Differentiating the angular momentum in time, we obtain the basic equation of the dynamics of rotational motion, known as Newton's second law for rotational motion, formulated as follows: the rate of change of the angular momentum L of a body rotating around a fixed point is equal to the resultant moment of all external forces M attached to the body, relative to this point:

dL /dt = M (14)

Since the angular momentum of a rotating body is directly proportional to the angular velocity  rotation, and the derivative d/ dt there is angular acceleration  , then this equation can be represented as

J = M (15)

where J - moment of inertia of the body.

Equations (14) and (15), describing the rotational motion of a body, are similar in content to Newton's second law for translational motion of bodies ( ma = F ). As you can see, during rotational motion as a force F the moment of force is used M , as an acceleration a - angular acceleration  , and the role of mass mcharacterizing the inertial properties of the body, the moment of inertia plays J.

Moment of inertia

The moment of inertia of a rigid body determines the spatial distribution of body mass and is a measure of the body's inertia during rotational motion. For a material point, or elementary mass  m i rotating around an axis, the concept of a moment of inertia was introduced, which is a scalar quantity numerically equal to the product of mass by the squared distance r i to the axis:

J i = r i 2 m i (16)

The moment of inertia of a volumetric solid is the sum of the moments of inertia of its constituent elementary masses:

For a homogeneous body with uniformly distributed density  \u003d  m i /V i (V i - elementary volume) can be written:

or, in integral form (the integral is taken over the entire volume):

J =  ∫ r 2 dV (19)

Using equation (19) allows you to calculate the moments of inertia of homogeneous bodies of various shapes with respect to any axes. The simplest result, however, is obtained by calculating the moments of inertia of homogeneous symmetric bodies about their geometric center, which in this case is the center of mass. The moments of inertia of some bodies of regular geometric shape with respect to the axes passing through the centers of mass, calculated in this way, are shown in Table 1.

The moment of inertia of a body relative to any axis can be found by knowing the body's own moment of inertia, i.e. moment of inertia about an axis passing through its center of mass, using Steiner's theorem. According to her, the moment of inertia Jabout an arbitrary axis is equal to the sum of the moment of inertia J 0 relative to the axis passing through the center of mass of the body parallel to the axis under consideration and the product of the body mass m per square distance r between axles:

J = J 0 + mr 2 (20)

The axis, when the body rotates around which, does not arise a moment of force that tends to change the position of the axis in space, is called the free axis of the given body. A body of any shape has three mutually perpendicular free axes passing through its center of mass, which are called the main axes of inertia of the body. The intrinsic moments of inertia of a body relative to the main axes of inertia are called the main moments of inertia.

Table 1.

Moments of inertia of some homogeneous bodies (with mass m) of the correct geometric shape with respect to the axes passing through the centers of mass

Body	Axis location (indicated by an arrow)	Moment of inertia
Ball radius r		2mr 2/5 (f1)
Radius hoop r		mr 2 (f2)
Radius disc r with a thickness that is negligible compared to the radius		mr 2/4 (ф3)
		mr 2/2 (ф4)
Solid cylinder of radius rwith height l		mr 2/2 (ф5)
		mr 2 /4 + ml 2/12 (ф6)
Hollow cylinder with inner radius r and wall thickness d		m [(r+ d) 2 + r 2] / 2 (f7)
Thin rod long l		ml 2/12 (f8)
Rectangular parallelepiped with sides a, b and c		m(a 2 + b 2) / 2 (ф9)
Cube with an edge length a		ma 2/6 (ф10)

Description of installation and measuring principle:

The installation used in this work to study the basic laws of the dynamics of the rotational motion of a rigid body around a fixed axis is called the Oberbeck pendulum. The general view of the installation is shown in Figure 4.

ABOUT the main element of the installation, which carries out a rotational movement around an axis perpendicular to the plane of the drawing, is the cross 1 consisting of four screwed into a pulley 2 rods (spokes) at right angles to each other, each of which is equipped with a cylindrical weight freely movable along the rod 3 mass secured in position with a screw 4 ... Along the entire length of the spokes with centimeter intervals, transverse cuts are applied, with which you can easily count the distances from the center of the weight location to the axis of rotation. By moving weights, a change in the moment of inertia is achieved J the entire crosspiece.

The rotation of the cross occurs under the action of the tension force (elastic force) of the thread 5 , fixed at one end in any one of the two pulleys ( 6 , or 7 ), on which it is wound when the cross is rotated. The other end of the thread with a weight attached to it P 0 8 variable mass m 0 is thrown over a fixed block 9 , which changes the direction of the rotational tension force coinciding with the tangent to the corresponding pulley. The use of one of two pulleys, differing in radii, allows you to change the shoulder of the rotating force, and, consequently, its moment M.

Verification of various patterns of rotational motion in this work is reduced to measuring time t lowering the load from a height h.

To determine the height of lowering the load on the Oberbeck pendulum, a millimeter scale is used 10 attached to an upright post 11 ... The quantity h corresponds to the distance between the risks, one of which is marked on the upper movable bracket 12 and the other is on the bottom bracket 13 secured motionless in a rack 11 ... The movable bracket can be moved along the rack and fixed in any desired position, setting the height of the lowering of the load.

Automatic measurement of the time of lowering the load is carried out using an electronic millisecond watch, the digital scale of which 14 located on the front panel, and two photoelectric sensors, one of which 15 fixed to the top bracket and the other 16 - on the lower fixed bracket. Sensor 15 gives a signal to start the electronic stopwatch when the load starts to move from its upper position, and the sensor 16 when the load reaches the lower position, it gives a signal that stops the stopwatch, fixing the time t the distance covered by the load h, and at the same time includes located behind the pulleys 6 and 7 a brake electromagnet that stops the rotation of the cross.

A simplified diagram of the pendulum is shown in Figure 5.

For cargo P 0 there are constant forces: gravity mg and thread tension T, under the influence of which the load moves downward uniformly with acceleration a... Pulley radius r 0 by thread tension T rotates with angular acceleration , while the tangential acceleration a t extreme points of the pulley will be equal to the acceleration a descending load. Acceleration a and  are related by:

a = a t \u003d  r 0 (21)

If the time of lowering the load P 0 denote by t, and the path he traveled through h, then according to the law of uniformly accelerated motion at an initial speed equal to 0, the acceleration a can be found from the ratio:

a = 2h/t 2 (22)

Measuring the diameter with a caliper d 0 of the corresponding pulley on which the thread is wound, and calculating its radius r o, from (21) and (22) you can calculate the angular acceleration of the crosspiece rotation:

 = a/r 0 = 2h/(r 0 t 2) (23)

When the weight attached to the thread is lowered, moving uniformly, the thread unwinds and drives the flywheel into uniformly accelerated rotational motion. The force that causes the body to rotate is the tension force of the thread. It can be determined from the following considerations. Since, according to Newton's second law, the product of the mass of a moving body and its acceleration is equal to the sum of the forces acting on the body, then in this case, by suspended on a thread and descending with uniform acceleration a body mass m 0 there are two forces at work: body weight m 0 gdownward and the thread tension Tpointing up. Therefore, the relation holds:

m 0 a = m 0 g – T (24)

T = m 0 (g – a) (25)

Therefore, the torque will be equal to:

M = Tr 0 = (m 0 g – m 0 a)r 0 (26)

where r 0 is the radius of the pulley.

If we neglect the friction force of the disc on the axis of the cross, then we can assume that only the moment acts on the cross M thread tension forces T... Therefore, using Newton's second law for rotational motion (13), one can calculate the moment of inertia J crosses with weights rotating on it taking into account (16) and (19) according to the formula:

J = M/ = m 0 (g – a)r 0 2 t 2 /2h (27)

or, substituting the expression for a (15):

J = m 0 r 0 2 (t 2 g/2h – 1) (28)

The resulting equation (28) is accurate. At the same time, having done experiments to determine the acceleration of the movement of the load P 0, one can make sure that a << g, and therefore in (27) the value ( g – a), neglecting the value a, can be taken equal to g... Then expression (27) will take the form:

J = M/ = m 0 r 0 2 t 2 g/2h (29)

If the quantities m 0 , r 0 and h do not change in the course of the experiments, then there is a simple quadratic dependence between the moment of inertia of the crosspiece and the time of lowering the load:

J = Kt 2 (30)

where K = m 0 r 0 2 g/2h... Thus, measuring the time t lowering a weight m 0, and knowing the height of its lowering h, you can calculate the moment of inertia of the cross, consisting of the spokes, the pulley in which they are fixed, and the weights located on the cross. Formula (30) allows you to check the basic laws of the dynamics of rotational motion.

If the moment of inertia of the body is constant, then different torques M 1 and M 2 will impart different angular accelerations ε 1 and ε 2 to the body, i.e. will have:

M 1 = Jε 1, M 2 = Jε 2 (31)

Comparing these expressions, we get:

M 1 /M 2 \u003d ε 1 / ε 2 (32)

On the other hand, the same torque will impart different angular accelerations to bodies with different moments of inertia. Really,

M = J 1 ε 1, M = J 2 ε 2 (33)

J 1 ε 1 \u003d J 2 ε 2, or J 1 /J 2 \u003d ε 1 / ε 2 (34)

Work order:

Exercise 1 . Determination of the moment of inertia of the cross and checking the dependence of the angular acceleration on the moment of the rotating force.

The task is performed with a crosspiece without weights put on it.

Select and set the height h lowering the load m 0 by moving the upper movable bracket 12 (height h can be set by the teacher). Value h enter in table 2.

Measure the diameter of the selected pulley with a caliper and find its radius r 0. Value r 0 enter in table 2.

By choosing the smallest mass value m 0, equal to the mass of the stand on which additional weights are put on, wind the thread on the selected pulley so that the load m 0 was raised to a height h... Measure three times the time t 0 lowering this weight. Record the data in table 2.

Repeat the previous experiment, for different (from three to five) masses m 0 of the descending weight, taking into account the weight of the stand on which the weights are put on. The weights of the stand and weights are indicated on them.

After each experiment, carry out the following calculations (entering their results in Table 2):

1. calculate the average time for lowering the load t 0 Wed and using it, by the formula (22) determine the linear acceleration of loads a... Points on the pulley surface move with the same acceleration;

   knowing the radius of the pulley r 0, by formula (23) find its angular acceleration ε;

   using the resulting linear acceleration a by the formula (26) find the torque M;

   based on the obtained values \u200b\u200bof ε and M calculate by the formula (29) the moment of inertia of the flywheel J 0 without weights on the rods.

Based on the results of all experiments, calculate and enter in table 2 the average value of the moment of inertia J 0, cf. ...

For the second and subsequent experiments, calculate, by entering the calculation results in Table 2, the ratios ε i / ε 1 and M i / M 1 (i is the number of the experiment). Check if the ratio is correct M i / M 1 \u003d ε 1 / ε 2.

According to table 2 for any one line, calculate the measurement errors of the moment of inertia using the formula:

 J = J 0 /J 0, cf. \u003d  m 0 /m 0 + 2r 0 /r 0 + 2t/t Wed +  h/h; J 0 =  J J 0, cf.

Values \u200b\u200bof absolute errors  r, t, h consider it to be equal to instrumental errors;  m 0 \u003d 0.5 g.

Table 2.

The plant parameters constant in this task used in the calculations:

r 0, m

m 0 , kg

t 0, s

t 0w. , from

a, m / s 2

J 0, kgm 2

J 0, cf. , kgm 2

J 0, kgm 2

M i / M 1

Assignment 2 . Checking the dependence of angular acceleration on the magnitude of the moment of inertia at a constant torque.

The crosspiece consists of four spokes (rods), four weights and two pulleys mounted on the axis of rotation. Since the masses of the pulleys are small and close to the axis of rotation, we can assume that the moment of inertia J of the entire crosspiece is equal to the sum of the moments of inertia of all rods (i.e. the moment of inertia of the crosspiece without weights J 0) and the moments of inertia of all weights on the rods J gr, i.e.

J = J 0 + J gr (35)

Then the moment of inertia of the weights about the axis of rotation is equal to:

J gr = J – J 0 (36)

Designating the moment of inertia of the crosspiece with weights located at a distance r 1 from the axis of rotation through J 1, and the corresponding moment of inertia of the loads themselves through J gr1, we rewrite (36) in the form:

J gr1 \u003d J 1 – J 0 (37)

Likewise for loads located at a distance r 2 from the axis of rotation:

J gr2 \u003d J 2 – J 0 (38)

Taking into account the approximate relation (30), we have:

J gr 1 \u003d Kt 1 2 – Kt 0 2 \u003d K(t 1 2 – t 0 2) and J gr 2 \u003d Kt 2 2 – Kt 0 2 \u003d K(t 2 2 – t 0 2) (39)

where t 1 - the time of lowering the load m 0 for the case when the weights on the rods are fixed at a distance r 1 from the axis of rotation; t 2 - time of lowering the load m 0 when securing loads on rods at a distance r 2 from the axis of rotation; t 0 - time of lowering the load m 0 when rotating the crosspiece without weights.

Hence it follows that the ratio of the moments of inertia of loads located at different distances from the axis of rotation is associated with the temporal characteristics of the process of lowering the load m 0 as:

J gr 1 / J gr 2 \u003d ( t 1 2 – t 0 2)/(t 2 2 – t 0 2) (40)

On the other hand, taking approximately 4 weights on the crosspiece as point masses m, we can assume that:

J gr 1 \u003d 4 mr 1 2 and J gr 2 \u003d 4 mr 2 2 , (41)

J gr1 / J gr2 \u003d r 1 2 /r 2 2 (42)

The coincidence of the right-hand sides of equations (40) and (42) could serve as an experimental confirmation of the presence of a direct proportional dependence of the moment of inertia of material points on the square of their distance to the axis of rotation. In fact, both relations (40) and (42) are approximate. The first of them was obtained under the assumption that the acceleration a lowering the load m 0 is negligible compared to gravitational acceleration g, and, in addition, in its derivation, the moment of friction forces of the pulleys on the axis and the moment of inertia of all pulleys relative to the axis of rotation are not taken into account. The second refers to point masses (i.e. masses of bodies whose dimensions can be neglected in comparison with their distance to the center of rotation), which cylindrical weights are not, and therefore, the further from the axis of rotation they are, the more accurately the relation (42 ). This can explain some discrepancy between the results obtained experimentally and theory.

To check the dependence (42), do the experiments in the following sequence:

Attach 4 weights to the rods closer to their ends at the same distance from the pulley. Determine and record in table 3 the distance r 1 from the axis of rotation to the centers of mass of the goods. It is determined by the formula: r 1 = r w + l + l c / 2, where r w is the radius of the pulley on which the rods are fixed, l - distance from load to pulley, l c is the length of the cylindrical load. Measure the pulley diameter and the length of the weights with a vernier caliper.

Measure three times the time t 1 lowering the load m 0 and calculate the average t 1w. ... Do the experiment for the same masses m 0, as in task 1. Record the data in table 3.

Shift the weights on the spokes to the center at an arbitrary distance, the same for all spokes r 2 < r 1 . Calculate this distance ( r 2) taking into account the comments in clause 1 and write it down in table 3.

Measure three times the time t 2 lowering the load m 0 for this case. Calculate the average t 2w. , repeat the experiment for the same masses m 0, as in item 2 and write down the obtained data in table 3.

Transfer from table 2 to table 3 values t 0w. obtained in the previous task for the corresponding values m 0 .

For all values m 0 using available averages t 0 , t 1 and t 2, using the formula (40), calculate the value bequal to the ratio of the moments of inertia of weights located at different distances from the axis of rotation: b= J group 1 / J gr. 2, and determine b Wed ... Record the results in Table 3.

Based on the data of any one line of Table 3, calculate the error allowed when determining the ratio (40), using the rules for finding errors in indirect measurements:

 b = b/b Wed \u003d 2 t (t 1 + t 0)/(t 1 2 – t 0 2) + 2t (t 2 + t 0)/(t 2 2 – t 0 2); b =  b b Wed

Calculate the value of the ratio r 1 2 /r 2 2 and write it down in table 3. Compare this ratio with the value b Wed and analyze some discrepancies within the margin of error of experience between the obtained results and theory.

Table 3.

m 0, kg

r 1m

t 1, s

t 1w. , from

r 2, m

t 2, s

t 2w. , from

t 0w. , from

r 1 /r 2

Assignment 3 . Checking the formulas for the moments of inertia of bodies of regular shape.

Theoretically calculated formulas for determining the proper moments of inertia of various uniform bodies of regular shape, i.e. moments of inertia about the axes passing through the centers of mass of these bodies are given in table 1. At the same time, using the experimental data obtained in tasks 1 and 2 (tables 2 and 3), it is possible to calculate the proper moments of inertia of such bodies of regular shape as weights, the crosses put on the rods, as well as the rods themselves, and compare the obtained values \u200b\u200bwith the theoretical values.

So, the moment of inertia of four weights located at a distance r 1 from the axis of rotation, can be calculated based on experimentally determined values t 1 and t 0 by the formula:

J gr1 = K(t 1 2 – t 0 2) (43)

Coefficient K in accordance with the notation introduced in (23) is

K = m 0 r 0 2 g/2h (44)

where m 0 - the mass of the descending load suspended on a thread; h - the height of its lowering; r 0 - radius of the pulley on which the thread is wound; g - acceleration of gravity ( g \u003d 9.8 m / s 2).

Considering the weights put on the spokes as homogeneous cylinders with a mass m c and taking into account the rule of additivity of moments of inertia, we can assume that the moment of inertia of one such cylinder rotating around an axis perpendicular to its axis of rotation and located at a distance r 1 from its center of mass is

J c1 \u003d K(t 1 2 – t 0 2)/4 (45)

According to Steiner's theorem, this moment of inertia is the sum of the moment of inertia of the cylinder relative to the axis passing through the center of mass of the cylinder perpendicular to its axis of rotation J μ0, and the values \u200b\u200bof the product m c r 1 2:

J c1 \u003d J c0 + m c r 1 2 (46)

J q 0 \u003d J C 1 - m c r 1 2 = K(t 1 2 – t 0 2)/4 – m c r 1 2 (47)

Thus, we have obtained a formula for the experimental determination of the proper moment of inertia of a cylinder relative to an axis perpendicular to its axis of rotation.

Similarly, the moment of inertia of the crosspiece, i.e. all spokes (rods) can be calculated using the formula:

J 0 = Kt 0 2 (48)

where the coefficient K is defined in the same way in the previous case.

For one rod, respectively:

J st \u003d Kt 0 2 /4 (49)

Using Steiner's theorem (here m st is the mass of the rod, r st is the distance from its middle to the axis of rotation and J st0 is the intrinsic moment of inertia of the rod relative to the axis perpendicular to it):

J st \u003d J st0 + m st r article 2 (50)

and considering that one of the ends of the rod is located on the axis of rotation, i.e. r st is half its length l st, we obtain a formula for the experimental determination of the moment of inertia of the rod relative to the axis perpendicular to it, passing through its center of mass:

J st0 \u003d J st - m st l st 2/4 \u003d ( Kt 0 2 – m st l st 2) / 4 (51)

To check the correspondence of the values \u200b\u200bof the proper moments of inertia of uniform bodies of regular shape, obtained experimentally and calculated theoretically, use the data of tasks 1 and 2 and perform the following operations:

In table 4, transfer the values \u200b\u200bfrom table 2 r 0 , h and m 0 .

For all, used in tasks 1 and 2, values m 0 calculate values K and write them down in table 4.

The values t 1w. and t 0w. from table 3 for the corresponding values m 0 transfer to table 4 (in columns t 1 and t 0).

Record in table 4 the weight of the cylinder weight m c (written on the load) and transfer into it from table 3 the value r 1 .

According to formula (47) for different values m 0 calculate the experimental values \u200b\u200bof the moment of inertia of the cylinder about the axis passing through the center of mass perpendicular to the axis of symmetry of the cylinder J c0 (e), and write them down in table 4. Calculate and write down the average J c0 (e-s) experimental value.

Measure the length with a caliper l q and diameter d ts of the load-cylinder. Write down 4 values \u200b\u200bin the table l c and r q \u003d d c / 2.

Using the values l c, r c, and m c, according to the formula (f6) from table 1, calculate J μ0 (t) is the theoretical value of the moment of inertia of the cylinder relative to the axis passing through the center of mass perpendicular to the axis of symmetry of the cylinder.

Measure the full length of the bar, bearing in mind that l st \u003d r w + lwhere r w is the radius of the pulley on which the rods are attached, and l - distance from the end of the rod to the pulley ( l st can be defined as half the measured distance between the ends of two oppositely directed rods). Write down the values l st and mass of the rod m st \u003d 0.053 kg in table 4.

According to formula (51) for different values m 0 calculate the experimental values \u200b\u200bof the moment of inertia of the bar about the axis passing through the center of mass perpendicular to the bar J st0 (e), and write them down in table 4. Calculate and write down the average J st0 (e-s) experimental value.

Using the values l st and m st, according to the formula (f8) from table 1, calculate J μ0 (t) is the theoretical value of the moment of inertia of the rod relative to the axis passing through the center of mass perpendicular to the rod.

Compare the experimentally and theoretically obtained values \u200b\u200bof the moments of inertia of the cylinder and the rod. Analyze the existing discrepancies.

Table 4.

For cylinder

For rod

J ts0 (e)

J c0 (e-c)

J c0 (t)

J st0 (e)

J st0 (e-s)

J st0 (t)

Test questions for preparation for work:

Formulate Newton's second law of rotational motion.

What is called the moment of inertia of an elementary mass and a rigid body? The physical meaning of the moment of inertia.

What is called the moment of force about a point and an axis of rotation? How to determine the direction of the vector of the moment of forces relative to a point?

What should be the relationship between the angular acceleration and the moment of the rotating force at a constant moment of inertia? How to check this dependence in practice?

How does the moment of inertia of a body depend on the distribution of mass in it or the distribution of mass in the system of rotating bodies? How to be convinced of this practically?

How to determine the moment of inertia of the crosspiece, the moment of inertia of rotating weights and spokes in the absence of friction?

Test questions for passing the test:

Output calculation formulas for all three tasks.

How the quantities  will change, J and M with a constant position of the weights on the spokes, if

a) increase the radius of the pulley r 0 with constant drop weight m 0 ?

b) increase m 0 at constant r 0 ?

How will the moment of inertia of the crosspiece with weights change if their distance from the axis of rotation is reduced by three times with a constant value m 0? Why?

What is the moment of inertia of the simplest bodies: a rod, a hoop, a disk.

Angular velocity and angular acceleration of a body: definition and meaning of these quantities.

EDUCATIONAL EDITION

Makarov Igor Evgenievich, professor, doctor of chemical sciences

Yurik Tamara Konstantinovna, Associate Professor, Candidate of Chemical Sciences

Study of the laws of rotation on the Oberbeck pendulum

(excluding frictional force)

Methodical instructions for laboratory work

Computer layout Skvortsov I.M.

Technical editor Kireev D.A.

Responsible for the issue R.V. Morozov

Offset paper. Printing on a risograph.

Service l. Circulation of copies Order

Information and publishing center MGUDT

A rigid body rotating around some axes passing through the center of mass, if it is freed from external influences, retains rotation indefinitely... (This conclusion is similar to Newton's first law of translational motion).

The occurrence of rotation of a rigid body is always caused by the action of external forces applied to individual points of the body. At the same time, the occurrence of deformations and the appearance of internal forces are inevitable, which ensure the practical preservation of its shape in the case of a solid body. When the action of external forces ceases, rotation is preserved: internal forces can neither cause nor destroy the rotation of a rigid body.

The result of the action of an external force on a body with a fixed axis of rotation is an accelerated rotational motion of the body... (This conclusion is similar to Newton's second law of translational motion).

The basic law of the dynamics of rotational motion: in the inertial reference frame, the angular acceleration acquired by a body rotating about a fixed axis is proportional to the total moment of all external forces acting on the body, and inversely proportional to the moment of inertia of the body relative to a given axis:

You can also give a simpler formulation the basic law of the dynamics of rotational motion (it is also called newton's second law for rotational motion): the torque is equal to the product of the moment of inertia and the angular acceleration:

Moment of impulse(moment of momentum, angular momentum) of a body is called the product of its moment of inertia by angular velocity:

The moment of impulse is a vector quantity. Its direction coincides with the direction of the angular velocity vector.

The change in angular momentum is determined as follows:

... (I.112)

A change in the angular momentum (with a constant moment of inertia of the body) can occur only due to a change in the angular velocity and is always caused by the action of the moment of force.

According to the formula, as well as formulas (I.110) and (I.112), the change in angular momentum can be represented as:

... (I.113)

The product in formula (I.113) is called moment of force or driving moment... It is equal to the change in angular momentum.

Formula (I.113) is valid provided that the moment of force does not change over time. If the moment of force depends on time, i.e. then

... (I.114)

Formula (I.114) shows that: the change in angular momentum is equal to the time integral of the moment of force... In addition, if this formula is presented in the form:, then the definition will follow from it moment of force: instantaneous moment of force is the first time derivative of the angular momentum,

Physics

The law of conservation of angular momentum. Equilibrium conditions for bodies

Newton's law for rotational motion. Newton's second law for a particle moving under the action of a force F, can be written as:

Where p \u003d mv - particle momentum. We multiply this equation vector by the radius vector of the particle r. Then

(18.1)

We now introduce new quantities - angular momentum L \u003d r p and moment of power N \u003d r F... Then the resulting equation takes the form:

For a particle moving in a circular motion in the plane (x, y), the angular momentum vector is directed along the axis z (i.e. along the angular velocity vector w) and is equal in modulus

(18.3)

Let's introduce the notation: I \u003d m r 2... The quantity I is called the moment of inertia of a material point relative to the axis passing through the origin. For a system of points rotating around an axis z with the same angular velocity, one can generalize the definition of the moment of inertia by taking the sum of the moments of inertia of all points relative to the common axis of rotation: I \u003d a m i r i 2... Using the concept of an integral, one can also define the moment of inertia of an arbitrary body about the axis of rotation. In any case, we can write that the angular momentum vector of a system of points or a body rotating with the same angular velocity around a common axis is

Then the equation of motion of a body rotating around a certain axis takes the form:

Here is a moment of power N - a vector directed along the axis of rotation and equal in magnitude to the product of the modulus of the force by the distance along the perpendicular from the point of application of the force to the axis of rotation (shoulder of the force).

Conservation of angular momentum in the field of central forces. If the force acting on a body from another body (located at the origin) is always directed along the radius vector rconnecting these bodies, it is called the central force. In this case, the cross product r F is equal to zero (as the cross product of collinear vectors). Therefore, the moment of force is equal to zero N and the equation of rotational motion takes the form dL / dt \u003d 0... This implies that the vector L does not depend on time. In other words, in the field of central forces the angular momentum is conserved.

The statement proved for one particle can be extended to a closed system containing an arbitrary number of particles. Thus, in a closed system, where central forces act, the total angular momentum of all particles is conserved.

So, in an arbitrary closed conservative mechanical system, there are generally seven conserved quantities - energy, three components of momentum and three components of angular momentum, which have the property that for a system of particles the values \u200b\u200bof these quantities represent the sum of values \u200b\u200btaken for individual particles. In other words, the total energy of the system is equal to the sum of the energies of individual particles, etc.

Statics. The section of mechanics that studies the conditions of equilibrium of extended, absolutely rigid bodies is called statics. The body is called absolutely solidif the distance between any pair of its points is unchanged. By definition, a body is in a state of static equilibrium if all points of the body are at rest in a certain inertial frame of reference.

The first equilibrium condition in IFR: the sum of all external forces applied to the body is zero.

In this case, the acceleration of the center of inertia (center of mass) of the body is equal to zero. You can always find such a frame of reference in which the center of inertia is at rest.

However, this condition does not mean that all points of the body are at rest. They can take part in a rotational movement around an axis. Therefore, there is a second equilibrium condition in IFR: the sum of the moments of all external forces about any axis is zero.

1. Write the basic equation for the dynamics of rotational motion (Newton's 2nd law for rotational motion).

This expression is called the basic equation of the dynamics of rotational motion and is formulated as follows: change in the angular momentum of a rigid body, is equal to the momentum of all external forces acting on this body.

2. What is the moment of force? (formula in vector and scalar form, pictures).

Moment strength(synonyms:torque; torque; torque ) - physical quantitycharacterizing the rotational actionforces on a solid.

Moment of force - vector quantity (М̅)

(vector view) М̅ \u003d | r̅ * F̅ |, r is the distance from the axis of rotation to the point of application of the force.

(kind of scalar) | M | \u003d | F | * d

The vector of the moment of force coincides with the axis О 1 О 2, its direction is determined by the right screw direction.
The moment of force is measured in newton meters... 1 Nm - moment of force which produces a force of 1 N on a 1 m long arm.

3. What is called a vector: rotation, angular velocity, angular acceleration. Where are they directed, how to determine this direction in practice?

Vectors Are pseudovectors or axial vectors that do not have a definite point of application: they are plotted on the axis of rotation from any of its points.

1. 
   Angular movement is a pseudo-vector, the modulus of which is equal to the angle of rotation, and the direction coincides with the axis around which the body rotates, and is determined by the rule of the right screw: the vector is directed in the direction from which the rotation of the body is visible counterclockwise (measured in radians)
2. 
   Angular velocity - a value characterizing the speed of rotation of a rigid body, equal to the ratio of the elementary angle of rotation and the elapsed time dt, for which this rotation took place.

Angular velocity vector is directed along the axis of rotation according to the right-hand screw rule, just like the vector.

1. 
   Angular acceleration - a value characterizing the speed of movement of the angular velocity.

The vector is directed along the axis of rotation towards the vector with accelerated rotation and opposite to the vector with decelerated rotation.

4. How does a polar vector differ from an axial one?

Polarvectorpossesses a pole, andaxial- no.

5. What is called the moment of inertia of a material point, a rigid body?

Moment inertiais a quantity characterizing the measure inertia material points when it rotates around the axis. Numerically, it is equal to the product of the mass by the square of the radius (distance to the axis of rotation).For solid body moment of inertia is equal to the sum of the moments of inertia its parts, and therefore can be expressed in integral form:

I \u003d ∫ r 2 dү.

6. What parameters does the moment of inertia of a rigid body depend on?

1. 
   From body weight
2. 
   From geometric dimensions
3. 
   From the choice of the axis of rotation

7. Steiner's theorem (illustrative figure).

Theorem: the moment of inertia of a body about an arbitrary axis is equal to the sum of the moment of inertia of this body about the axis parallel to it, passing through the center of mass of the body, and the product of the body mass by the square of the distance between the axes:
- required moment of inertia about a parallel axis

Known moment of inertia about an axis passing through the body's center of mass

Body mass

- distance between specified axes

8. Moment of inertia of a ball, cylinder, rod, disk.

Moment of inertia m.t. relative to the pole is called a scalar value equal to the product of this mass. points per square of the distance to the pole.

Moment of inertia m.t. can be found by the formula

The axis goes through the center of the ball

Cylinder axis

The axis is perpendicular to the cylinder and passes through its center of mass
9.How to determine the direction of the moment of force?

The moment of force relative to some point is the vector product strength on shortest distance from this point to the line of action of the force.

[M] = Newton meter

M - moment of force (Newton meter), F - Applied force (Newton), r - distance from the center of rotation to the place of application of the force (meter), l - the length of the perpendicular lowered from the center of rotation to the line of action of the force (meter), α - angle between the force vector F and position vector r

M = F l = F r sin(α )

(m, F, r-vector quantities)

Moment of power - axial vector... It is directed along the axis of rotation. The direction of the vector of the moment of force is determined by the gimlet rule, and its magnitude is M.
10. How do the moment of forces, angular velocities, moments of impulse add up?

Moment of power

If several forces act simultaneously on a body that can rotate around a point, then the rule of adding the moments of forces should be used to add the moments of these forces.

The rule of addition of the moments of forces says - The resulting vector of the moment of force is equal to the geometric sum of the components of the vectors of moments with

For the rule of addition of moments of forces, two cases are distinguished

1. The moments of forces lie in the same plane, the axes of rotation are parallel... Their sum is determined by algebraic addition. Right-hand screws are included in the amount with a sign minus... Left-hand screws - with a sign a plus

2. The moments of forces lie in different planes, the axes of rotation are not parallel... The sum of the moments is determined by geometric addition of vectors.

Angular Velocities

Angular velocity (rad / s) is a physical quantity that is an axial vector and characterizes the speed of rotation of a material point around the center of rotation. The angular velocity vector is equal in magnitude to the angle of rotation of a point around the center of rotation per unit of time

is directed along the axis of rotation according to the rule of the gimbal, that is, in the direction into which the gimbal with a right-hand thread would be screwed if it rotated in the same direction.

Angular velocities are plotted on the axis of rotation and can add up if they are directed in one direction, in the opposite direction - they are subtracted

Moment of impulse

In the International System of Units (SI), impulse is measured in kilogram-meter per second (kg m / s).

The momentum of the pulse characterizes the amount of rotational movement. A quantity that depends on how much mass rotates, how it is distributed about the axis of rotation, and at what speed it rotates.

If there is a material point with mass, moving with speed and located at a point described by the radius vector, then the angular momentum is calculated by the formula:
where is the sign of the vector product

To calculate the angular momentum of a body, it must be broken into infinitely small pieces and vector sum their moments as moments of momentum of material points, that is, take the integral:
11.Formulate the law of conservation of total mechanical energy as applied to a body rotating around a fixed axis.
MgH \u003d (IоW ^ 2) / 2

potential energy is maximum at the starting point of the pendulum's motion. The potential energy MgH is converted into kinetic energy, which is maximum at the moment the pendulum lands on the ground.
Io-moment of inertia about the axis for one weight (we have 4 of them)

I \u003d 4Io \u003d 4ml ^ 2 (Io \u003d ml ^ 2)

hence

MgH \u003d 2ml ^ 2W ^ 2
12.Formulate the law of conservation of total mechanical energy in relation to a body rotating around a fixed axis.
The moment of impulse of a rotating body is directly proportional to the speed of rotation of the body, its mass and linear extent. The higher any of these values, the higher the angular momentum.

In mathematical representation, the angular momentum L a body rotating at an angular velocity ω , is equal L \u003d Iω, where the quantity Icalled moment of inertia

Moment of impulse of a rotating body

where is the body weight; - speed; - the radius of the orbit along which the body moves; - moment of inertia; Is the angular velocity of a rotating body.

The angular momentum conservation law:

- for rotary motion

13.What expression determines the work of the moment of forces

= FORCE_MOMENT * ANGLE

In the SI system, work is measured in Joules, moment of force in Newton * meter, and ANGLE in radians

Usually the angular velocity in radians per second and the time of action of the MOMENT are known.

Then the WORK perfect by the MOMENT of force is calculated as:

= MOMENT OF POWER * *

14. Get the formula that determines the power developed by the moment of forces.
If the force performs an action at any distance, then it performs mechanical work. Also, if a moment of force performs an action across an angular distance, it does work.

= MOMENT_FORCE * ANGULAR_SPEED

In the SI system, power is measured in watts, moment of force in newton meters, and ANGULAR SPEED in radians per second.

15. Get the formula that determines the power developed by the moment of forces.

Forces and moments of forces act on the links of the mechanism, developing the corresponding powers. Thus, the power of all assigned forces consists of two parts:
,
where N R - power developed by forces applied at various points of the links performing translational or complex plane motion; N M - the power developed by the moments of forces applied to the rotating links.

Then , PowerN M calculated by the formula:
,
whereM k - the moment acting onk -e rotating links; w k - angular velocities of these links.
16. What is the kinetic energy of a rolling body?

During the rotational motion of a rolling body, each point participates in 2 movements - translational and rotational.

17. in my opinion the moment of force will increase / decrease by 2 times (direct dependence)

moment of inertia is the same
18. the moment of force will increase / decrease by 2 times

moment of inertia increase / decrease by 4 times

22. Why is the laboratory installation No. 4 called Oberbeck's PENDULUM?

A weight hangs on the back of the thread. Under the influence of gravity, this weight pulls the block. And because of this, the pendulum starts spinning. When the thread ends, stretches, and the load falls, the pendulum continues to rotate due to inertia until it stops. If the thread is fixed, then when it ends and is stretched, the pendulum continues to rotate by inertia, thus, the thread begins to wind again, and the load, accordingly, rises. And then it will stop and start descending again. And this process of raising and lowering is the meaning of the pendulum.
23. How does the inclusion of friction forces affect the result of measuring the moment of inertia of the Oberbeck pendulum? In which case is the measured value of the moment of inertia of the Oberbeck pendulum higher (with or without friction forces)? Justify the answer.

If we take into account the friction force, then the equation looks like this: .

That is, (if we deduce I from this formula) the friction force helps to reduce the moment of inertia of a rigid body. Consequently, the measured value of the moment of inertia of the pendulum without taking into account the friction forces will be greater than taking them into account.

24. What forces act on the falling weight of the Oberbek pendulum? What are they equal to.

For cargo actshim powergravity ([mg] \u003d 1 Newton) and powerthread tension ([T] \u003d 1 Newton).

The load in the downward direction is acted upon by the force of gravity F

where m is the mass of the load, and g is the acceleration of gravity (9.8 m / (s ^ 2).

Since the load is motionless, and besides the force of gravity and the tension force of the thread, other forces do not act on it, then, according to Newton's second law, T \u003d Ftyach \u003d mg, where T is the tension force of the thread.

If the load moves uniformly, that is, without acceleration, then T is also equal to mg according to Newton's first law.

If the load with mass m moves downward with acceleration a.

Then, according to Newton's second law, Ftyazh-T \u003d mg-T \u003d ma. Thus, T \u003d mg-a.
25. A man stands in the center of the rotating platform (carousel). How will the rotation speed of the platform change if a person moves to the edge of the platform?

The vector of the (instantaneous) velocity of any point of an (absolutely) rigid body rotating with angular velocity is determined by the formula:

where is the radius vector to a given point from the origin located on the axis of rotation of the body, and the square brackets denote the cross product.

Linear velocity (coinciding with the modulus of the velocity vector) of a point at a certain distance (radius)from the axis of rotation can be considered as follows:

Therefore, the greater the distance, the greater the speed. This means the carousels will spin faster.
26. The hoop and the solid cylinder have the same masses and radii. Determine their kinetic energies if they are rolling at the same speeds.

Kinetic energyrotary motion - energy body associated with its rotation.

For absolutely solidthe total kinetic energy can be written as the sum of the kinetic energy of translational and rotational motion:

Axial moments of inertia

Cylinder

Speed \u200b\u200b\u003d R * ω

In the photo, the formulas W are the formulas of T. Found them to. Energy and the ratio of energies.
27. What is the moment of force if the direction of action of the force is: a / perpendicular to the axis of rotation, b / parallel to the axis of rotation, c / passes through the axis of rotation.
A. M \u003d +/- Fh

B. The moment of force about the axis is zero if the force is parallel to the axis. In this case, the projection of the force on the plane perpendicular to the axis is equal to zero.

B. The moment of force about an axis is zero if the line of action of the force intersects this axis. In this case, the line of action of the force on the plane perpendicular to the axis passes through the point of intersection of the axis with the plane and, therefore, the shoulder of the force relative to point O is equal to zero.

28. ???

29. What is called the center of gravity of a rigid body?

Center of gravitya rigid body is called a point invariably associated with this bodyFROM, through which the line of action of the resultant of the gravity forces of the given body passes, for any position of the body in space.

30. In what two ways can you change the moment of force driving the Oberebek pendulum into rotation?

31. In what two ways can the moment of force be changed without changing the point of application of the force?

Change force value or radius

32. What formula can theoretically calculate the total moment of inertia of the weights on the spokes of the Oberbeck pendulum? Explain the values \u200b\u200bincluded in it.

− weighti -th material point

- the distance of the material point to the considered axis

33. Specify the direction of the vector of angular acceleration of a rotating body with a fixed axis of rotation relative to the vector of angular velocity.

When the body rotates around a fixed axis, the angular acceleration vector is directed along the rotation axis towards the vector of the elementary increment of the angular velocity. With accelerated motion, the vectorE co-directional with vectorW, in slow motion - opposite to it.

E - vector of angular acceleration

W Is the angular velocity vector

34. Using the measurement data, calculate the work of the frictional forces during the rotation of the Oberbeck pendulum at the moment the falling weight hits the floor.
35. Using the measurement data, calculate the kinetic energy of rotation of the Oberbeck pendulum at the moment the falling weight hits the floor.

E vr - kinetic energy of a rotating flywheel with a load.

I is the moment of inertia of the flywheel (together with weights);  is the angular velocity of rotation of the flywheel at the moment the weight hits the floor.

36. Using the measurement data, calculate the potential energy of the falling weight of the Oberbeck pendulum before the system starts to move.

m is the mass of the load, h is its height above the floor level

37. What is called a "pair of forces", write the formula, determine the moment of the "pair of forces", where is it directed?

A pair of forces is a system of two equal in magnitude, opposite in direction and not lying on one straight line.A pair of forces has a rotating effect, which can be estimated by the moment of the pair:

M (F 1, F 2) \u003d F 1 h \u003d F 2 h

where h is the shoulder of the pair, i.e. the distance between the lines of action of the forces of the pair.

The moment of the pair of forces M is perpendicular to the plane of action of the pair ( the plane in which the vectors of the pair of forces are located) and directed according to the rule of the right screw.The vector moment of a pair of forces can be applied at any point in space, i.e. is an free vector.

38. What types of energy does the potential energy of the falling weight transfer into when the Oberbeck pendulum rotates?

The potential energy of the falling weight is converted into the kinetic energy of the translational motion of this weight and the kinetic energy of the rotational motion of the pendulum.

39. What types of energy does the kinetic energy of the Oberbeck pendulum transfer into when it rotates?

Potential?

40. Draw the forces acting on the falling weight, what are they equal to? What is the nature of the movement of the falling weight?

T - thread tension, mg - gravity

The falling weight moves uniformly.

LITERATURE

Main

Sotskiy N.B. Biomechanics. - Minsk: BGUFK, 2005.

Nazarov V.T. Athlete movement. M., Polymya 1976

Donskoy D.D. Zatsiorsky V.M. Biomechanics: A Textbook for Institutes of Physical Culture.- M., Physical Culture and Sport, 1979.

Zagrevsky V.I. Biomechanics of physical exercise. Tutorial. - Mogilev: Moscow State University named after A.A. Kuleshova, 2002.

Additional

Nazarov V.T. Biomechanical stimulation: reality and hope.-Mn., Polymya, 1986.

Utkin V.L. Biomechanics of physical exercises. - M., Education, 1989.

Sotskiy N.B., Kozlovskaya O.N., Korneeva Zh.V. Laboratory course in biomechanics. Minsk: BGUFK, 2007.

Newton's laws for translational and rotational motions.

The formulation of Newton's laws depends on the nature of the motion of bodies, which can be represented as a combination of translational and rotational movements.

When describing the laws of the dynamics of translational motion, it should be borne in mind that all points of a physical body move in the same way, and to describe the laws of this movement, you can replace the entire body with one point containing the amount of substance corresponding to the entire body. In this case, the law of motion of the body as a whole in space will not differ from the law of motion of the specified point.

Newton's first law establishes the cause that causes the movement or changes its speed. This reason is the interaction of the body with other bodies. This is noted in one of the formulations of Newton's first law: "If other bodies do not act on a body, then it retains a state of rest or uniform rectilinear motion."

The measure of the interaction of bodies, as a result of which the nature of their movement changes, is force. Thus, if any physical body, for example the body of an athlete, has acquired acceleration, then the cause should be sought in the action of force from another body.

Using the concept of force, it is possible to formulate Newton's first law in another way: "If the body is not acted upon by forces, then it maintains a state of rest or uniform rectilinear motion."

Newton's second law establishes a quantitative relationship between the force of interaction of bodies and the acquired acceleration. So, during translational motion, the acceleration acquired by the body is directly proportional to the force acting on the body. The greater the indicated force, the more acceleration the body acquires.

To take into account the properties of interacting bodies that manifest themselves when imparting acceleration to them, a proportionality coefficient is introduced between force and acceleration, which is called the mass of the body. The introduction of mass allows us to write Newton's second law in the form:

a = -- (2.1)

where and - acceleration vector; F - force vector; m is body weight.

It should be noted that in the above formula, acceleration and force are vectors, therefore, they are not only related by proportional dependence, but also coincide in direction.

The mass of a body, introduced by Newton's second law, is associated with such a property of bodies as inertia. It is a measure of this property. The body's inertia is its ability to resist a change in speed. Thus, a body with a large mass and, accordingly, inertness is difficult to disperse and no less difficult to stop.

Newton's third law gives an answer to the question of how exactly bodies interact. He claims that when bodies interact, the force of action from one body to another is equal in magnitude and opposite in direction to the force acting from the other body on the first.

For example, a shot pusher, accelerating its projectile, acts on it with a certain force F, at the same time the same magnitude, but opposite in direction force acts on the athlete's hand and through it on the whole body as a whole. If this is not taken into account, the athlete may not keep within the throwing sector and the attempt will not be valid.

If a physical body interacts simultaneously with several bodies, all the acting forces are added according to the vector addition rule. In this case, the first and second Newton's laws mean the resultant of all forces acting on the body.

Dynamic characteristics of translational motion (force, mass).

The measure of the interaction of bodies, as a result of which the nature of their movement changes, is force. Thus, if any physical body, for example the body of an athlete, has acquired acceleration, then the cause should be sought in the action of force from another body. For example, when performing a high jump, the vertical speed of the athlete's body after taking off from the support until reaching the highest position decreases all the time. The reason for this is the force of interaction between the athlete's body and the earth - the force of gravity. In rowing, both the acceleration of the boat and the reason for its deceleration is the drag force of the water. In one case, it, acting on the hull of the boat, slows down the movement, and in the other, interacting with the oar, increases the speed of the vessel. As can be seen from the above examples, forces can act both at a distance and in direct contact of interacting objects.

It is known that the same force acting on different bodies leads to different results. For example, if a middleweight wrestler tries to push an opponent in his weight class and then a heavyweight athlete, then the accelerations acquired in both cases will differ markedly. Thus, the body of a middleweight opponent will acquire greater acceleration than in the case of a heavyweight opponent.

To take into account the properties of interacting bodies, manifested when imparting acceleration to them, a proportionality coefficient is introduced between the force and acceleration, which is called the mass of the body.

Strictly speaking, if different bodies are acted upon by the same force, then the fastest change in speed for the same period of time will be observed in the least massive body, and the slowest in the most massive.

Dynamic characteristics of rotary motion (moment of force, moment of inertia).

In the case of rotational motion of a body, the formulated laws of dynamics are also valid, but they use slightly different concepts. In particular, “force” is replaced by “moment of force” and “mass” is replaced by moment of inertia.

Moment of power is a measure of the interaction of bodies during rotational motion. It is determined by the product of the magnitude of the force by the shoulder of this force relative to the axis of rotation. The shoulder of the force is the shortest distance from the axis of rotation to the line of action of the force. So, when performing a large turn on the crossbar in the situation shown in Fig. 13, the athlete makes a rotational motion by gravity. The magnitude of the moment of force is determined by the force of gravity mg and the shoulder of this force relative to the axis of rotation d. In the process of performing a large turn, the rotational action of gravity changes in accordance with the change in the magnitude of the force arm.

Figure: 13. The moment of gravity when performing a large turn on the bar

So, the minimum value of the moment of force will be observed in the upper and lower positions, and the maximum - when the body is close to horizontal. The moment of force is a vector. Its direction is perpendicular to the plane of rotation and is determined by the "gimbal" rule. In particular, for the situation shown in Fig., The vector of the moment of force is directed "from the observer" and has a "minus" sign.

In the case of plane motions, it is convenient to determine the sign of the moment of force from the following considerations: if the force acts on the shoulder, trying to turn it in the "counterclockwise" direction, then such a moment of force has a "plus" sign, and if "clockwise", then the sign "minus".

According to the first law of the dynamics of rotational motion, the body maintains a state of rest (with respect to rotational motion) or uniform rotation in the absence of forces acting on it or the total moment is zero.

Newton's second law for rotational motion is:

e = --- (2.2)

where e - angular acceleration; M - moment of power; J is the moment of inertia of the body.

According to this law, the angular acceleration of a body is directly proportional to the moment of force acting on it and inversely proportional to its moment of inertia.

Moment of inertia is a measure of the body's inertia during rotational motion. For a material point of mass m located at a distance r from the axis of rotation, the moment of inertia is defined as J \u003d mr 2 ... In the case of a rigid body, the total moment of inertia is defined as the sum of the moments of inertia of its constituent points and is found using a mathematical integration operation.

The main forces that take place during exercise.

The force of gravity of a body located near the earth's surface can be determined by the mass of the body m and the acceleration of gravity g:

F \u003d m g (2.30)

The force of gravity acting on a physical body from the side of the earth is always directed vertically downward and is applied in the general center of gravity of the body.

Support reaction force acts on a physical body from the side of the support surface and can be decomposed into two components - vertical and horizontal. In most cases, horizontal is the friction force, the patterns of which will be discussed below. The vertical reaction of the support is numerically determined by the following relationship:

R \u003d ma + mg (2.31)

where a is the acceleration of the center of mass of the body in contact with the support.

Friction force... The frictional force can manifest itself in two ways. This can be the frictional force that occurs when walking and running, as a horizontal reaction of the support. In this case, as a rule, the link of the body interacting with the support does not move relative to the latter, and the friction force is called the "frictional force". In other cases, there is a relative movement of the interacting links, and the resulting force is the friction-sliding force. It should be noted that there is a frictional force acting on a rolled object, for example, a ball or a wheel - rolling friction, however, the numerical ratios that determine the magnitude of such a force are similar to those in sliding friction, and we will not consider them separately.

The amount of friction-rest is equal to the amount of the applied force tending to move the body. This situation is most typical for bobsleigh. If the projectile being moved is at rest, then a certain force must be applied to start moving it. In this case, the projectile will begin to move only when this force reaches a certain limiting value. The latter depends on the state of the contacting surfaces and on the force of pressure of the body on the support.

When the shear force exceeds the limit value, the body begins to move, slide. Here, the sliding-friction force becomes somewhat less than the limiting value of static friction at which movement begins. In the future, it depends to some extent on the relative speed of surfaces moved relative to each other, but for most sports movements it can be considered approximately constant, determined by the following relationship:

where k is the coefficient of friction, and R is the normal (perpendicular to the surface) component of the support reaction.

Friction forces in sports movements, as a rule, play both positive and negative roles. On the one hand, without friction force, it is impossible to provide horizontal movement of the athlete's body. For example, in all disciplines related to running, jumping, sports games and martial arts, they seek to increase the coefficient of friction between sports shoes and the support surface. On the other hand, during competitions in skiing, ski jumping, luge, bobsleigh, downhill, the first task that ensures a high sports result is to reduce the amount of friction. Here, this is achieved by selecting the appropriate materials for the skis and sled runners or by providing appropriate lubrication.

The friction force is the basis for the creation of a whole class of training devices, for the development of specific qualities of an athlete, such as strength and endurance. For example, in some very common designs of bicycle ergometers, the friction force quite accurately sets the load for the trainee.

Environmental resistance forces... When performing sports exercises, the human body is always under the influence of the environment. This action can be manifested both in the difficulty of movement, and ensure the possibility of the latter.

The force acting from the side of the flow incident on a moving body can be represented as consisting of two terms. It - drag forcedirected in the direction opposite to the movement of the body, and liftacting perpendicular to the direction of travel. When performing sports movements, the resistance forces depend on the density of the medium r, the speed of the body V relative to the medium, the area of \u200b\u200bthe body S (Fig. 24), perpendicular to the incident flow of the medium and the coefficient C, which depends on the shape of the body:

F res \u003d СSrV 2 (2.33)

Figure: 24. The area perpendicular to the incident flow, which determines the magnitude of the force

resistance.

Elastic forces... Elastic forces arise when the shape (deformation) of various physical bodies changes, restoring the original state after the elimination of the deforming factor. The athlete meets such bodies when performing trampoline jumping, pole vaulting, when performing exercises with rubber or spring shock absorbers. The elastic force depends on the properties of the deformable body, expressed by the elastic coefficient K, and the magnitude of the change in its shape Dl:

F ex. \u003d - КDl (2.35)

The buoyancy force depends on the volume V of the body or its part immersed in the medium - air, water or any other liquid, the density of the medium r and the acceleration of gravity g.