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Chapter 7: Conservation of Energy with Frictional Forces Present

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Chapter 7: Conservation of Energy with Frictional Forces Present Lecture 12: Potential Energy; Momentum and Collisions 1 Chapter 7: Conservation of Energy with Frictional Forces Present If there are frictional forces present, then the work done against the frictional (non{conservative) forces is equal to the change (decrease) in the total Mechan- ical Energy. The total Mechanical Energy is not constant when frictional forces are present. The Mechanical Energy will decrease because of the work done against the frictional forces. Wfriction = (Kf + Uf ) (Ki + Ui) − Worked Example A 3 kg block slides down a rough incline 1 m in length. The block starts from rest at the top of the inclined plane, and experiences a constant force of friction of magnitude 5 N. The angle of the incline is 30o. Using conservation of energy, determine the speed of the block when it reaches the bottom of the plane. Wfriction = (Kf + Uf ) (Ki + Ui) − 1 fs = ( mv2 + 0) (0 + mgh) − 2 f − The force of friction f is given as 5 N, the length s over which it acts is given as 1.0 m, and the initial height of the block h may be found from simple trigonometry to be 0.5 m. 1 5 1 = 3v2 3 9:8 0:5 − · 2 f − · · 2 2 2 v = 6:47 m =s = vf = 2:54 m/s f ) In this example, the total Mechanical Energy was not conserved because of the non{conservative frictional forces. The decrease in the Mechanical Energy went into doing work against friction, and that work actually would show up as heat. Lecture 12: Potential Energy; Momentum and Collisions 2 Chapter 7: Conservation of Mechanical Energy in Spring Problems The principle of conservation of Mechanical Energy can also be applied to sys- tems involving springs. First take a simple case of a mass traveling in a horizontal direction at constant speed. The mass strikes a spring and the spring begins to compress slowing down the mass. Eventually the mass stops and the spring is at its maximum compression. At this point the mass has zero kinetic energy and the spring has a maximum of potential energy. Of course, the spring will rebound and the mass will finally be accelerated to the same speed but opposite in direction. The mass has the same kinetic energy as before, and the spring returns to zero potential energy. Spring Potential Energy If a spring is compressed (or stretched) a distance x from its normal length, then the spring acquires a potential energy U spring(x): 1 U spring(x) = kx2 (k = force constant of the spring) 2 Worked Example A mass of 0.80 kg is given an initial velocity vi = 1.2 m/s to the right, and then collides with a spring of force constant k = 50 N/m. Calculate the maximum compression of the spring. Solution by Conservation of Energy Initial Mechanical Energy = Final Mechanical Energy Ki + Ui = Kf + Uf 1 1 mv2 + 0 = 0 + kx2 2 i 2 m v0:8 s u = x = vi = 1:2u = 0:152 m ) k t 50 Lecture 12: Potential Energy; Momentum and Collisions 3 Chapter 7: Conservation of Energy and the Motion of a Pendulum A simple pendulum consists of a sphere of mass m (called a \bob") attached to a very light (massless) string of length L. Initially the mass m can be hanging motionless, straight down from the string. Then the mass m is displaced upward such that the string makes an angle θ0 with the vertical direction. The mass is then released. The mass goes back to the vertical position and acquires kinetic energy. Because the mass is moving it does not simply stop when the string is at the vertical position, but rather continues to the other side until it reaches the same height at which it was first released. It then repeats the motion in the opposite direction. In the limit that there is no friction, the mass of a pendulum will constantly swing back and forth. What is really happening is that there is a continuous transformation of potential energy into kinetic energy, and then kinetic energy back into potential energy. As long as none of the mechanical energy is lost to friction, the motion should continue forever. The motion of the pendulum goes by the name \oscillation" or \simple harmonic motion", and will be studied in greater detail in Chapter 12. For the moment, we can calculate the maximum speed of the mass. The maximum speed of the mass occurs when the mass is at the lowest point of the motion. The zero of potential energy is defined conveniently at the top of the string. Initial Mechanical Energy = Final Mechanical Energy Ka + Ua = Kb + Ub 1 2 0 mgL cos θ0 = mv mgL − 2 b − vb = q2gL(1 cos θ0) (independent of the mass) − Lecture 12: Potential Energy; Momentum and Collisions 4 Chapter 8: LINEAR MOMENTUM and COLLISIONS The first new physical quantity introduced in Chapter 8 is Linear Momentum Linear Momentum can be defined first for a particle and then for a system of particles or an extended body. It is just the product of mass and velocity, and is a vector in the same direction as the velocity: ~p = m~v particle P~ = M~vcm system of particles Why have this quantity? In fact Newton himself introduced the quantity in his version of Newton's Second Law. For the case of a particle one has: d~p F~ = dt d(m~v) d~v = F~ = = m = m~a ) dt dt Here we are making use of the fact that the mass m of a particle does not change with time. The same derivation can be made for a system of particles, or an extended body, as long as we always include all the mass. dP~ F~ = ext dt d(M~vcm) d~vcm = F~ ext = = M = M~acm ) dt dt Conservation of Linear Momentum The important use of Linear Momentum comes about when we consider the special case when there is no net force acting. This defines an isolated system. In that case, the left hand sides of the two above equations are zero. Therefore, the linear momentum of the particle, or of the system of particles, is constant. F = 0 = ~p = constant or ~pi = ~pf ) Fext = 0 = P~ = CONSTANT or P~ i = P~ f ) THE CONSERVATION OF ENERGY LAW AND THE CONSERVATION OF MOMENTUM LAW ARE THE TWO MOST IMPORTANT LAWS OF PHYSICS. Lecture 12: Potential Energy; Momentum and Collisions 5 EXAMPLE of LINEAR MOMENTUM CONSERVATION One example of linear momentum conservation involves the recoil of a cannon (or a rifle) when a shell is fired. A cannon of mass M = 3000 kg fires a shell of mass m = 30 kg in the horizontal direction. The cannon recoils with a velocity of 1.8 m/s in the +^i direction. What is the velocity of the velocity of the shell just after it leaves the cannon ball? Remember that we have to deal with isolated or self-contained systems. In this example the isolated system is the cannon plus the shell, not just the cannon by itself of the shell by itself. The explosion which fires the shell is an INTERNAL force, so it does not enter into the problem. There are no EXTERNAL forces acting in the horizontal direction, so linear momentum is conserved in the horizontal direction P~ i = P~ f The initial linear momentum P~ i = 0 because nothing is moving. The final linear momentum P~ f = 0 also, but it can be expressed as the sum of the linear momenta of the cannon and the shell: P~ i = 0 = P~ f = MV~ + m~v Here V~ is the velocity of the cannon and ~v is the velocity of the shell. Clearly V~ and ~v are in opposite directions, and MV~ 3000 1:8 ~v = = ~v = · = 180^i m/s − m ) − 30 − Decay of Subatomic Particles (Example 8.3, page 247) Another example of conservation of momentum is the decay of an isolated sub- atomic particle such as a neutral kaon written symbolically as K0. A neutral kaon decays into two other subatomic particles called charged pions, symbolized + as π and π−. The decay equation is written as 0 + K π + π− ! By conservation of momentum (see pages 247{248) we can easily prove that the two pions have equal and opposite momenta. Lecture 12: Potential Energy; Momentum and Collisions 6 Impulse of a Force We define another vector physical quantity called the Impulse of a Force. In the simplest case, if a constant force F acts over a short period of time ∆t, then the impulse of that force is equal to the product of the force and the length of time over which it acts. The impulse is denoted by the symbol I ~I = F~ ∆t (constant force F) ~I = ∆~p If the force is not constant, then the definition of impulse requires an integral ~I Z F~ (t)dt = ∆~p ≡ The impulse calculation is useful in determining how much force or momentum is involved in violent collisions lasting very short periods of time. The impulse of a force is a useful vector quantity for determining how much force or momentum is involved in violent collisions lasting very short periods of time. By definition, the impulse ~I is given as the product of the average force and the time over which the force was exerted t0=t+∆t ~I F~ ∆t = ~I = Z F~ dt ≡ ) t0=t However, by Newton's Second Law the average force can be written as: ∆~p ∆~p F~ = = ~I = ∆t = ∆~p ∆t ) ∆t The book calls this equality the Impulse-Linear-Momentum-Theorem but it is just a simple consequence of Newton's Second Law of motion.
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