uniform or uneven.
DEFINITION
Uniform movement– this is movement with a constant modulus speed.
In case of uneven movement we speak of:
Examples of solving problems on the topic “Speed”
EXAMPLE 1
| Exercise | The car covered the first half of the journey between two settlements at a speed of 90 km/h, and the second half at a speed of 54 km/h. Determine the average speed of the car. |
| Solution | It would be incorrect to calculate the average speed of a car as the arithmetic mean of the two indicated speeds. Let's use the definition of average speed: Since rectilinear uniform motion is assumed, the signs of the vectors can be omitted. Time spent by the car to cover the entire distance: where is the time spent on completing the first half of the path, and is the time spent on completing the second half of the path.
The total movement is equal to the distance between populated areas, i.e. . Substituting these ratios into the formula for average speed, we get: Let's convert the speeds in individual sections to the SI system: Then the average speed of the car is:
|
| Answer | The average speed of the car is 18.8 m/s |
(m/s)EXAMPLE 2
| Exercise | A car travels for 10 seconds at a speed of 10 m/s and then drives for another 2 minutes at a speed of 25 m/s. Determine the average speed of the car. |
| Solution | Let's make a drawing. |
3.1. Uniform motion in a straight line.
3.1.1. Uniform motion in a straight line- movement in a straight line with acceleration constant in magnitude and direction:
3.1.2. Acceleration()- a physical vector quantity showing how much the speed will change in 1 s.
In vector form:
where is the initial speed of the body, is the speed of the body at the moment of time t.
In projection onto the axis Ox:
where is the projection of the initial velocity onto the axis Ox, - projection of the body velocity onto the axis Ox at a point in time t.
The signs of the projections depend on the direction of the vectors and the axis Ox.
3.1.3. Projection graph of acceleration versus time.
With uniformly alternating motion, the acceleration is constant, therefore it will appear as straight lines parallel to the time axis (see figure):
3.1.4. Speed during uniform motion.
In vector form:
In projection onto the axis Ox:
For uniformly accelerated motion:
For uniform slow motion:
3.1.5. Projection graph of speed versus time.
The graph of the projection of speed versus time is a straight line.
Direction of movement: if the graph (or part of it) is above the time axis, then the body is moving in the positive direction of the axis Ox.
Acceleration value: the greater the tangent of the angle of inclination (the steeper it goes up or down), the greater the acceleration module; where is the change in speed over time
Intersection with the time axis: if the graph intersects the time axis, then before the intersection point the body slowed down (uniformly slow motion), and after the intersection point it began to accelerate in the opposite direction (uniformly accelerated motion).
3.1.6. Geometric meaning of the area under the graph in the axes
Area under the graph when on the axis Oy the speed is delayed, and on the axis Ox- time is the path traveled by the body.
In Fig. 3.5 shows the case of uniformly accelerated motion. The path in this case will be equal to the area of the trapezoid: (3.9)
3.1.7. Formulas for calculating path
| Uniformly accelerated motion | Equal slow motion |
|---|---|
| (3.10) | (3.12) |
| (3.11) | (3.13) |
| (3.14) | |
All formulas presented in the table work only when the direction of movement is maintained, that is, until the straight line intersects with the time axis on the graph of the velocity projection versus time.
If the intersection has occurred, then the movement is easier to divide into two stages:
before crossing (braking):
After the intersection (acceleration, movement in the opposite direction)
In the formulas above - the time from the beginning of movement to the intersection with the time axis (time before stopping), - the path that the body has traveled from the beginning of movement to the intersection with the time axis, - the time elapsed from the moment of crossing the time axis to this moment t, - the path that the body has traveled in the opposite direction during the time elapsed from the moment of crossing the time axis to this moment t, - the module of the displacement vector for the entire time of movement, L- the path traveled by the body during the entire movement.
3.1.8. Movement in the th second.
During this time the body will travel the following distance:
During this time the body will travel the following distance:
Then during the th interval the body will travel the following distance:
Any period of time can be taken as an interval. Most often with.
Then in 1 second the body travels the following distance:
In 2 seconds:
In 3 seconds:
If we look carefully, we will see that, etc.
Thus, we arrive at the formula:
In words: the paths traversed by a body over successive periods of time are related to each other as a series of odd numbers, and this does not depend on the acceleration with which the body moves. We emphasize that this relation is valid for
3.1.9. Equation of body coordinates for uniform motion
Coordinate equation
The signs of the projections of the initial velocity and acceleration depend on the relative position of the corresponding vectors and the axis Ox.
To solve problems, it is necessary to add to the equation the equation for changing the velocity projection onto the axis:
3.2. Graphs of kinematic quantities for rectilinear motion
3.3. Free fall body
By free fall we mean the following physical model:
1) The fall occurs under the influence of gravity:
2) There is no air resistance (in problems they sometimes write “neglect air resistance”);
3) All bodies, regardless of mass, fall with the same acceleration (sometimes they add “regardless of the shape of the body,” but we are considering the movement of only a material point, so the shape of the body is no longer taken into account);
4) The acceleration of gravity is directed strictly downwards and is equal on the surface of the Earth (in problems we often assume for convenience of calculations);
3.3.1. Equations of motion in projection onto the axis Oy
Unlike movement along a horizontal straight line, when not all tasks involve a change in direction of movement, in free fall it is best to immediately use the equations written in projections onto the axis Oy.
Body coordinate equation:
Velocity projection equation:
As a rule, in problems it is convenient to select the axis Oy in the following way:
Axis Oy directed vertically upward;
The origin coincides with the level of the Earth or the lowest point of the trajectory.
With this choice, the equations and will be rewritten in the following form:
3.4. Movement in a plane Oxy.
We considered the motion of a body with acceleration along a straight line. However, the uniformly variable motion is not limited to this. For example, a body thrown at an angle to the horizontal. In such problems, it is necessary to take into account movement along two axes at once:
Or in vector form:
And changing the projection of speed on both axes:
3.5. Application of the concept of derivative and integral
We will not provide a detailed definition of the derivative and integral here. To solve problems we need only a small set of formulas.
Derivative:
Where A, B and that is, constant values.
Integral:
Now let's see how the concepts of derivative and integral apply to physical quantities. In mathematics, the derivative is denoted by """, in physics, the derivative with respect to time is denoted by "∙" above the function.
Speed:
that is, the speed is a derivative of the radius vector.
For velocity projection:
Acceleration:
that is, acceleration is a derivative of speed.
For acceleration projection:
Thus, if the law of motion is known, then we can easily find both the speed and acceleration of the body.
Now let's use the concept of integral.
Speed:
that is, the speed can be found as the time integral of the acceleration.
Radius vector:
that is, the radius vector can be found by taking the integral of the velocity function.
Thus, if the function is known, we can easily find both the speed and the law of motion of the body.
The constants in the formulas are determined from the initial conditions - values and at the moment of time
3.6. Velocity triangle and displacement triangle
3.6.1. Speed triangle
In vector form with constant acceleration, the law of speed change has the form (3.5):
This formula means that a vector is equal to the vector sum of vectors and the vector sum can always be depicted in a figure (see figure).
In each problem, depending on the conditions, the velocity triangle will have its own form. This representation allows the use of geometric considerations in the solution, which often simplifies the solution of the problem.
3.6.2. Triangle of movements
In vector form, the law of motion with constant acceleration has the form:
When solving a problem, you can choose the reference system in the most convenient way, therefore, without losing generality, we can choose the reference system in such a way that, that is, we place the origin of the coordinate system at the point where the body is located at the initial moment. Then
that is, the vector is equal to the vector sum of the vectors and Let us depict it in the figure (see figure).
As in the previous case, depending on the conditions, the displacement triangle will have its own shape. This representation allows the use of geometric considerations in the solution, which often simplifies the solution of the problem.
1.2. Straight-line movement
1.2.3. Graphic calculation of kinematic quantities
Some kinematic characteristics of movement can be calculated graphically.
Definition of Projected Velocity
Using graphs of the dependence of the coordinate on time x (t) (or the distance traveled on time S (t)), you can calculate the corresponding velocity projection v x at a certain point in time (Fig. 1.11), for example t = t 1.
To do this you should:
1) mark on the time axis the indicated value of the moment of time t 1;
2) restore the perpendicular to the intersection with the graph x (t);
5) determine the projection of the velocity onto the Ox axis as the tangent of the tangent angle to the positive direction of the time axis:
v x (t 1) = tan α 1 .
It should be noted that the projection of velocity v x is
- positive if the tangent to the graph forms an acute angle with the direction of the t axis (see Fig. 1.11);
- negative if the tangent to the graph forms an obtuse angle with the direction of the t axis (Fig. 1.12).
In Fig. Figure 1.12 shows a graph of the coordinate versus time x (t). To determine the projection of velocity onto the Ox axis at time t 3, a perpendicular t = t 3 is drawn. At the point of intersection of the perpendicular with the dependence x (t) a tangent line is drawn. It forms an obtuse angle with the t axis. Therefore, the projection of the velocity v x onto the Ox axis at the indicated time is a negative value:
v x (t 3) = − | tan α 3 | .
Rice. 1.12
Definition of Acceleration Projection
Using the graph of the velocity projection versus time v x (t), you can calculate the acceleration projection a x on the corresponding axis at a certain point in time (Fig. 1.13), for example t = t 2.
To do this you should:
1) mark on the time axis the indicated value of the moment of time t 2;
2) restore the perpendicular to the intersection with the graph v x (t);
3) draw a tangent line to the graph at the point of its intersection with the perpendicular;
5) determine the projection of acceleration onto the Ox axis as the tangent of the tangent angle to the positive direction of the time axis:
a x (t 2) = tan α 2 .
It should be noted that the projection of acceleration a x is
- positive if the tangent to the graph forms an acute angle with the direction of the t axis (see Fig. 1.13);
Rice. 1.13
- negative if the tangent to the graph forms an obtuse angle with the direction of the t axis (Fig. 1.14).
Rice. 1.14
Explanation of the use of the algorithm. In Fig. Figure 1.14 shows a graph of the velocity projection versus time v x (t). To determine the projection of acceleration onto the Ox axis at time t 4, a perpendicular t = t 4 is drawn. At the point of intersection of the perpendicular with the dependence v x (t) a tangent line is drawn. It forms an obtuse angle with the t axis. Therefore, the projection of acceleration a x onto the Ox axis at the specified time is a negative value:
a x (t 4) = − | tg α 4 | .
Determination of distance traveled and displacement module (combination of uniform and uniformly accelerated motion)
Using the graph of the velocity projection as a function of time v x (t), you can calculate the distance traveled and travel module material point (body) for a certain period of time ∆t = t 2 − t 1 .
To calculate the specified characteristics using a graph containing sections only uniformly accelerated and uniform motion, it follows:
4) calculate the distance traveled S and the displacement module ∆r as sums:
∆r = S 1 + S 2 + ... + S n,
where S 1, S 2, ..., S n are the paths traversed by the material point in each of the sections of uniformly accelerated and uniform motion.
In Fig. Figure 1.15 shows the dependence of the velocity projection on time for a material point (body) moving uniformly accelerated in section AB, uniformly in section BC, uniformly accelerated in section CD, but with an acceleration different from the acceleration in section AB.
Rice. 1.15
In this case, the distance traveled S and the displacement module ∆r coincide and are calculated using the formulas:
S = S 1 + S 2 + S 3,
∆r = S 1 + S 2 + S 3,
where S 1 is the path traveled by a material point (body) in section AB; S 2 - path traveled on section BC; S 3 - path traveled in section CD; S 1 , S 2 , S 3 are calculated according to the algorithm given above.
Determination of distance traveled and displacement module (combination of uniform, uniformly accelerated and uniformly decelerated motion)
To calculate the indicated characteristics using the graph v x (t), containing sections of not only uniformly accelerated and uniform, but also equally slow movement, you should:
1) mark the specified time interval ∆t on the time axis;
2) restore perpendiculars from points t = t 1 and t = t 2 until they intersect with the graph v x (t);
4) calculate the distance traveled S as the sum:
S = S 1 + S 2 + ... + S n,
where S 1, S 2, ..., S n are the paths traversed by the material point in each of the sections;
5) calculate travel module as the difference between the total path traveled by the material point to the stopping point and the path traveled by the material point after stopping.
Explanation of the use of the algorithm. In Fig. Figure 1.16 shows the dependence of speed on time for a material point (body) moving uniformly accelerated in section AB, uniformly in section BC, uniformly slow in section CF.
Rice. 1.16
In the case when there is a section of uniformly slow motion (including a stopping point - point D), the distance traveled S and the displacement module ∆r do not coincide. The distance traveled is calculated using the formula
S = S 1 + S 2 + S 3 + S 4,
where S 1 is the path traveled by a material point (body) in section AB; S 2 - path traveled on section BC; S 3 - path traveled in section CD; S 4 - path traveled in section DF; S 1 , S 2 , S 3 , S 4 are calculated according to the algorithm given above; It should be noted that the value of S 4 is positive.
The displacement module is calculated using the formula
∆r = S 1 + S 2 + S 3 − S 4,
subtracting the path traveled by the material point (body) after the rotation.
Determination of modulus of speed change
From the graph of the projection of acceleration versus time a x (t) one can find speed change module∆v of a material point (body) for a certain time interval ∆t = t 2 − t 1 (Fig. 1.17).
To do this you should:
1) mark the specified time interval ∆t on the time axis;
2) restore perpendiculars from points t = t 1 and t = t 2 until they intersect with the graph a x (t);
4) calculate the modulus of change in speed for the specified time interval as an area.
Example 4. The graph of the projection of the velocity of the first body onto the Ox axis versus time is depicted by a straight line passing through the points (0; 6) and (3; 0), the second - through the points (0; 0) and (8; 4), where the velocity is given in meters per second, time - in seconds. How many times do the acceleration modules of the first and second bodies differ?
Solution. Graphs of velocity projections versus time for both bodies are shown in the figure.
The acceleration projection of the first body is defined as the tangent of the obtuse angle α 1 ; its module is calculated by the formula
| a x 1 | = | tan α 1 | = | tg (180 − α 3) | = 6 3 = 2 m/s 2.
The first body moves equally slow; the magnitude of its acceleration is a 1 = = 2 m/s 2.
The acceleration projection of the second body is defined as the tangent of the acute angle α 2 ; its module is calculated by the formula
a x 2 = tan α 2 = 4 8 = 0.5 m/s 2.
The second body moves with uniform acceleration; the magnitude of its acceleration is a 2 = 0.5 m/s 2.
The required ratio of the acceleration modules of the first and second bodies is equal to:
a 1 a 2 = 2 0.5 = 4 .
The acceleration of the first body is 4 times greater than the acceleration of the second body.
Example 5. The graph of the y-coordinate versus time for the first body is depicted as a straight line passing through the points (0; 0) and (5; 3), the second - through the points (3; 0) and (6; 6), where the coordinate is given in meters, time - in seconds. Determine the ratio of the modules of the velocity projections of the indicated bodies.
Solution. Graphs of the y-coordinate versus time for both bodies are shown in the figure.
The projection of the velocity of the first body is defined as the tangent of the angle α 1; its module is calculated by the formula
v y 1 = tan α 1 = 3 5 = 0.6 m/s.
The projection of the velocity of the second body is defined as the tangent of the angle α 2; its module is calculated by the formula
v y 2 = tan α 2 = 6 3 = 2 m/s.
Both velocity projections have a positive sign; therefore, both bodies move with uniform acceleration.
The ratio of the modules of the velocity projections of the indicated bodies is:
| v y 2 | | v y 1 | = 2 0.6 ≈ 3 .
The magnitude of the projection of the velocity of the second body is approximately 3 times greater than the magnitude of the projection of the velocity of the second body.
Example 6. The graph of the dependence of the speed of a body on time is depicted as a straight line passing through the points (0; 4.0) and (2.5; 0), where the speed is given in meters per second, time - in seconds. How many times is the distance traveled by the body greater than the module of displacement in 6.0 s of movement?
Solution. A graph of body speed versus time is shown in the figure. The stopping point τ rest = 2.5 s falls in the interval from 0 s to 6.0 s.
Therefore, the distance traveled is the sum
S = S 1 + S 2,
and the displacement module is the difference
| Δ r → | = | S 1 − S 2 | ,
where S 1 is the path traveled by the body during the time interval from 0 s to 2.5 s; S 2 is the path traveled by the body in a time interval from 2.5 s to 6.0 s.
We calculate the values of S 1 and S 2 graphically as the areas of the triangles shown in the figure:
S 1 = 1 2 ⋅ 4.0 ⋅ 2.5 = 5.0 m;
S 2 = 1 2 ⋅ (6.0 − 2.5) ⋅ 5.6 = 9.8 m.
Note: the value of speed v = 5.6 m/s at time t = 6.0 s is obtained from the similarity of triangles, i.e. from attitude
v 4.0 = 6.0 − 2.5 2.5 − 0 .
Let's calculate the distance traveled:
S = S 1 + S 2 = 5.0 + 9.8 = 14.8 m
and the amount of movement:
| Δ r → | = | S 1 − S 2 | = | 5.0 − 9.8 | = 4.8 m.
Let us find the required ratio of the distance traveled and the displacement module:
S | Δ r → | = 14.8 4.8 ≈ 3.1.
The distance traveled is approximately 3.1 times the displacement.
