2. DynamicsSavchenko Formulas, chapter 2 of 14, 60 formulas
Sections follow the book, and within a section the most used formulas come first. An italic problem number means the formula appears in its statement. Rest the cursor on a number to see the statement.
$\displaystyle \vec F = \frac{d\vec p}{dt}$$\displaystyle a_x = \frac{\sum F_x}{m}$
m
mass of the body
$\vec a$
its acceleration
$\sum\vec F$
the sum of all forces on the body
A body's acceleration is proportional to the sum of the forces on it and inversely proportional to its mass. The equation is a vector one and is written in projections on well chosen axes, for each body of a system separately. It holds in an inertial frame.
$\displaystyle F_{\text{тр}} \le \mu N\ \text{(покой)}$$\displaystyle a = \mu g\ \text{(торможение на горизонтали)}$
$\mu$
coefficient of friction
N
normal force
$F_{\text{тр}}$
friction force
While sliding, friction equals $\mu N$ and points against the relative velocity. While the body rests, static friction takes any value up to $\mu N$ and follows from equilibrium, not from the formula. On a horizontal surface $N = mg$ and the braking deceleration is $\mu g$.
$\displaystyle F = k\,\Delta l$$\displaystyle E_{\text{упр}} = \frac{kx^2}{2}$
k
spring constant
x
extension or compression from the natural length
The elastic force is proportional to the deformation and points to the equilibrium position. The minus sign marks a restoring force. A stretched spring pulls both its ends with the same force $kx$.
$\displaystyle m = \frac{4}{3}\pi r^3\rho\ \text{(шар)}$$\displaystyle m = \rho S l\ \text{(стержень, столб жидкости)}$
$\rho$
density
V
volume
S
cross-section
The mass of a uniform body is density times volume, $\frac{4}{3}\pi r^3\rho$ for a sphere, $\rho S l$ for a rod or a column of liquid. The mass of an element is introduced the same way, $dm = \rho\,dV$.
Along the plane act $mg\sin\alpha$ and friction, across it $N = mg\cos\alpha$. The body slides when $\tan\alpha > \mu$, with acceleration $g(\sin\alpha - \mu\cos\alpha)$. Moving up, friction adds and the deceleration is $g(\sin\alpha + \mu\cos\alpha)$.
The rope is inextensible, so its length ties the loads' coordinates and differentiating that tie links the accelerations. A movable pulley moves half as fast as the rope's end. Along sloping surfaces the accelerations are linked through the sine or tangent of the angle.
A load on a string moves on a circle, the horizontal part of the tension supplies the centripetal acceleration and the vertical part balances gravity. This gives the angle of the string and the lean of a cyclist, an aeroplane or a banked road on a turn. A solution exists only when $\omega^2 l \ge g$.
$\displaystyle a = \frac{F}{m_1 + m_2}\ \text{(тела толкают вместе)}$$\displaystyle T = m_1(g + a)$
m1, m2
masses of the bodies
T
tension of the rope
a
common acceleration
Newton's second law is written for each body with its own forces, the rope gives equal tension at both ends and accelerations equal in magnitude. Adding the equations gives the system's acceleration, substituting back gives the tension. This solves Atwood's machine, blocks on a table and loads over pulleys.
$\displaystyle N = 0\ \text{(невесомость при }a = g)$$\displaystyle N = m\left(g + \frac{v^2}{R}\right)\ \text{(внизу окружности)}$
N
force on the support or the tension of the suspension
a
acceleration of the body, plus when upward
Weight is the force with which a body presses on its support, and it differs from $mg$ when the support accelerates. Accelerating up it exceeds $mg$, accelerating down it is less, and with the support in free fall it is zero. The same holds for the tension of a rope a body hangs on.
Two bodies act on each other with forces equal in magnitude and opposite in direction. The forces act on different bodies and so never cancel. The tension of one rope pulls both ends equally, and the push on a support equals the support's reaction.
$\displaystyle v_{\text{уст}} = \frac{mg}{\beta}\ \text{(установившееся падение)}$$\displaystyle F = \frac{1}{2}C\rho S v^2$
$\beta$
viscous drag coefficient, $6\pi\eta r$ for a sphere
$\alpha$
quadratic drag coefficient
v
speed relative to the medium
Slow motion in a viscous medium is resisted by a force proportional to speed, fast motion by one proportional to the square of the speed and to the cross-section. A falling body speeds up until drag equals its weight, then the speed stays constant.
$\displaystyle T = m\omega^2 L\ \text{(вращающийся стержень, у оси)}$
F
force applied at the end
x
distance from that end
l
length of the rope or rod
In a rope with mass the tension varies along it, because each piece accelerates everything beyond it. One takes an element $dm = m\,dx/l$ or a whole segment and writes Newton's second law for it. At the free end the tension is zero.
In series both springs carry the same force and their extensions add, in parallel they share one extension and their forces add. Half of a spring is twice as stiff as the whole.
$\displaystyle T = \frac{m\omega^2 R}{2\pi}\ \text{(натяжение кольца)}$
T
tension of the ring or the thread
$\Delta\varphi$
angle subtended by the element
$\Delta m$
its mass
A small element of a ring is pulled by its two neighbours, whose resultant at a small angle is $T\Delta\varphi$ towards the centre. It supplies the element's centripetal acceleration, and with friction and gravity present those enter the equation too.
$\displaystyle v^2 = gR\tan\alpha\ \text{(наклонная дорога без трения)}$$\displaystyle v_{\max}^2 = gR\,\frac{\mu + \tan\alpha}{1 - \mu\tan\alpha}$
$\mu$
coefficient of friction of the wheels on the road
R
radius of the turn
$\alpha$
banking angle of the road
On a turn the centripetal acceleration is supplied by static friction, which cannot exceed $\mu N$. Hence the top speed $\sqrt{\mu g R}$ on a level road. Banking adds a component of the normal force towards the centre and raises the limit.
$\displaystyle \sum\vec p = \text{const}$$\displaystyle m_1 v_1 = m_2 v_2\ \text{(разлёт из покоя)}$
m1, m2
masses of the bodies
$\vec v_1, \vec v_2$
velocities before the interaction
$\vec v_1\,', \vec v_2\,'$
velocities after
When external forces are absent or their impulse over the interaction is small, the total momentum of the system does not change. The law is a vector one, written in projections, and it may hold along one axis while an external force acts along another. In an impact, an explosion or a shot, gravity and friction have no time to change the momentum.
$$\vec p = m\vec v, \qquad \Delta\vec p = \vec F\,\Delta t$$
$\displaystyle \vec F = \frac{d\vec p}{dt}$$\displaystyle p = \sqrt{2mE_k}$
$\vec p$
momentum of the body
$\vec F$
mean force over the time $\Delta t$
$\Delta t$
duration of the force
The change of a body's momentum equals the impulse of the force. That is how the mean force of an impact follows from the momentum change and the time, and the other way round. For a time-dependent force the impulse is the integral, the area under $F(t)$.
The centre of mass of a system moves as a point of the system's whole mass would under the external forces alone. Internal forces, tensions and impacts between the parts, do not affect it. With no external force along an axis, the centre of mass coordinate along it is conserved.
$\displaystyle \vec v_c = \frac{\sum m_i\vec v_i}{\sum m_i}$$\displaystyle m_1 r_1 = m_2 r_2\ \text{(два тела, от центра масс)}$
$\vec r_i$
position vectors of the bodies
mi
their masses
$\vec r_c$
position of the centre of mass
The centre of mass divides the segment between two bodies inversely to their masses. Its velocity is the total momentum over the total mass, and with no external forces the centre of mass moves uniformly or rests. For uniform shapes it is at the centre of symmetry, for a triangle at the intersection of the medians.
$\displaystyle \frac{dm}{dt} = \rho S v$$\displaystyle F = \rho S v(u - v)\ \text{(струя догоняет стенку)}$
$\rho$
density of the stream
S
cross-section of the jet
v
speed of the stream
In a time $dt$ a mass $\rho S v\,dt$ with momentum $\rho S v^2\,dt$ reaches the obstacle, and when the stream stops there the force is $\rho S v^2$. On elastic reflection the force doubles, and for a moving obstacle the relative speed enters.
$\displaystyle \Delta p = 2mv\ \text{(упругий удар о стенку по нормали)}$$\displaystyle \Delta p = 2mv\cos\alpha\ \text{(под углом }\alpha\text{ к нормали)}$
p
magnitude of the momentum, unchanged
$\alpha$
angle through which the velocity turns
When the velocity only turns, the momentum change is the base of an isosceles triangle with sides $p$ and angle $\alpha$. On elastic reflection from a wall only the normal component changes, and the wall receives $2mv\cos\alpha$.
In a short time the rocket ejects a mass $dm$ at speed $u$ relative to itself, and momentum conservation gives $M\,dv = -u\,dM$. Integrating gives Tsiolkovsky's formula. The thrust equals the mass flow rate times the exhaust speed, and the rocket lifts off when thrust exceeds weight.
The force on the table is the weight of the part lying there, $\rho g x$, plus the force stopping the arriving links. In $dt$ a mass $\rho v\,dt$ at $v = \sqrt{2gx}$ is stopped, giving another $\rho v^2 = 2\rho g x$. Together three times the weight of the fallen part.
Kinetic energy depends on the square of the speed and hence on the frame of reference. In terms of momentum it is $p^2/2m$, which is handy in collisions and decays. For a system the kinetic energies add.
$\displaystyle A = \int_{x_1}^{x_2} F_x\,dx\ \text{(площадь под графиком силы)}$$\displaystyle A_{\text{упр}} = \frac{kx^2}{2}$
$\vec F$
force
$d\vec r$
displacement of the point of application
$\alpha$
angle between the force and the displacement
Work is the dot product of force and displacement, summed over small pieces of the path. A force perpendicular to the displacement does no work, friction does negative work. For a variable force the work is the area under $F(x)$.
The work of gravity does not depend on the path and equals the drop in $mgh$, which is why that quantity is the potential energy. The zero level is arbitrary, only differences matter. For an extended body one takes the height of its centre of mass.
$\displaystyle A_{\text{упр}} = \frac{kx_1^2}{2} - \frac{kx_2^2}{2}$$\displaystyle E = \frac{F^2}{2k}\ \text{(растянута силой }F)$
k
spring constant
x
deformation from the natural length
The work to stretch a spring is the area under $F = kx$, a triangle of area $kx^2/2$. That energy is returned on unloading. In terms of the tension it is $F^2/2k$.
When only gravity and elastic forces do work, with no friction or inelastic impacts, the sum of kinetic and potential energy is conserved. The normal reaction and the tension of an inextensible rope do no work. The equation is written for two positions, usually the initial one and the one asked about.
work of all forces, friction and reactions included
$\Delta E_k$
change of kinetic energy
The change of a body's kinetic energy equals the work of all forces on it. It follows from Newton's second law and is handy when the force is known as a function of position. Friction's work over a distance $s$ is $-\mu mgs$, hence the braking distance $v_0^2/2\mu g$.
$\displaystyle v_{\text{низ}} \ge \sqrt{5gR}\ \text{(чтобы пройти верхнюю точку на нити)}$$\displaystyle h = \frac{2R}{3},\ \cos\alpha = \frac{2}{3}\ \text{(отрыв от гладкой сферы)}$
T
tension of the string or normal reaction
$\alpha$
angle from the lowest point
R
radius
Along the normal the tension together with a component of gravity supplies the centripetal acceleration, and energy conservation gives the speed at each point. At the top the string stays taut while $v^2 \ge gR$, so $\sqrt{5gR}$ is needed at the bottom. A body on a smooth sphere leaves it when the reaction vanishes, at a height $2R/3$ above the centre.
$\displaystyle \vec F = -\nabla U$$\displaystyle E \ge U(x)\ \text{(область движения)}$
U
potential energy as a function of position
Fx
component of the force
The force is minus the derivative of the potential energy and points where $U$ decreases. A minimum of $U$ is a stable equilibrium. A body with total energy $E$ moves where $E \ge U$ and stops at the turning points, where $E = U$.
angles of the velocity to the normal of the boundary
When a force acts only across a certain line, the velocity component along that line is unchanged, and the speed follows from energy conservation. The result is a refraction law for particles, as for light with index $v_2/v_1$.
The loss of mechanical energy in friction or an inelastic collision becomes internal energy, heat. It is the difference of the energies before and after, and in a perfectly inelastic collision it equals the kinetic energy of the relative motion. In a nuclear reaction the same difference is the reaction energy.
kinetic energy of motion relative to the centre of mass
$\mu$
reduced mass
A system's kinetic energy is the energy of its centre of mass motion plus the energy of motion relative to it. Internal forces and impacts change only the second part, and an inelastic collision loses exactly that. For two bodies it is half the reduced mass times the relative speed squared.
$\displaystyle A = Nt\ \text{(постоянная мощность)}$$\displaystyle N_{\text{струи}} = \frac{\mu u^2}{2}$
N
power
A
work done in time $t$
$\vec F$
driving force
$\vec v$
velocity of the point where the force acts
Power is work per unit time, for a force its product with the velocity. At constant power the driving force falls as the speed grows, and the top speed is reached when the drive equals the resistance. The power of a jet is the kinetic energy carried away per second.
Efficiency is the share of the energy spent that goes to the intended job. The rest is lost to friction, heat and the ejected mass. For a jet drive the useful work is that done on the craft, while part of the energy leaves with the stream.
In an elastic collision both momentum and kinetic energy are conserved. For a head-on collision the two equations give that the relative velocity flips sign, hence the velocities after. Equal masses exchange velocities, and in an oblique collision of equal masses they fly apart at a right angle, $p_0^2 = p_1^2 + p_2^2$.
In a perfectly inelastic collision the bodies move on together, momentum is conserved and the kinetic energy drops by the energy of the relative motion. Balls stick, a bullet lodges in a block, wagons couple this way.
The vector equality $\vec p_0 = \vec p_1 + \vec p_2$ is a triangle, and the cosine law links the momenta to the angle of separation. Together with energy conservation written through $p^2/2m$ it solves decays and oblique collisions without projections.
In the centre-of-mass frame the total momentum is zero, the bodies approach with momenta equal in magnitude, and in an elastic collision their velocities only turn, keeping their size. Laboratory velocities follow by adding $\vec v_c$. This gives the angles of separation and the largest scattering angles.
$\displaystyle \vec F = -G\frac{Mm}{r^3}\vec r$$\displaystyle G = 6{,}67\cdot 10^{-11}\ \text{Н}\cdot\text{м}^2/\text{кг}^2$
M, m
masses of the bodies
r
distance between their centres
G
gravitational constant
Two point masses attract with a force proportional to the product of the masses and inverse to the square of the distance. A uniform sphere attracts from outside as a point at its centre, and inside a spherical shell the force is zero. Inside a uniform sphere the force grows in proportion to the distance from the centre.
$\displaystyle \Delta U = \frac{GMmh}{R(R + h)}\ \text{(подъём на }h)$$\displaystyle \varphi = -\frac{GM}{r}\ \text{(потенциал)}$
r
distance from the centre of the attracting body
M, m
masses
The zero of potential energy is taken at infinity, so near a body it is negative and rises with distance. For small heights $h \ll R$ the difference reduces to $mgh$. The work to carry a body to infinity is $GMm/r$.
$\displaystyle \frac{mv^2}{R} = G\frac{Mm}{R^2}$$\displaystyle T = 2\pi\sqrt{\frac{R^3}{GM}}$
v1
circular speed at the surface, about 7.9 km/s for the Earth
M, R
mass of the planet and the radius of the orbit
On a circular orbit gravity supplies the whole centripetal acceleration, hence the speed $\sqrt{GM/r}$ and the period $2\pi\sqrt{r^3/GM}$. The speed falls with the orbit's radius while the period grows. At the surface it is the first cosmic velocity.
The sum of kinetic and potential energy is conserved along the whole orbit and fixed by the semi-major axis alone. On a circular orbit the kinetic energy is half the size of the potential one. The energy equation for two points of the orbit together with angular momentum conservation gives the speeds at perigee and apogee.
The central body sits at a focus of the ellipse, and the perigee and apogee distances add up to the major axis. The speeds at those points follow from conservation of angular momentum and energy. The total energy depends on the semi-major axis alone, so moving between orbits means changing that.
$\displaystyle T = 2\pi\sqrt{\frac{a^3}{G(M + m)}}$$\displaystyle \frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3}$
T
orbital period
a
semi-major axis, the radius for a circle
M
mass of the central body
The squares of the periods are as the cubes of the semi-major axes. For a circle it follows at once from $mv^2/r = GMm/r^2$, for an ellipse the same holds with the semi-major axis in place of the radius. A straight fall to the centre is a degenerate ellipse with semi-axis $r/2$, and its time is half that orbit's period.
$\displaystyle M = \frac{gR^2}{G}$$\displaystyle g(h) = g\frac{R^2}{(R + h)^2}$
g
free-fall acceleration at the surface
M, R
mass and radius of the planet
At a planet's surface the gravitational force is $mg$, hence $g = GM/R^2$. This lets one replace $GM$ by $gR^2$, known more precisely, and find a planet's mass from $g$ and its radius or from a satellite's orbit. With altitude $g$ falls as $1/r^2$.
Gravity points to the centre and has no torque about it, so the angular momentum is conserved and the radius vector sweeps equal areas in equal times. At perigee and apogee the velocity is perpendicular to the radius, and $v_p r_p = v_a r_a$.
the least speed to reach infinity, about 11.2 km/s for the Earth
M, R
mass and radius of the planet
A body escapes to infinity when its total energy is non-negative, that is its kinetic energy is at least $GMm/R$. The speed at infinity follows from the same energy balance. A body with negative total energy stays on a bound orbit.
The moment of inertia measures inertia in rotation, the sum of masses times squared distances to the axis. It depends on the axis chosen. For a composite body the moments of inertia of its parts about one axis add.
$\displaystyle M = \frac{dL}{dt}$$\displaystyle a = \varepsilon R\ \text{(точка на ободе)}$
I
moment of inertia about the axis
$\varepsilon$
angular acceleration
M
net torque about the axis
A body's angular acceleration about a fixed axis is the torque over the moment of inertia, as a point's acceleration is force over mass. Torque is force times lever arm. For a rolling body the rotation equation is joined by the centre of mass equation and the tie $a = \varepsilon R$.
Rotational energy is the sum of the energies of all points, $m_i(\omega r_i)^2/2$. A rolling body has both translational and rotational energy, so a ball and a ring with the same centre speed carry different energy and roll down with different accelerations.
When the external torque about the axis is zero the angular momentum is conserved, and a change of the moment of inertia changes the angular velocity. In an impact on a free body both momentum and angular momentum about the centre of mass are conserved, and the impulse $mvh$ goes into rotation.
Rotation with constant angular acceleration follows the same formulas as uniformly accelerated motion, with angle in place of distance. The number of turns before stopping is $\omega_0^2/4\pi\varepsilon$.
Rolling without slipping, static friction does no work, and the acceleration follows from the centre of mass and rotation equations or from energy conservation. A body with a larger moment of inertia rolls down more slowly. Rolling lasts while the friction required stays below $\mu N$.
$$v = \mu g t, \qquad \omega = \omega_0 - \frac{\mu g t}{R}$$
$\displaystyle v = \omega R\ \text{(момент, когда скольжение прекращается)}$
$\mu$
coefficient of friction
$\omega_0$
initial angular velocity
R
radius
While the body slips, sliding friction $\mu mg$ speeds up the centre and slows the spin until the centre's speed matches $\omega R$. From then on friction is static and the motion uniform. The energy lost becomes heat.
moment of inertia about the axis through the centre of mass
d
distance between the parallel axes
m
mass of the body
The moment of inertia about any axis equals that about the parallel axis through the centre of mass plus $md^2$. So $ml^2/12$ for a rod's middle gives $ml^2/3$ for its end, and a rolling wheel has $I_c + mR^2$ about the contact point.
The work of a torque during a turn is torque times angle, as a force's work is force times distance. The power of a rotating engine is torque times angular velocity. That is how the work of friction in a bearing or on a brake shoe is found.
$\displaystyle F_1 l_1 = F_2 l_2\ \text{(правило рычага)}$$\displaystyle M = F l\sin\alpha\ \text{(момент силы)}$
$\sum\vec F$
sum of all forces on the body
$\sum M$
sum of torques about any point
l
lever arm, the distance from the axis to the line of action
A body rests when both the sum of the forces and the sum of their torques about any point are zero. The point for torques is chosen so that unknown forces pass through it and drop out. Three forces that hold a body meet at one point or are parallel.
An element of the rope is pressed to the cylinder with $F\,d\theta$, and the friction on it is $\mu F\,d\theta$, so the tension grows along the wrap geometrically. One turn at $\mu = 0.5$ holds a force $e^{\pi}$, about 23, times larger.