- 1. Algebraic angular momentum about the center. Algebraic ABOUT-- scalar quantity taken with the sign (+) or (-) and equal to the product of the modulus of momentum m to a distance h(perpendicular) from this center to the line along which the vector is directed m:
- 2. Vector moment of momentum relative to the center.
Vector angular momentum material point relative to some center ABOUT -- vector applied at this center and directed perpendicular to the plane of vectors m And in the direction from which the point’s movement is visible counterclockwise. This definition satisfies the vector equality
Momentum material point relative to some axis z is a scalar quantity taken with the sign (+) or (-) and equal to the product of the modulus projection vector momentum per plane perpendicular to this axis perpendicular h, lowered from the point of intersection of the axis with the plane to the line along which the indicated projection is directed:
Kinetic moment of a mechanical system relative to the center and axis
1. Momentum relative to the center.
Kinetic moment or the main moment of the quantities of motion of a mechanical system relative to some center called geometric sum moments of momentum of all material points of the system relative to the same center.
2. Kinetic moment about the axis.
The kinetic moment or principal moment of the quantities of motion of a mechanical system relative to a certain axis is the algebraic sum of the moments of the quantities of motion of all material points of the system relative to the same axis.
3. Kinetic moment solid, rotating around fixed axis z with angular velocity.
Theorem on the change in angular momentum of a material point relative to the center and axis
1. Theorem of moments about the center.
Derivative in time from the moment of momentum of a material point relative to some fixed center is equal to the moment of force acting on the point relative to the same center
2. Theorem of moments about an axis.
Derivative in time from the moment of momentum of a material point relative to a certain axis is equal to the moment of force acting on the point relative to the same axis
Theorem on the change in the angular momentum of a mechanical system relative to the center and axis
Theorem of moments about the center.
Derivative in time from the kinetic moment of a mechanical system relative to some fixed center is equal to the geometric sum of the moments of all external forces, acting on the system, relative to the same center;
Consequence. If main point external forces relative to some center is zero, then the angular momentum of the system relative to this center does not change (the law of conservation of angular momentum).
2. Theorem of moments about an axis.
Derivative in time from the kinetic moment of a mechanical system relative to some fixed axis is equal to the sum of the moments of all external forces acting on the system relative to this axis
Consequence. If the main moment of external forces relative to a certain axis is zero, then the kinetic moment of the system relative to this axis does not change.
For example, = 0, then L z = const.
Work and power of forces
Work of force-- scalar measure of the action of force.
1. Elementary work of force.
Elementary the work done by a force is an infinitesimal scalar quantity equal to scalar product force vector to the vector of infinite small displacement of the force application point: ; - radius vector increment the point of application of force, the hodograph of which is the trajectory of this point. Elementary movement points along the trajectory coincides with due to their small size. That's why
if then dA > 0;if, then dA = 0;if , That dA< 0.
2. Analytical expression elementary work.
Let's imagine the vectors And d through their projections on the Cartesian coordinate axes:
, . We get (4.40)
3. The work done by a force on a final displacement is equal to the integral sum basic work on this move
If the force is constant and the point of its application moves linearly,
4. Work of gravity. We use the formula: Fx = Fy = 0; Fz = -G = -mg;
Where h- moving the point of application of force vertically downwards (height).
When moving the point of application of gravity upward A 12 = -mgh(dot M 1 -- down, M 2 - at the top).
So, . The work done by gravity does not depend on the shape of the trajectory. When moving along a closed path ( M 2 matches M 1 ) work is zero.
5. The work of the elastic force of the spring.
The spring stretches only along its axis X:
F y = F z = ABOUT, F x = = -сх;
where is the magnitude of the spring deformation.
When the point of application of force moves from the lower position to the upper position, the direction of force and the direction of movement coincide, then
Therefore, the work of the elastic force
Work of forces on final displacement; If = const, then
where is the final angle of rotation; , Where p -- the number of revolutions of a body around an axis.
Kinetic energy of a material point and a mechanical system. Koenig's theorem
Kinetic energy- scalar measure of mechanical motion.
Kinetic energy of a material point - a scalar positive quantity equal to half the product of the mass of a point and the square of its speed,
Kinetic energy of a mechanical system -- the arithmetic sum of the kinetic energies of all material points of this system:
Kinetic energy of a system consisting of n interconnected bodies is equal to arithmetic sum kinetic energies of all bodies of this system:
Koenig's theorem
Kinetic energy of a mechanical system in general, its motion is equal to the sum kinetic energy motion of the system together with the center of mass and kinetic energy of the system when it moves relative to the center of mass:
Where Vkc -- speed k- th points of the system relative to the center of mass.
Kinetic energy of a rigid body under various motions
Forward movement.
Rotation of a body around a fixed axis . ,Where -- moment of inertia of a body relative to the axis of rotation.
3. Plane-parallel motion. , where is the moment of inertia flat figure relative to an axis passing through the center of mass.
When moving flat body kinetic energy consists of the kinetic energy of the translational motion of the body with the speed of the center of mass and kinetic energy rotational movement around an axis passing through the center of mass, ;
Theorem on the change in kinetic energy of a material point
The theorem in differential form.
Differential from the kinetic energy of a material point is equal to the elementary work of the force acting on the point,
The theorem in integral (finite) form.
Change kinetic energy of a material point at a certain displacement is equal to the work of the force acting on the point at the same displacement.
Theorem on the change in kinetic energy of a mechanical system
The theorem in differential form.
Differential from the kinetic energy of a mechanical system is equal to the sum of elementary works of external and internal forces, acting on the system.
The theorem in integral (finite) form.
Change kinetic energy of a mechanical system at a certain displacement is equal to the sum of the work of external and internal forces applied to the system at the same displacement. ; For a system of solid bodies = 0 (according to the property of internal forces). Then
Law of conservation of mechanical energy of a material point and mechanical system
If for material point or mechanical system If only conservative forces act, then at any position of a point or system the sum of kinetic and potential energies remains constant.
For a material point
For mechanical system T+ P= const
Where T+ P -- total mechanical energy of the system.
Rigid body dynamics
Differential equations of motion of a rigid body
These equations can be obtained from general theorems of the dynamics of a mechanical system.
1. Equations of translational motion of a body - from the theorem on the movement of the center of mass of a mechanical system In projections on the axes of Cartesian coordinates
2. The equation for the rotation of a rigid body around a fixed axis - from the theorem on the change in the kinetic moment of a mechanical system relative to an axis, for example, relative to an axis
Since the kinetic moment L z rigid body relative to the axis, then if
Since or, the equation can be written as or, the form of writing the equation depends on what needs to be determined in a particular problem.
Differential equations of plane-parallel the motions of a rigid body are a set of equations progressive movement of a flat figure together with the center of mass and rotational movement relative to an axis passing through the center of mass:
Physical pendulum
Physical pendulum is a rigid body that rotates around a horizontal axis that does not pass through the center of mass of the body and moves under the influence of gravity.
Differential rotation equation
In case of small fluctuations.
Then where
Solution of this homogeneous equation.
Let at t=0 Then
-- equation of harmonic vibrations.
Period of oscillation of a pendulum
Given length a physical pendulum is the length of a mathematical pendulum whose period of oscillation equal to the period oscillations of a physical pendulum.
The direction and magnitude of the moment of momentum is determined in exactly the same way as in the case of estimating the moment of force (section 1.2.2).
At the same time we define ( main) angular momentum How vector sum moments of the number of movements of the points of the system under consideration. It also has a second name - kinetic moment :
Let us find the time derivative of expression (3.40) using the rules for differentiating the product of two functions, and also the fact that the derivative of a sum is equal to the sum of derivatives (i.e., the sign of the sum can be moved as a coefficient during differentiation):
.
Let us take into account the obvious kinematic equalities: 
. Then: 
. We use the average equation from formulas (3.26) 
, and also that vector product two collinear vectors ( and ) is equal to zero, we get:
Applying the property of internal forces (3.36) to the 2nd term, we obtain an expression for the theorem on the change in the main moment of momentum of a mechanical system:
 .
| (3.42) |
The time derivative of the kinetic moment is equal to the sum of the moments of all external forces acting in the system.
This formulation is often called briefly: moment theorem .
It should be noted that the theorem of moments is formulated in a fixed frame of reference relative to a certain fixed center O. If a rigid body is considered as a mechanical system, then it is convenient to choose the center O on the axis of rotation of the body.
One important property of the moment theorem should be noted (we present it without derivation). The theorem of moments is also true in a translationally moving reference system if the center of mass (point C) of the body (mechanical system) is chosen as its center:
The formulation of the theorem in this case remains practically the same.
Corollary 1
Let the right side of expression (3.42) be equal to zero =0, - the system is isolated. Then from equation (3.42) it follows that .
For an isolated mechanical system, the vector of the kinetic moment of the system does not change either in direction or in magnitude over time.
Corollary 2
If the right side of any of the expressions (3.44) is equal to zero, for example, for the Oz axis: =0 (partially isolated system), then from equations (3.44) it follows: =const.
Consequently, if the sum of the moments of external forces relative to any axis is zero, then the axial kinetic moment of the system along this axis does not change over time.
The formulations given above in the corollaries are the expressions law of conservation of angular momentum in isolated systems .
Momentum of a rigid body
Let's consider special case– rotation of a rigid body around the Oz axis (Fig. 3.4).
Fig.3.4
A point on a body separated from the axis of rotation by a distance h k, rotates in a plane parallel to Oxy at a speed of . In accordance with the definition of the axial moment, we use expression (1.19), replacing the projection F XY force on this plane by the amount of motion of the point 
. Let us estimate the axial kinetic moment of the body:
According to the Pythagorean theorem 
, therefore (3.46) can be written as follows:
 | (3.47) |
Then expression (3.45) will take the form:
| (3.48) |
If we use the law of conservation of angular momentum for a partially isolated system (Corollary 2) in relation to a solid body (3.48), we obtain . In this case, you can consider two options:
QUESTIONS FOR SELF-CONTROL
1. How is the angular momentum of a rotating rigid body determined?
2. How does the axial moment of inertia differ from the axial kinetic moment?
3. How does the rotation speed of a rigid body change over time in the absence of external forces?
Axial moment of inertia of a rigid body
As we will see later, the axial moment of inertia of a body has the same significance for the rotational motion of a body as the mass of the body during its forward movement. This is one of the the most important characteristics body, which determines the inertia of the body during its rotation. As can be seen from definition (3.45), this is a positive scalar quantity, which depends on the masses of the points of the system, but to a greater extent on the distance of the points from the axis of rotation.
For continuous homogeneous bodies of simple shapes, the value of the axial moment of inertia, as in the case of estimating the position of the center of mass (3.8), is calculated by the integration method, using the mass of an elementary volume instead of a discrete mass dm=ρdV:
 | (3.49) |
For reference, we present the values of the moments of inertia for some simple bodies:
m and length l relative to the axis passing perpendicular to the rod through its middle (Fig. 3.5).
Fig.3.5
The moment of inertia of a thin homogeneous rod with a mass m and length l relative to the axis passing perpendicular to the rod through its end (Fig. 3.6).
Fig.3.6
The moment of inertia of a thin homogeneous ring of mass m and radius R relative to the axis passing through its center perpendicular to the plane of the ring (Fig. 3.7).
Fig.3.7
The moment of inertia of a thin homogeneous disk with a mass m and radius R relative to the axis passing through its center perpendicular to the plane of the disk (Fig. 3.7).
Fig.3.8
· Moment of inertia of a body of arbitrary shape.
For bodies of arbitrary shape, the moment of inertia is written in the following form:
Where ρ - so-called radius of gyration body, or the radius of a certain conventional ring with mass m, the axial moment of inertia of which is equal to the moment of inertia of the given body.
Huygens–Steiner theorem
Fig.3.9
Let us associate two parallel coordinate systems with the body. The first Cx"y"z", with the origin at the center of mass, is called central, and the second Oxyz, with the center O, lying on the Cx" axis at a distance CO = d(Fig. 3.9). It is easy to establish connections between the coordinates of body points in these systems:
In accordance with formula (3.47), the moment of inertia of the body relative to the Oz axis:
Here the factors 2 are constant for all terms of the 2nd and 3rd sums of the right side d And d taken out of the corresponding amounts. The sum of the masses in the third term is the body mass. The second sum, in accordance with (3.7), determines the coordinate of the center of mass C on the axis Cx" (), and the equality is obvious: . Taking into account that the 1st term, by definition, is the moment of inertia of the body relative to the central axis Cz" (or Z C ) 
, we obtain the formulation of the Huygens-Steiner theorem:
 | (3.50) |
The moment of inertia of a body relative to a certain axis is equal to the sum of the moment of inertia of the body relative to a parallel central axis and the product of the mass of the body by the square of the distance between these axes.
QUESTIONS FOR SELF-CONTROL
1. Give formulas for axial moments inertia of a rod, ring, disk.
2. Find the radius of gyration of a round solid cylinder relative to its central axis.
Kinetic moment of a point and a mechanical system
Rice. 3.14
One of the dynamic characteristics of the motion of a material point and a mechanical system is the kinetic moment or angular momentum.
For a material point kinetic moment relative to any center O is called the angular momentum of a point relative to this center (Fig. 3.14),
The kinetic moment of a material point relative to an axis is the projection onto this axis of the kinetic moment of the point relative to any center on this axis:
The kinetic moment of a mechanical system relative to the center O is the geometric sum of the kinetic moments of all points of the system relative to the same center (Fig. 3.15):
(3.20)
The kinetic moment is applied to the point ABOUT, relative to which it is calculated.
If we project (3.20) onto the axes of the Cartesian coordinate system, we obtain projections of the kinetic moment onto these axes, or kinetic moments relative to the coordinate axes:
Let us determine the kinetic moment of the body relative to its fixed axis of rotation z(Fig. 3.16).
According to formulas (3.21), we have
But when the body rotates with angular velocity w, the speed 
and the amount of motion of the point 
perpendicular to the segment dk and lies in the plane perpendicular to the axis of rotation Oz, hence,
| Rice. 3.15 | Rice. 3.16 |
For the whole body:
Where Jz– moment of inertia relative to the axis of rotation.
Consequently, the angular momentum of a rigid body relative to the axis of rotation is equal to the product of the moment of inertia of the body relative to a given axis and the angular velocity of the body.
2. Theorem on the change in angular momentum
mechanical system
Kinetic moment of the system relative to the stationary center O(Fig. 3.15)
Let us take the derivative with respect to time from the left and right sides of this equality:
(3.22)
Let's take into account that 
then expression (3.22) will take the form
Or, given that 
– the sum of the moments of external forces relative to the center O, we finally have:
(3.23)
Equality (3.23) expresses the theorem about the change in angular momentum.
Theorem on the change in angular momentum. The time derivative of the kinetic moment of a mechanical system relative to a fixed center is equal to the principal moment of the external forces of the system relative to the same center.
Having projected equality (3.23) onto the fixed axes of Cartesian coordinates, we obtain a representation of the theorem in projections onto these axes:
From (3.23) it follows that if the main moment of external forces relative to any fixed center is zero, then the kinetic moment relative to this center remains constant, i.e. If
(3.24)
If the sum of the moments of the external forces of the system relative to any fixed axis is zero, then the corresponding projection of the kinetic moment remains constant,
(3.25)
Statements (3.24) and (3.25) represent the law of conservation of angular momentum of the system.
Let us obtain a theorem about the change in the kinetic moment of the system by choosing the point as a point when calculating the kinetic moment A, moving relative to the inertial reference frame with speed
Kinetic moment of the system relative to the point A(Fig. 3.17)
| Rice. 3.17 |
because 
That
Considering that 
where is the speed of the center of mass of the system, we obtain
Let's calculate the time derivative of the angular momentum
In the resulting expression:
Combining the second and third terms, and considering that 
finally we get
If the point coincides with the center of mass of the system C, That 
and the theorem takes the form
those. it has the same shape as for a fixed point ABOUT.
3. Differential equation of rotation of a rigid body
around a fixed axis
Let a rigid body rotate around a fixed axis Az(Fig. 3.18) under the influence of a system of external forces 
Let us write the equation of the theorem on the change in the angular momentum of the system in projection onto the axis of rotation:
| Rice. 3.18 |
For the case of rotation of a rigid body around a fixed axis:
Where Jz– constant moment of inertia relative to the axis of rotation; w – angular velocity.
Taking this into account, we get:
If we introduce the angle of rotation of the body j, then, taking into account the equality 
we have
(3.26)
Expression (3.26) is differential equation rotation of a rigid body around a fixed axis.
4. Theorem on the change in the angular momentum of the system
in relative motion relative to the center of mass
To study a mechanical system, we choose a fixed coordinate system Ox 1 y 1 z 1 and movable Cxyz with origin at the center of mass C, moving forward (Fig. 3.19).
From a vector triangle:
| Rice. 3.19 |
Differentiating this equality with respect to time, we obtain
or 
where is the absolute speed of the point Mk, - absolute speed of the center of mass WITH, 
- relative speed of the point Mk, because 
Momentum about a point ABOUT
Substituting the values and , we get
In this expression: 
– mass of the system; 
; 
– kinetic moment of the system relative to the center of mass for relative motion in the coordinate system Сxyz.
The kinetic moment takes the form
Theorem on the change in angular momentum relative to a point ABOUT looks like
Let's substitute the values and 
we get
Let us transform this expression taking into account that
or 
This formula expresses the theorem on the change in the angular momentum of a system relative to the center of mass for the relative motion of the system with respect to a coordinate system moving translationally with the center of mass. It is formulated in the same way as if the center of mass were a fixed point.
First, let's consider the case of one material point. Let be the mass of the material point M, be its speed, and be the amount of motion.
Let us select a point O in the surrounding space and construct the moment of the vector relative to this point according to the same rules by which the moment of force is calculated in statics. We get the vector quantity
which is called the angular momentum of the material point relative to the center O (Fig. 31).
Let us construct a Cartesian O with the beginning at the center rectangular system coordinates Oxyz and project the vector ko onto these axes. Its projections on these axes, equal to moments vectors relative to the corresponding coordinate axes are called angular momentum of a material point relative to the coordinate axes:
Let us now have a mechanical system consisting of N material points. In this case, the angular momentum can be determined for each point of the system:
The geometric sum of the angular momentum of all material points included in the system is called the principal angular momentum or kinetic moment of the system.
Dynamics:
Dynamics of a material point
§ 28. Theorem on the change in momentum of a material point. Theorem on the change in angular momentum of a material point
Problems with solutions
28.1 A railway train moves along a horizontal and straight section of track. When braking, a resistance force equal to 0.1 of the train's weight develops. At the moment of braking, the speed of the train is 20 m/s. Find the braking time and braking distance.
SOLUTION
28.2 A heavy body descends along a rough inclined plane making an angle α=30° with the horizon. initial speed. Determine how long it will take T for the body to travel a path of length l=39.2 m if the coefficient of friction f=0.2.
SOLUTION
28.3 A train of mass 4*10^5 kg enters an ascent i=tg α=0.006 (where α is the angle of ascent) at a speed of 15 m/s. The friction coefficient (total resistance coefficient) when the train moves is 0.005. 50 s after the train enters the rise, its speed drops to 12.5 m/s. Find the traction force of the diesel locomotive.
SOLUTION
28.4 A weight M is attached to the end of an inextensible thread MOA, part of which OA is passed through a vertical tube; the weight moves around the axis of the tube along a circle of radius MC=R, making 120 rpm. Slowly drawing the thread OA into the tube, shorten the outer part of the thread to length OM1, at which the weight describes a circle of radius R/2. How many revolutions per minute does the weight make around this circle?
SOLUTION
28.5 To determine the mass of a loaded train, a dynamometer was installed between diesel locomotives and cars. The average dynamometer reading for 2 minutes turned out to be 10^6 N. During the same time, the train picked up a speed of 16 m/s (at first the train stood still). Find the mass of the composition if the friction coefficient is f=0.02.
SOLUTION
28.6 What should be the coefficient of friction f of the wheels of a braked car on the road, if at a driving speed v=20 m/s it stops 6 s after the start of braking?
SOLUTION
28.7 A bullet of mass 20 g flies out of a rifle barrel at a speed v=650 m/s, traveling through the barrel in time t=0.00095 s. Determine the average pressure of gases ejecting a bullet if the cross-sectional area of the channel is σ=150 mm^2.
SOLUTION
28.8 Point M moves around a fixed center under the influence of the force of attraction towards this center. Find the speed v2 at the point of the trajectory farthest from the center if the speed of the point at the position closest to it is v1=30 cm/s, and r2 is five times greater than r1.
SOLUTION
28.9 Find the impulse of the resultant of all forces acting on the projectile during the time when the projectile goes from the initial position O to highest position M. Given: v0=500 m/s; α0=60°; v1=200 m/s; projectile mass 100 kg.
SOLUTION
28.10 Two asteroids M1 and M2 describe the same ellipse, at the focus S of which is the Sun. The distance between them is so small that arc M1M2 of the ellipse can be considered a straight line segment. It is known that the length of the arc M1M2 was equal to a when its middle was at perihelion P. Assuming that the asteroids move with equal sectorial velocities, determine the length of the arc M1M2 when its middle passes through aphelion A, if it is known that SP = R1 and SA =R2.
SOLUTION
28.11 A boy of mass 40 kg stands on the runners of a sports sled, the mass of which is 20 kg, and pushes every second with an impulse of 20 N*s. Find the speed acquired by the sled in 15 s if the coefficient of friction is f=0.01.
SOLUTION
28.12 The point commits uniform motion along the circle at a speed v=0.2 m/s, making full turn in time T=4 s. Find the impulse S of the forces acting on the point during one half-cycle, if the mass of the point is m=5 kg. Determine the average value of force F.
SOLUTION
28.13 Two mathematical pendulums suspended on threads of lengths l1 and l2 (l1>l2) oscillate with the same amplitude. Both pendulums simultaneously began to move in the same direction from their extreme deflected positions. Find the condition that the lengths l1 and l2 must satisfy in order for the pendulums to simultaneously return to the equilibrium position after a certain period of time. Determine the shortest time interval T.
SOLUTION
28.14 A ball of mass m, tied to an inextensible thread, slides along a smooth horizontal plane; the other end of the thread is pulled in from constant speed a into a hole made on the plane. Determine the motion of the ball and the tension of the thread T, if it is known that at the initial moment the thread is located in a straight line, the distance between the ball and the hole is equal to R, and the projection of the initial velocity of the ball perpendicular to the direction of the thread is equal to v0.
SOLUTION
28.15 Determine the mass M of the Sun, given the following data: radius of the Earth R=6.37*106 m, average density 5.5 t/m3, semi-major axis Earth's orbit a=1.49*10^11 m, time of revolution of the Earth around the Sun T=365.25 days. Strength universal gravity between two masses equal to 1 kg at a distance of 1 m we consider equal to gR2/m Н, where m is the mass of the Earth; From Kepler's laws it follows that the force of attraction of the Earth by the Sun is equal to 4π2a3m/(T2r2), where r is the distance of the Earth from the Sun.
SOLUTION
28.16 A point of mass m, subject to the action of a central force F, describes the lemniscate r2=a cos 2φ, where a is a constant value, r is the distance of the point from the force center; at the initial moment r=r0, the speed of the point is equal to v0 and makes an angle α with the straight line connecting the point with the force center. Determine the magnitude of the force F, knowing that it depends only on the distance r. By Binet's formula F =-(mc2/r2)(d2(1/r)/dφ2+1/r), where c is the double sector velocity of the point.
SOLUTION
28.17 A point M, the mass of which is m, moves near a fixed center O under the influence of a force F emanating from this center and depending only on the distance MO=r. Knowing that the speed of the point v=a/r, where a is a constant value, find the magnitude of the force F and the trajectory of the point.
SOLUTION
28.18 Determine the movement of a point whose mass is 1 kg under the action of a central force of attraction, inversely proportional to the cube of the distance of the point from the center of attraction, given the following data: at a distance of 1 m, the force is 1 N. At the initial moment, the distance of the point from the center of gravity is 2 m, speed v0=0.5 m/s and makes an angle of 45° with the direction of the straight line drawn from the center to the point.
SOLUTION
28.19 A particle M of mass 1 kg is attracted to a fixed center O by a force inversely proportional to the fifth power of the distance. This force is equal to 8 N at a distance of 1 m. At the initial moment, the particle is at a distance OM0 = 2 m and has a speed perpendicular to OM0 and equal to 0.5 m/s. Determine the trajectory of the particle.
SOLUTION
28.20 A point of mass 0.2 kg, moving under the influence of an attractive force to a stationary center according to Newton’s law of gravity, describes a complete ellipse with semi-axes 0.1 m and 0.08 m for 50 s. Determine the largest and the smallest value attractive force F during this movement.
SOLUTION
28.21 A mathematical pendulum, each swing of which lasts one second, is called a seconds pendulum and is used to count time. Find the length l of this pendulum, assuming the acceleration due to gravity to be 981 cm/s2. What time will this pendulum show on the Moon, where the acceleration of gravity is 6 times less than on Earth? What length l1 should the second lunar pendulum have?
SOLUTION
28.22 At some point on the Earth, the seconds pendulum counts time correctly. Being moved to another place, it lags behind by T seconds per day. Determine the acceleration due to gravity in the new position of the seconds pendulum.
