Summary of the physical kinematics formulas in the game

 

I,Linear Motion, Free fallII,Curve movement and universal gravitation,Force (synthesis and decomposition of common force and force) 4,Dynamics (motion and force) 5,Vibration and wave (propagation of mechanical vibration and mechanical vibration)6. Impulse and momentum (changes in force and momentum of an object)

1. Particle Motion (1) ------ linear motion

1) uniformly variable speed linear movement

1. Average speed v = S/T (Definition)

2. Useful inferences Vt * Vt-vo * vo = 2as

3. Intermediate time speed VT/2 = V flat = (vt + VO)/2

4. Last speed VT = VO +

5. Center speed VS/2 = [(VO * VO + Vt * Vt)/2] 1/2

6. Shift S = V flat T = vot + at * t/2 = Vt/2 t

7. Acceleration A = (Vt-vo)/t {in the vo positive direction, a and VO In the same direction (acceleration) a> 0; in the reverse direction, a <0}

8. The experiment uses inferences △s = at * t {△s is the deviation of continuous adjacent equal time (T) Inner displacement}

9. main Physical Quantity and unit: initial velocity (VO): m/s; acceleration (a): M/S * s; last velocity (VT): m/s; time (t) second (s); displacement (s): meter (m); distance: meter; speed unit conversion: 1 m/s = 3.6 km/h.

Note:

(1) The average speed is a vector;

(2) The object speed is large, and the acceleration is not necessarily large;

(3) A = (Vt-vo)/T is a measurement, not a decision;

(4) other related content: particle, displacement and distance, reference system, Time and time (see P19 in volume 1)/S-T diagram, V-T diagram/speed and speed, instantaneous velocity (see p24 in volume 1 〕.

2) free fall

1. The initial speed Vo = 0?

2. Last speed VT = gt

3. Falling height H = gt * T/2 (calculated from the VO position down)

4. Deduce Vt * Vt = 2gh

Note:

(1) The free falling motion is a uniformly accelerated linear motion with zero initial speed, following the linear motion rule of uniformly variable speed;

(2) A = G = 9.8 m/s * s ≈ 10 m/s * s (the acceleration of gravity is smaller near the equator and smaller than the flat area in the mountains, vertical ).

(3) Vertical throwing

1. Shift S = vot-GT * T/2

2. Final speed VT = Vo-GT (G = 9.8 m/s * s ≈ 10 m/s * s)

3. Useful inferences Vt * Vt-vo * vo =-2gs

4. maximum height of rise Hm = Vo * vo/2G (counted from the throw point)

5. Round-trip time t = 2vo/g (from the time when the throw falls back to the original location)

Note:

(1) handling of the whole process: the whole process is moving uniformly and slowing down in a straight line. The process is in the upward and positive directions, and the acceleration value is negative;

(2) segment processing: moving up to a uniform deceleration line and down to a free falling body with symmetry;

(3) The process of rise and fall is symmetric, such as the inverse speed at the same point.

Ii. Particle Motion (2) ---- curve motion and gravitation

1) flat throwing

1. horizontal speed: vx = Vo

2. Vertical Speed: Vy = gt

3. Horizontal Displacement: x = vot

4. Vertical Displacement: Y = gt * T/2

5. Exercise time t = (2y/g) 1/2 (usually expressed as (2 h/g) 1/2)

6. clamping speed VT = (VX * VX + Vy * Vy) 1/2 = [VO * VO + (GT) * (GT)] 1/2 clamping speed direction and horizontal angle β: TG β = Vy/VX = gt/V0

7. Joint Displacement: S = (x * x + y * Y) 1/2, displacement direction and horizontal angle α: tg α = y/x = gt/2vo

8. horizontal acceleration: AX = 0; vertical acceleration: Ay = G

Note:

(1) flat throwing is a uniformly variable speed curve movement with an acceleration of G. It can generally be seen as a synthesis of a horizontal moving of a uniform straight line and a vertical moving of a free falling body;

(2) The moving time is determined by the falling height H (Y) and has nothing to do with the horizontal throw speed;

(3) The relationship between θ and β is TG β = 2tg α;

(4) In the flat throwing motion, time t is the key to solving the problem. (5) an object in curve motion must have an acceleration. When the velocity direction and the Force (acceleration) when the direction is not in the same line, the object performs curve motion.

2) Uniform Circular Motion

1. wire speed v = S/T = 2 π R/T

2. angular velocity ω = PHI/t = 2 π/t = 2 π F

3. centripetal acceleration A = v2/r = ω 2R = (2 π/T) 2R

4. centripetal force F heart = mv2/r = M ω 2R = MR (2 π/T) 2 = M ω v = f combination

5. Period and frequency: T = 1/F

6. Relationship between angular velocity and linear velocity: V = ω R

7. Relationship between angular velocity and rotation speed ω = 2π N (here the frequency is the same as the rotation speed)

8. main Physical Quantity and unit: arc length (s): meter (m); angle (PHI): radian (RAD); frequency (F): He (HZ); cycle (t ): second (s); speed (n): R/S; radius (R): meter (m); line speed (V): m/s; angular velocity (ω ): rad/s; centripetal acceleration: M/S * s.

Note:

(1) The centripetal force can be provided by a specific force, or by a combination of force, or by a split force. The direction is always vertical to the center of the velocity;

(2) The centripetal force of an object moving at a uniform speed in a circular direction is equal to the force of force, and the centripetal force only changes the speed direction without changing the speed. Therefore, the kinetic energy of the object remains unchanged, and the centripetal force does not do work, but the momentum keeps changing.

3) Universal Gravitation

1. kepler's third law: T2/R3 = K (= 4 π 2/GM) {R: orbital radius, T: Cycle, K: constant (unrelated to planetary mass, depends on the quality of the central celestial body )}

2. law of universal gravitation: F = GM 1m2/R2 (G = 6.67 × 10-11nm2/kg2, directed to their connections)

3. Gravity and gravity acceleration on the celestial body: GMM/r2 = Mg; G = GM/R2 {R: astronomical radius (M), M: astronomical mass (kg )}

4. satellite bypass speed, angular velocity, period: V = (GM/R) 1/2; ω = (GM/R3) 1/2; t = 2 π (R3/GM) 1/2 {M: central celestial body quality}

5. 1 (2 and 3) cosmic velocity V1 = (G location R location) 1/2 = (GM/R location) 1/2 = 7.9 km/s; v2 = 11.2 km/s; V3 = 16.7 km/s

6. GMM/(R + H) 2 = M4 π 2 (R + H)/T2 {H ≈ 36000 km, H: The height from the Earth's surface, r location: Earth's radius}

Note:

(1) The centripetal force required by the motion of the celestial body is provided by the universal gravitation, and the F direction is = F;

(2) the law of universal gravitation can be used to estimate the mass density of a celestial body;

(3) The Earth's synchronous satellite can only run over the equator, and its operation cycle is the same as that of the earth's rotation;

(4) the satellite's orbital radius changes by hour, the potential energy decreases, the kinetic energy increases, the speed increases, and the cycle decreases (three inversion together );

(5) the maximum and minimum transmission speeds of the earth's satellites are both 7.9 km/s.

Iii. Force (synthesis and decomposition of common force and force) 1) Common force

1. Gravity G = Mg (vertical downward direction, G = 9.8 m/s * s ≈ 10 m/s * s, attention points in the center of gravity, suitable for the vicinity of the earth's surface)

2. Hu Ke's law F = kx {direction along the direction of recovery deformation, K: stiffness coefficient (N/m), X: Shape variable (m )}

3. Sliding Friction F = μ FN {opposite to the relative movement direction of the object, μ: Friction Factor, FN: Positive Pressure (n )}

4. Static Friction: 0 ≤ F static ≤ FM (opposite to the relative movement trend of the object, FM is the maximum static friction)

5. Universal Gravitation F = GM 1m2/R2 (G = 6.67 × 10-11nm2/kg2, directed to their connections)

Note:

(1) The stiffness coefficient K is determined by the spring itself;

(2) The friction factor μ is independent of the pressure and contact area, and is determined by the contact surface material characteristics and surface conditions;

(3) FM is slightly greater than μ FN, which is generally regarded as FM ≈ μ FN;

2) force synthesis and Decomposition

1. The synthesis of the Force in the same line is the same direction: F = F1 + F2, reverse: F = F1-F2 (F1> F2)

2. Synthesis of mutual angular force:

F = (F12 + f22 + 2f1f2cos α) 1/2 (cosine theorem) F1 when F2: F = (F12 + f22) 1/2

3. force size range: | F1-F2 | ≤ F ≤ | F1 + F2 |

4. Orthogonal Decomposition of force: FX = fcos beta, FY = FSIN beta (beta is the angle between the Force and the X axis TG Beta = Fy/FX)

Note:

(1) The Synthesis and decomposition of force (vector) follow the parallelogram rule;

(2) The relationship between the Force and the Force splitting is an equivalent substitution relationship. The force can be used to replace the force splitting, and vice versa;

(3) In addition to the formula method, it can also be used as a method to solve the problem. In this case, you must select a scale and strictly plot the problem;

(4) When the F1 and F2 values are fixed, the greater the angle (α angle) between F1 and F2, the smaller the combination force;

(5) The combination of forces on the same straight line can take the positive direction along the straight line and use the positive and negative signs to represent the direction of the force, which is reduced to algebra.

Iv. Dynamics (motion and force)

1. Newton's first law of motion (Inertial Law): An object has inertia and always keeps moving or static at a constant speed until an external force forces it to change this state.

2. Newton's second law of motion: F combination = ma or a = f combination/Ma {determined by the combination of external force, consistent with the direction of the combination of external force}

3. Newton's third law of motion: F =-F' {negative signs indicate the opposite direction. F and F' respectively act on the other side. The equilibrium force differs from the reaction force. Practical application: the reverse force}

4. Balanced F-sum = 0 of common force, promotion {Orthogonal Decomposition Method, triplicate principle}

5. overweight: FN> G, weightless: FN <G {downward direction of acceleration, all weightless, upward direction of acceleration, all overweight}

6. Applicable Conditions of Newton's law of motion: suitable for solving low-speed Motion Problems, suitable for macro objects, not suitable for handling high-speed problems, not suitable for micro particles

Note: The Equilibrium State indicates that the object is in a static or uniform straight line state or rotating at a uniform speed.

V. Vibration and wave (propagation of mechanical vibration and mechanical vibration)

1. Simple Harmonic Vibration F =-kx {f: recovery force, K: proportional coefficient, X: displacement, negative signs indicate that the direction of F is always opposite to that of x}

2. single Pendulum cycle T = 2 π (L/G) 1/2 {L: Pendulum Length (M), G: Local Gravity acceleration value, condition: pendulum angle θ <100; l> r}

6. Impulse and momentum (changes in force and momentum of an object)

1. Momentum: P = MV {P: momentum (kg/s), M: Quality (kg), V: Speed (m/s), the direction is the same as the speed direction}

3. Impulse: I = ft {I: impulse (NS), F: constant force (N), T: force action time (s), direction determined by f}

4. Momentum Theorem: I = △p or Ft = MVT-MVO {△p: momentum change △p = MVT-MVO, vector}

5. Conservation of momentum law: the sum before P = the sum after P or P = p'' can also be m1v1 + m2v2 = m1v1 '+ m2v2 ′

6. elastic collision: △p = 0; △ek = 0 {conservation of momentum and kinetic energy of the system}

7. Non-elastic collision △p = 0; 0 <△ek <△ekm {△ek: kinetic energy of loss, ekm: Maximum kinetic energy of loss}

8. Completely non-elastic collision △p = 0; △ek = △ekm {the collision is connected together to form a whole}

9. An object M1 has an elastic positive collision with an object m2 at the initial velocity of V1:

V1 '= (M1-M2) V1/(M1 + m2) v2' = 2m1v1/(M1 + m2)

10. deduction from 9 ----- exchange speed (conservation of kinetic energy and conservation of momentum) when mass elasticity is being touched)

11. the bullet m horizontal velocity vo is injected into the long block m that is still placed on the horizontal smooth ground, and the mechanical energy loss is embedded in it. e loss = MVO * vo/2-(m + M) vt * Vt/2 = FS relative {vt: Common speed, F: resistance, s relative displacement of the relatively long block of the bullet}