Mechanics 4. Work, Power, Energy Linear Momentum Conservation Laws. Work W

Transcription

1 Mechanics 4 Work, Power, Energy Linear Momentum Conservation Laws Work W If a force F acts on a body and the body moves a distance s in the direction of the force, the work W done by the force F is defined by W = Fs If the angle between the force F and the distance vector s is a, then W = F. s = Fs cosa If the force F changes with the position, the work is defined as the area in s-f curve: W = s 2 F ds s 1 The unit of work is 1N*1m = 1 Joule = 1 J 1

2 Power P Power it the ratio of work and time used to do it: W P = t The unit of power is 1 J/1s = 1 Watt = 1 W Efficiency h of an engine = power input/power output P in Engine thermal power loss P out η = P P out in tariff unit: kwh Another energy unit = kwh From the definition of power we get W = P*t. If we express power in kilowatts and time in hours we get another energy unit 1 kwh =1000W*3600s = J = 3.6 MJ kwh is the unit in which energy is sold on the energy market. The consumer price in Finland is in 2004 about 8 cnt/kwh. Another formula for power: From P = W/t we get also P = F*s/t = F v P = F v power = force * velocity 2

3 Example: Electric stairs take during peek hours 100 persons/min from a metro station to the street level 30m above. The engine of the stairs has an efficiency of Calculate the power consumption of the stairs. Let s assume that the average weight of people is 75 kg. Solution The power output is: P out = mgh/t = 7500 kg*9.81 m/s 2 *30m/60s = W The consumed electric power is P in = P out /h = 36788W/0.7 = 52.5 kw Potential energy E p If a body (mass m) is lifted up to height h, the amount of work done is W = m g h The gravitational force is called a conservative force, because a work done against it is conserved and can be transformed back to other forms of energy. We say that the body has energy due to it s position in the gravitational field. This energy is called potential energy E p = m g h The zero level of potential energy can be chosen freely. (floor, sea level, ) Only the differences in potential energy have significance. 3

4 Hydroelectric Power Plant HPP transforms potential energy of water into electric power. Example: Let the difference between water levels in a HPP be 14 m and the water flow through the turbines is 500 m 3 /s. Calculate the power output of the power plant and the energy produced in one day. The efficiency of turbines is Solution: P = h mgh/t = 0.92*500000kg*9.81m/s 2 *14m/1s = W = 63 MW The tariff unit of energy is kwh. The daily energy production is kw*24 h = kwh = 1.5 GWh In Finland the consumer price is 8 cnt/kwh, so the market value of one day s production is about

5 Kinetic Energy E k Let F be the accelerating force and let initial velocity be zero. Then the work done by the accelerating force F W = Fs = ma* ½ a t 2 = ½ m (at) 2 = ½ m v 2 This work is conserved to the motion of the body. This form of energy is called the kinetic energy of the body E k = ½ m v 2 Wind Power Plant This plant transforms kinetic energy of the wind to electric energy with efficiency less than 0.45 Example: Let the length of the three rotating wings be 8.0 m and the speed of the wind 7.0 m/s. Calculate the electric power produced by the windmill, if it s efficiency is 0.4. The density of air is 1.25 kg/m 3. Solution: In one second a volume V = pr 2 *v*t = p*8 2 *7*1 m 3 = 1407 m 3 is going through the mill. It s mass m = rv = 1.25kg/m 3 *1407 m 3 = 1759 kg. The produced power P =h ½ m v 2 /t = 0.4* ½*1759*7 2 /1 W = W = 17.2 kw 5

6 The energy conservation laws Definition: The sum of potential energy and kinetic energy is called the mechanical energy E m = ½ m v 2 + m g h 1) The conservation of mechanical energy: If there is no external forces (except gravitation) acting on a body, the mechanical energy of the body is conserved 6

7 Example: A pendulum consist of a mass m and a 80 cm long rope. The mass is moving around a vertical circle. Calculate the minimum velocity that the mass at least should have at the lowest point of the circle to be able to move as described. v 2 Solution: At the top there must be a balance between the centrifugal force and gravitation: mg = mv 2 2 /r On the other hand from the energy principle we get ½ mv 1 2 =½ mv mgh r=80cm v 1 h From the first equation we get v 2 = (gr)= (9.81*0.8)= 2.8 m/s From the second equation we get v 1 = (v gh)= ( *9.81*1.6)=6.3 m/s 2) The general conservation law: If the work done by external forces on a body is W, the work is equal to the change of mechanical energy of the body ½ m v 12 + m g h 1 + W = ½ m v 22 + m g h 2 Remember to put a right sign for the work W. If the external force is friction, W is negative, because the directions of force F and distance s are opposite. 7

8 Example: A body starts gliding from the top of a house down the roof and finally falls to the ground. Calculate the velocity the body has when it hits the ground. The coefficient of friction between the body and the roof is 0,3. 2m s 2.5m 4 m From Energy principle: mgh Fms = 0 + ½ mv 2 h = 4,5 m s = ( ) = 2.83 m Fm= - m mg cosa mgh m mg cosa s = 0 + ½ mv 2 => gh - m g s cos a = ½ v 2 => v = ( 2g h - m g s cos a ) = (2*9,81* *9.81*2)= 9.1 m/s A classic exam problem P h R zero level In an amusement park a car with passengers glides due to gravity frictionless down a track (picture), goes once around a circle with radius R to the other side. Calculate minimum starting height h as a function of R. Solution: At point P the centrifugal force must exceed gravity => Limiting case: mv 2 /R = mg => v = (gr) From energy principle: mgh = mg(2r) + ½ m(gr) h = 2R + ½ R = 2 ½ R Answer: The initial height h must exceed 2½ R 8

9 Energy principle in the gravitational field of the Earth Near the surface of the Earth the work needed to lift up a mass m to a height of h can be expressed W = F*s = mg*h * This work increases the body s energy by mgh, which is called the potential energy E p. The zero level of E p is often agreed to be the Earth s surface. More generally the gravitational force on a mass m at the radius of r of the Earth s centre is F = g mm/r 2 g = coefficient of gravitation * The work needed to move a body of mass m from distance r a to distance r b from Earth s centre is an integral W mm mm mm ( γ ) r rb γ = γ Earth = r 2 a r r ra b a W is equal to the potential energy change of the body. The zero level of potential energy is chosen in the way that E p = 0 at infinity r =, where the gravitation vanishes So the potential energy of the body at radius r of the Earth s centre is defined by E p mm == γ r One could say that the Earth is at a centre of a potential hole. The potential energy of a mass m at the surface of the Earth is E p mm == γ R where R is the radius of the Earth 6380 km 9

10 Example: Calculate the second escape velocity, the velocity a rocket must have to get totally out from the Earth s gravitational field. Solution: The sum of the kinetic energy and potential energy on the Earth must be greater than the potential energy at the infinity. E k + E p = mv mm R 1 2 γ > 2 0 v > (2gM/R) = (2*6.67*10-11 *6.0*10 24 /6.38*10 6 ) = m/s = 11.2 km/s The second escape velocity is about 11.2 km/s If you want to send a rocket to the Moon, that is about the speed needed to launch the rocket. The mechanical energy of a body in the gravitational field of the Earth is E m = E k + E p = ½ mv 2 - g mm/r If E k + E p > 0, the body has enough energy to leave the gravitational field of the Earth. The orbit is a hyperbola If E k + E p < 0, the body cannot leave the Earth s influence. The orbit is an ellipse. If E k + E p = 0, which is the limiting case of the former two, a body has the minimum energy to leave the gravitational field. The orbit is parabola. 10

11 Linear momentum p = mv 1. Impulse of the force 2. Conservation law 3. Recoil problems 4. Collisions Definitions: 1)The product of the mass and velocity of the body is called the linear momentum of the body. It is a vector quantity. p = m v (linear momentum) 2)The product of the force F and the time interval Dt is called the impulse of the force. I = F Dt (impulse) 11

12 1. Impulse principle The original form of Newton s second law is: F = Dp/ Dt = D(mv)/ Dt The resultant force acting on the body is equal to the time rate of change of the linear momentum of the body If the mass of the body is constant this becomes F = D(mv)/ Dt = mdv/ Dt = ma The last equation can be multiplied by Dt, and we get F Dt = D (mv) (impulse principle) The impulse of the force on the body is equal to the change of the linear momentum of the body Example: A riffle fires a bullet of mass 15g at the speed of 500 m/s. Assuming that the bullet accelerates uniformly inside the barrel of the riffle which is 60 cm long, find the average force acting on the bullet inside the gun. Solution: The time the bullet spends in the barrel is approximately s/v a = 0.6m/(250m/s) = 2.4*10-3 s From impulse principle: F Dt = D (mv) = m D v => F = m D v/ D t = (0.015kg*500m/s )/2.4*10-3 s = 3125 N = 3.1 kn 12

13 2. The conservation of linear momentum Let s assume that our system consists of two masses: m1 and m2, which collide together linearly -F F m1 v1 v2 m2 m1m2 u1 m1 m2 u2 Before the collision the velocities are v1 and v2 During the collision, the forces are equal but opposite After the collision the velocities are u1 and u2 From impulse principle we get for the mass m1: -F Dt = (u 1 -v 1 ) for the mass m2: F Dt = m 2 (u 2 -v 2 ) Adding these two equations together, we get u 1 v 1 +m 2 u 2 m 2 v 2 = 0 => v 1 +m 2 v 2 = u 1 + m 2 u 2 The obtained result is very important: It means that the total linear momentum (the sum of the linear momentums of the bodies) is conserved in the collision. This result can be generalized for many isolated many particle systems: Their total linear momentum is a constant. Actually this equation is also true in vector form, because the linear momentum is a vector quantity. The conservation law is a direct consequence of the laws of Newton. 13

14 Applications 1. Recoil problems A case when a body divides to two or more bodies by explosion or under action of internal forces and the parts start to move into different directions, is a recoil problem. Example: A bullet (15 g) is shot at the speed of 500 m/s from a riffle with mass 3000 g. What is the recoil speed of the gun? Solution: The total linear momentum before the shot is 0, so it must also be zero after the shot: 0 = v 1 + m 2 v 2 (m1 = 3000 g, m2=15 g, v2=500m/s). Hence the recoil velocity v 1 = -m 2 / *v 2 = 15/3000*500m/s = 2.5 m/s The recoil formula: v 1 = -m 2 / *v 2 Example: A machine gun sends 120 bullets of mass 15 g and velocity 500 m/s in a minute. The mass of the gun is 5000 g. Calculate the average recoil force on the gun. Solution: The recoil velocity during one shot v1= -m2/m1*v2 = - 15/5000*500 m/s = -1,5 m/s The time interval for this velocity change is 60s/120 = 0.5 s The average force F = Dv 1 / Dt = 5.0 kg *1.5 m/s / 0.5 s = 15 N 14

15 Collisions: Inelastic collision A collision, where the colliding bodies continue together with the same velocity after the collision, is called inelastic collision. Example: A bullet ( =15 g) is shot at a speed of 300 m/s to a piece of wood (m 2 = 200 g) hanging from a rope. The bullet stays inside the wood. Calculate the velocity of the system just after the hit. Solution: According to the conservation law v = ( +m 2 )*u, where u is the asked velocity Hence u = /( +m 2 ) v 1 = 15/215*300 m/s = 21 m/s More generally: the equation for inelastic collisions is v 1 + m 2 v 2 = ( +m 2 ) u m1 and m2 are the colliding masses, v1 and v2 are their velocities before the collision, u is the common velocity after the collision. 15

16 Collisions: Completely elastic collision A completely elastic collision can be defined as a collision where the total kinetic energy is conserved. Some collision of elementary particles are such. Also the collisions of balls on the billiard table can sometimes be regarded as completely elastic. The completely elastic collisions obey the following pair of equations: v 1 + m 2 v 2 = u 1 + m 2 u 2 ½ v 12 + ½ m 2 v 22 = ½ u 12 + ½ m 2 u 2 2 Example: On a billiard board a ball of mass =110 g and speed 2.0 m/s hits another ball (m 2 =100g), which is at rest on the table. The hit is central. Calculate the velocities of both balls after the hit. v1 u1 u2 m1 m2 m1 m2 Equations: 110*2 = 110 u u 2 /100 ½ 110*2 2 = ½ 110*u 12 + ½ 100*u 2 2 /100 * 2 1.1*u 1 + u 2 = *u 12 + u 22 = 4.4 => u 2 = u1 (substitution) 1.1*u 12 + ( u 1 ) 2 = 4.4 => 2.31*u *u = 0 => u 1 = m/s tai u1 = 2m/s ( a miss) U 2 = *0.095 = m/s Answer: The hitting ball carries on with velocity m/s and the target ball gets a velocity m/s 16

17 If you look at the result, you might notice that the velocity difference between the balls is the same before and after the collision. Is it possible to prove this for all completely elastic collisions? That would certainly make calculations easier. Hypothesis: The velocity difference of the two bodies is conserved in a completely elastic collision. Proof: v 1 + m 2 v 2 = u 1 + m 2 u 2 ½ v 12 + ½ m 2 v 22 = ½ u 12 + ½ m 2 u 2 2 v 12 + m 2 v 22 = u 12 + m 2 u 2 2 v 12 - u 12 = m 2 u 22 - m 2 v 2 2 (v 1 -u 1 )(v 1 +u 1 )=m 2 (u 2 -v 2 )(u 2 +v 2 ) v 1 +u 1 = v 2 + u 2 v 1 -v 2 = u 2 -u 1 From the 1 st eq we get: v 1 - u 1 = m 2 u 2 -m 2 v 2 => (v 1 -u 1 )= m 2 (u 2 -v 2 ) which is use in the left. The last form indicates that the velocity difference is conserved. Let s solve our billiard problem using this new information about completely elastic collisions: We need only one equations 110*2 = 110 u (u 1 +2) Dividing by 100 we get Taking into account that the velocity difference is 2.0, u 2 = u = 1.1 u 1 + (u 1 +2) = 2.1 u = 2.1 *u 1 u 1 = 0.2 / 2.1 = m/s u 2 = = m/s 17

18 Collisions Partially elastic collisions Most of the real life collisions are somewhere between inelastic and completely elastic collisions. To describe the degree of elasticity we define Coefficient of restitution e = (u 2 -u 1 )/(v 1 -v 2 ) = Du/Dv e is a ratio of velocity differences after and before collision In inelastic case u 2 =u 1 => e = 0 In completely elastic case e = 1 In most collisions 0 < e < 1 Example: A rubber ball is dropped from the height of 100 cm on the floor. The coefficient of restitution when the ball hits the floor is 0.8. Calculate how high the ball jumps from the floor at the first and second jump. Solution: The velocity if the ball hitting the floor the first time can be obtained from mgh = ½ mv 2 (energy principle) v = (2gh) The floor is so massive that it doesn t move in the collision, so the velocity of the ball leaving the floor for the first time is v1= e*v = e (2gh) It rises to the height h1: whence h1 = e 2 *h mgh 1 = ½ m(ev) 2 = ½ m*e 2 *2gh So the height of the first jump is e 2 *the original height = *100cm = 64 cm Similarly we conclude that the height if the second jump Would be e 2 *h1 = *64cm = 41 cm 18

19 Problem: A car with mass m1 = 1400 kg collides with another car (m2=1300 kg) to the back. The cars have velocities 15 m/s and 10 m/s at the moment of collision. The collision has a coefficient of restitution 0.7. a) Calculate the velocities of the cars just after the collision. b) Calculate the amount of energy that is lost in the collision to the form changes and production of thermal energy Solution: The velocity difference before is = 5 m/s and after the collision 0.7*5 = 3.5 m/s Equation: 1400* *10=1400*u1+1300*(u1+3.5) u1 = 10.9 m/s and u2 = u1+3.5 = 14.4 m/s The energies: Before ½*1400* ½ *1300*10 2 = J =222.5 kj After ½*1400* ½ *1300* =217951J =218.0 kj Only 4.5 kj is lost during the collision. Collisions: Non-linear collisions m1 v1 m2 m1 u1 m2 a1 u2 a2 Let m1=0.11kg and v1 = 2.0 m/s and m2= 0.1kg. Calculate u1 and angle a1 if we know that u2=2.3 m/s and a2=30 o. Solution: The conservation law for linear momentum can be represented as a vector triangle: m 2 *u 2 *u 1 a2 a1 *v 1 => 0.23 p 30 o a

20 From the cosine law: p 2 = *0.22*0.23*cos(30 o ) p = Now p = u 1 => u 1 = p/ = 0.117/0.110= 1.06 m/s From the law of sines: sin(a1)/0.23 = sin(30 o )/0.117 => a1 = 79.4 degrees 20