PHY183-Lecture17 - Physics for Scientists& Engineers 1 Spring Semester 2011 Lecture 17 Spring Example Non-conservative Forces and Equilibrium

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Unformatted text preview: Physics for Scientists & Engineers 1 Spring Semester 2011 Lecture 17 Spring Example, Non-conservative Forces and Equilibrium February 18, 2011 Physics for Scientists&Engineers 1 1 1 The Human Cannonball A favorite circus act is the “human cannonball”, in which a person is shot from a long barrel Before the Italian Zacchini brothers invented the compressed air powered cannon to shoot human cannonballs in the 1920s, the Englishman George Farini used a springloaded cannon to do the same since the 1870s Suppose we want to recreate the spring-loaded human cannonball cannonball act with a spring inside of a barrel Suppose the barrel is 4.00 m long, and you have room to compress the spring by 3.30 m Further, let’s point the barrel vertically towards the ceiling of our auditorium, where at a height of 7.50 m above the head is a bar that our human cannonball of height 1.75 m and mass 68.4 kg is supposed to grab on to at the top of his trajectory February 18, 2011 Physics for Scientists&Engineers 1 2 2 The Human Cannonball PROBLEM • What is the value of the spring constant needed to accomplish this stunt? THINK • Apply energy conservation February 18, 2011 Physics for Scientists&Engineers 1 3 3 The Human Cannonball SKETCH February 18, 2011 Physics for Scientists&Engineers 1 4 4 The Human Cannonball PROBLEM • What is the value of the spring constant needed to accomplish this stunt? THINK • Apply energy conservation • Potential energy is stored in the spring initially and then converted to gravitational potential energy at the top of the human cannonball’s cannonball’s flight • We choose the top of the barrel as the zero point for gravitational potential energy • Enough energy must be stored in the spring to elevate the human cannonball to a height of 7.50 m above the zero point • His feet need to be elevated by h = 7.50 m – 1.75 m = 5.75 m • We specify the y-coordinate as by bottom of his feet February 18, 2011 Physics for Scientists&Engineers 1 5 5 The Human Cannonball SKETCH February 18, 2011 Physics for Scientists&Engineers 1 6 6 The Human Cannonball RESEARCH • When the spring is initially compressed, the human cannonball has zero kinetic energy and potential energies from the spring force and the gravitational force, giving him a total energy of E= 12 kyb + mgyb 2 12 mvc 2 • When the human cannonball passes yb = 0, he has only kinetic energy energy E= • After leaving the spring, the human cannonball flies through the air and reaches the top, where he has only gravitational potential energy E = mgye February 18, 2011 Physics for Scientists&Engineers 1 7 7 The Human Cannonball SIMPLIFY • Energy conservation gives us 12 kyb + mgyb = mgye 2 • We rearrange to get the spring constant k = 2 mg ye − yb 2 yb 5.50 m − ( −3.30 m ) CALCULATE k = 2 ( 68.4 kg ) ( 9.81 m/s2 ) ROUND ( 3.30 m ) 2 = 1115.26 N/m k = 1120 N/m February 18, 2011 Physics for Scientists&Engineers 1 8 8 The Human Cannonball DOUBLE-CHECK • When the spring is compressed initially, the potential energy stored is U= 121 2 kyb = (1120 N/m )( 3.30 m ) = 6.07 kJ 2 2 • The gravitation potential energy gained by the human cannonball is U = mg ∆y = ( 68.4 kg ) ( 9.81 m/s2 ) ( 9.05 m ) = 6.07 kJ • So we feel that our answer is reasonable • Note that the mass of the human cannonball enters, so a cannon with the same spring will shot people of different masses to different heights February 18, 2011 Physics for Scientists&Engineers 1 9 9 Clicker Quiz What is the maximum acceleration (in terms of the acceleration of gravity g) that the human cannonball experiences? • • • • • a) 1.00g b) 2.14g c) 3.25g d) 4.48g e) 7.30g k = 1115 N/m yb = 3.30 m m = 68.4 kg February 18, 2011 Physics for Scientists&Engineers 1 10 10 Clicker Quiz What is the maximum acceleration (in terms of the acceleration of gravity g) that the human cannonball experiences? • • • • • a) 1.00g b) 2.14g c) 3.25g d) 4.48g e) 7.30g k = 1115 N/m yb = 3.30 m m = 68.4 kg February 18, 2011 Physics for Scientists&Engineers 1 11 11 Nonconservative Forces In the presence of nonconservative forces, the total mechanical energy is not conserved Where does the energy go? In our sliding box example, we showed that friction does not do work but instead dissipates mechanical energy into internal excitation energy We defined Wf to be the total energy dissipated by nonconservative forces into internal energy and then into other energy forms besides mechanical energy If we add this type of energy to the total mechanical energy, we get the total energy Etotal = Emechanical + Eother = K + U + Eother Here Eother stands for all forms of energy other than kinetic or potential energy February 18, 2011 Physics for Scientists&Engineers 1 12 12 Nonconservative Forces The total energy is conserved, even for nonconservative forces The total energy – the sum of all forms of energy, mechanical or other – is always conserved in an isolated system. We can write this law of energy conservation as ∆Etotal = 0 February 18, 2011 Physics for Scientists&Engineers 1 13 13 Example: Sledding on Mickey Mouse Hill PROBLEM • A boy on a sled starts from rest down snowcovered Mickey Mouse Hill • Together the boy and sled have a mass of 23.0 kg • Mickey Mouse Hill makes an angle of 35.0° with the horizontal • The surface of the hill is 25.0 m long. When the boy and the sled reach the bottom of the hill, they continue sliding on a horizontal snowcovered field • The coefficient of kinetic friction between the sled and the snow is 0.100 • How far do the boy and sled go on the horizontal field before they stop? February 18, 2011 Physics for Scientists&Engineers 1 14 14 Sledding SKETCH February 18, 2011 Physics for Scientists&Engineers 1 15 15 Sledding THINK • The boy and sled start with zero kinetic energy and finish with zero kinetic energy • The boy and sled have gravitational potential energy at the top of Mickey Mouse Hill • As the boy and sled go down the hill, they gain kinetic energy • At the bottom of the hill, the potential energy is zero, and the boy and and sled have kinetic energy • However, the boy and sled are continuously losing energy to friction • Thus, the change in potential energy will equal the energy lost to friction • We must take into account the fact that the force of friction will be different when the sled is on Mickey Mouse Hill than when it is on the flat field February 18, 2011 Physics for Scientists&Engineers 1 16 16 Sledding SKETCH February 18, 2011 Physics for Scientists&Engineers 1 17 17 Sledding RESEARCH • The boy and sled start with zero kinetic energy and finish with zero kinetic energy • We define the length of Mickey Mouse Hill to be d1 and the distance that the boy and sled travel on the flat field to be d2 • Assuming that the gravitational potential energy of the boy and sled is zero at the bottom of the hill, the change in the gravitational potential energy from the top of Mickey Mouse Hill to the flat field is ∆U = mgh (m is the mass of the boy/sled ) February 18, 2011 Physics for Scientists&Engineers 1 18 18 Sledding • The force of friction will be different on the slope of Mickey Mouse Hill and on the flat field because the normal force is different • The force of friction on Mickey Mouse Hill is fk1 = µk N1 = µk mg cosθ • The force of friction on the flat field is fk 2 = µk N 2 = µk mg • The energy dissipated by friction is equal to the energy dissipated sliding on the hill plus the energy dissipated on the flat surface Ef = E1 + E2 = fk1d1 + fk 2 d2 February 18, 2011 Physics for Scientists&Engineers 1 19 19 Sledding SIMPLIFY • The energy dissipated by friction is then Ef = f k1d1 + f k 2 d 2 = ( µ k mg cos θ ) d1 + ( µ k mg ) d 2 • The change in potential energy is ∆U = mgd1 sin θ • Because we have ∆K = 0, mgd1 sin θ = ( µk mg cos θ ) d1 + ( µk mg ) d 2 or d2 = d1 ( sin θ − µk cos θ ) µk February 18, 2011 Physics for Scientists&Engineers 1 20 20 Sledding CALCULATE • Putting in our numerical values we get d2 = ( 25.0 m ) ( sin 35.0° − 0.100 ⋅ cos35.0° ) = 122.9153 m 0.100 ROUND d2 = 123 m DOUBLE-CHECK • Our units are correct for a distance • The boy and sled travel a little more than the length of a football field, which seems reasonable February 18, 2011 Physics for Scientists&Engineers 1 21 21 Clicker Quiz A sled with mass m initially has speed v and slides to a stop on a level surface with coefficient of friction µk after travelling a distance d. Which of the following expressions is correct for the initial speed? A) v = 2 µ k gd B) v = µ k mgd C) v = 2 µ k gd / m D) v = (1 / 2)µ k mgd 2 E) v = 2 m / µ k dg February 18, 2011 12 ∆K = Fd = mv = µ k mgd 2 12 v = µ k gd 2 v = 2 µ k gd 22 Physics for Scientists&Engineers 1 22 Power Produced by Niagara Falls PROBLEM • Niagara Falls pours an average of 5520 m3 of water over a drop of 49.0 m every second • If you could convert all that potential energy to electrical energy, how much electrical power can Niagara Falls generate? THINK • The mass of one m3 of water is 1000 kg • The work done by the falling water is equal to the change in gravitational potential energy of the water • The average power is the work per unit time February 18, 2011 Physics for Scientists&Engineers 1 23 23 Power Produced by Niagara Falls SKETCH • Define the y-axis and the height h February 18, 2011 Physics for Scientists&Engineers 1 24 24 Power Produced by Niagara Falls RESEARCH • The average power is given by the work per unit time W t • The work that is done by the water going over Niagara Falls is equal to the change in gravitational potential energy energy, P= ∆U = W • The change in gravitational potential energy of a given mass m of water falling a distance h is given by ∆U = mgh February 18, 2011 Physics for Scientists&Engineers 1 25 25 Power Produced by Niagra Falls SIMPLIFY • We can combine the preceding three equations to get W mgh m P= = = gh t t t CALCULATE m m3 1000 kg 6 = 5520 = 5.52 ⋅10 kg/s 3 t s m P = ( 5.52 ⋅106 kg/s ) ( 9.81 m/s 2 ) ( 49.0 m ) P = 2653.4088 MW ROUND P = 2650 MW = 2.65 GW February 18, 2011 Physics for Scientists&Engineers 1 26 26 Power Produced by Niagara Falls DOUBLE-CHECK • Our result is comparable to the output of large electrical power plants on the order of 1000 MW (= 1 GW) • If you look up the combined power generation capability of all of the hydroelectric power stations at Niagara Falls, you can see a peak capability of 4.4 GW • The falling water doesn’t produce power simply by falling over Niagara Falls • A large fraction of the water of the Niagara River is diverted upstream from the Falls and sent through tunnels, where it drives generators to generate the power • The water that actually makes it across the Falls during daytime and in the summer tourist season is only about 50% of the flow of the Niagara River • This flow is reduced even further, down to 10%, and more water is diverted for power generation, during nighttime and in the winter February 18, 2011 Physics for Scientists&Engineers 1 27 27 Potential Energy and Stability E, K, and U for a roller coaster February 18, 2011 Physics for Scientists&Engineers 1 28 28 Potential Energy and Force Special points on these plots are marked by vertical gray lines • x1 and x3 represent minima in U • Zero in F and a • x2 represents maximum in U • Zero in F and a • x1, x2 and x3 represent equilibrium points • x1 and x3 represent stable equilibrium points • x2 represents an unstable equilibrium point February 18, 2011 Physics for Scientists&Engineers 1 29 29 Stable and Unstable Equilibrium Definition At stable equilibrium points, small perturbations result in small oscillations around the equilibrium point. At unstable equilibrium points, small perturbations result in an accelerating movement away from the equilibrium point. Mathematically, what makes an equilibrium point stable or unstable is the second derivative of U(x) • Positive derivative (curvature) means stable • Negative derivative (curvature) means unstable February 18, 2011 Physics for Scientists&Engineers 1 30 30 Clicker Quiz Which of the four drawings represents a stable equilibrium point for the ball on its supporting surface? February 18, 2011 Physics for Scientists&Engineers 1 31 31 ...
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This note was uploaded on 04/23/2011 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.

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