Dynamics What You Will Learn How to relate position, velocity, acceleration and time. How to relate force and motion. Introduction The study of dynamics is the study of motion, including the speed and acceleration of a part, the path it takes, and the orientation of the part during its journey. Position, velocity, acceleration and time can be related through the techniques of calculus, and will be the focus of the first part of this section. The second half of the section deals with how to relate force to motion. There are several methods available to engineers, and none will be discussed in detail. Instead, simple problems will be solved by each method so that we can compare and contrast the benefits of using each so that you will be able to pick the best method for a problem. A typical industrial application might be setting requirements for motors for a robot based on the type of work it will do, and the speed at which the task must be accomplished. Another area where dynamics concepts play a large role is in the design of space missions, whether a simple orbit, or the meeting up of a shuttle with the space station, the trajectories must be carefully choreographed using the tools introduced in this section. Relating Position, Velocity, Acceleration and Time Often times, position as a function of time may be known, but not acceleration or velocity. In other situations acceleration may be known as a function of velocity, but it is desired to know the position over time. The following examples may be useful in demonstrating that this is not just a theoretical problem. Examples 1. Braking Distance a. Given: A top FSAE student competition car going 6kph can decelerate at about 1g. b. Find: The distance needed to stop. c. Mathematical rephrasing of question: Given a constant acceleration, what is the change in position, x if velocity changes from 6kpm to zero? This will require knowing velocity as a function of position, v(x). 2. Mars Landing a. Given: A scientific mission to Mars will be out of communication during descent. b. Find: The time to touchdown, so that the mission can be planned accordingly. c. Mathematical rephrasing of question: The gravitational force will increase as the lander approaches the surface, so the acceleration will be a function of position, a(x). We want to know v(x), where x is the height of the surface. 1 D y n a m i c s
3. Human Cannonball a. Given: You are pursuing a career as a human cannonball. You know the launch velocity, and can assume gravity is a constant. b. Find: how far away to put landing pad. c. Mathematical rephrasing of question: You have vertical acceleration as a function of time a(t), but you need to get distance as a function of time x(t). How can we solve these problems? We must start with the most basic of definitions, first that velocity is the change in distance over time: change in distance velocity = change in time In the language of calculus, we can write this as dx t v t = Similarly, acceleration can be defined as the change in velocity over the change in time change in velocity acceleration = change in time In the language of calculus, we can write this as v t a t = We can combine these equations to get acceleration in terms of position: 2 D y n a m i c s a t = v t = d dx t = d2 x t 2 In order to reverse these calculations, we integrate, which is the opposite of differentiation: dx t v t = dx = v(t) x t = v t Similarly, we can find velocity by integrating acceleration: dv t a t = dv = a(t) v t = a t There is one more relationship that we need in order to be able to work with any combination of dependent and independent variables. Start with the definition of acceleration a = dv Now we will multiply by a clever form of one. a = dv dx dx 1 Now rearrange the numerator on the right hand side. a = dx dv dv = v dx dx v This relationship allows us to relate acceleration to velocity when velocity is a function of position. The chart below is a graphical representation of the relationship between position velocity acceleration and time. The arrows in front of each equation indicate which way the equation works
a(x) òadx=òvdv a = vdv/dx v(x) ò=ò(1/v)dx v = dx/ * a(v) òdx=ò(v/a)dv a = vdv/dx* ò=ò(1/a)dv a = dv/* * These three are tricky. Substitute original known back into equation after differentiation to remove unwanted variable. x(t) a(t) òa=òdv a = dv/ v(t) òv=òdx v = dx/ Examples 1. Sprinting Given: A sprinter accelerates from rest at a constant 9 m/s 2 for 1.5 seconds before reaching maximum speed. Find: What is the maximum speed? Solution: v = a = 1.5 9 = 9t 1.5 = 13. 5 m/s 2. Braking Distance Given: Given an initial velocity of 6kph and a constant acceleration of 1g = 9.81 m/s 2. Find: How far does the car go before stopping? Solution: Acceleration is a constant, and we want to know v(x), so we can start in the upper left box, a(x), and move to the right, to v(x) using the relationship adx = vdv Including the limits of velocity of 6kph (16.7m/s) to kph, and the limits on displacement of at the starting point, and the unknown final distance x, we get: x adx = vdv 6 We know acceleration, so we can integrate both sides. x x adx = 9.81dx = 9.81 x x = 9.81 x = 9.81x Thus 16.7 vdv = 1 2 v2 16.7 = 1 2 2 16.7 2 = 278.9 9.81x = 278.9 x = 278.9 9.81 = 28. 4 m 3 D y n a m i c s
3. Pendulum a. Given: The horizontal position of a pendulum is given as x t = 3 cos t [m] b. Find: The maximum acceleration. c. Solution: We see that we must differentiate twice to get from x(t) to a(t): a t = d2 dx 2 x t = 3 cos t [m/s2 ] But at what time is the acceleration maximum? This is an important point, so think this over carefully. In this case, the acceleration is at maximum when velocity is zero. We could plot sine and cosine curves to show you graphically, but it might be more helpful to simply imagine yourself on a swing, being pushed very high up. Your stomach will churn most not at the bottom, where velocity is high, but at the extremes of height, where you stop to swing back in the other direction. To find when velocity is zero, we can differentiate x(t) once: v t = d x t = 3 sin t [m/s] dx And then set velocity equal to zero, and solve for the time: v t = = 3 sin t t = asin t =, [s] Of course there will be many times when v =, but we only need one. Substitute this time into the acceleration equation to find out what the maximum acceleration is: a Max = a t = = 3 cos = 3 m/s 2 Of course, with another time where v=, we could find that the value of a max is positive. It depends on the direction the pendulum is swinging. Relating Force and Motion There are several equally fundamental ways to relate force and motion. They are: 1. Newton s Second Law (Force equals mass times acceleration) 2. Tracking the energy added, removed, and stored in the system as it moves a certain distance. 3. Tracking the change in momentum of a system over time Pay careful attention to the wording above. Momentum and time are related, as are energy and distance. Newton s second law is the only one that incorporates acceleration directly. We saw in the previous section that all these parameters can be related, so it shouldn t be surprising that we say that these approaches are equally valid. Based on the known quantities, and the goals of the analysis, one of these methods may be more direct than another. All the methods require knowledge of force and mass, but the parameters of motion are different. This is summarized in the table below: 4 D y n a m i c s Method Acceleration Velocity Displacement Time F=ma X Energy X X Momentum X X Proficient use of these methods will require more time than we have, and much more practice, but we will solve a single problem with each method, just to demonstrate the idea.
Example Box Being Pushed Given: a 3kg box at rest on a flat conveyor belt. The belt starts to move, and a friction force of 5N is applied to the box. Goal: Apply each of the methods to this problem, and see what information they provide. The first step with any of these methods (and most engineering problems!) is to draw a Free Body Diagram (FBD), which we first encountered when discussing Statics The picture above does a good job showing the situation, but the forces and masses aren t obvious. A FBD is an idealized sketch of the problem, and has all forces and masses clearly labeled. It is a useful stepping stone between a physical problem and a mathematical equation. F weig ht y F Friction = 1N x Method 1: Newton s Second Law Review of the Method The mathematical expression of Newton s Second Law should be familiar from physics class. acceleration of a body is proportional to the force, and inversely proportional to the mass. F = ma The Example We can apply Newton s 2 nd Law in order to find the acceleration of the box in the x direction. Note that we include a summation sign. If there were more than one force in the x direction, they would simply be added, just as we say in statics: ΣF x = ma x 1 N = 3 kg a 1 N a = 3 kg = 1 m 3 s 2 Using the techniques of the previous section, we could now find velocity as a function of time or distance, but extra steps run counter to the engineering ethos of optimization of effort. If we need to know velocity, distance or time, it might be easier to use another method. 5 D y n a m i c s
Method 2: Energy and Work Review of the Method The mathematical expression Kinetic Energy (K) should be familiar from physics class. K = 1 2 mv2 The quantity of Work as used by engineers is equal to the magnitude of an applied force, times the distance the part moves: Work = Force Distance Work and energy are related by the following equation: K 1 + W 1 2 = K 2 Where K 1 and K 2 are the kinetic energy at the beginning and end of the move, and W 1 2 is the work done as the part moves over a certain distance between points 1 and 2. Example Continuing with our example of the box on the conveyor belt, we weren t really given enough info about the problem. Let s say that we want to find out the velocity after the box has moved 5cm. We can start the solution by finding the kinetic energy at x= and x=5cm: K 1 = 1 2 mv 1 2 = 1 2 m 2 = K 2 = 1 2 mv 2 2 = 1 2 1kg v 2 2 = 5v 2 2 J And the work done on the box will be: W 1 2 = F d = 1N.5m =.5[ J] Substituting these values into the governing equation gives: +.5 = 5v 2 2 v 2 =.5 5 =.31 m s There are several important things we should emphasize. First, we never needed to find acceleration. Second, we never needed to consider any vectors. This may not be as apparent with such a simple example. Work and energy are both scalars, and so we did not need to be concerned with vector algebra. That is to say, the math is relatively simple, even for more complicated problems. 6 D y n a m i c s
Method 3: Impulse and Momentum Review of the Method The quantities of impulse and momentum are likely familiar to you, but it is likely that you have not seen a rigorous mathematical definition of them before. The definition of momentum aligns closely with the everyday use: Momentum = Mass Velocity The higher the mass of an object, and the higher the speed, the less you want to get in its way. For example, Indiana Jones is rightly afraid of the giant rolling stone. It has both high mass and high velocity. Impulse is simply the change in momentum over time, which is the same as integrating the applied force over time. It can be calculated as t 2 Imp 1 2 = F t 1 The momentum at time 1 and time 2 are related by an equation that is very similar in form to the equation for work and energy: mv 1 + Imp 1 2 = mv 2 Where mv 1 and mv 2 are the momentum at the beginning and end of a time interval, and Imp 1 2 is the impulse imparted on the part between time 1 and 2. Example Returning for a final go on our example of the box on the conveyor belt, again, we were not given enough info about the problem. Let s say that we want to find out the velocity after the belt has been moving for.5 seconds. We can start the solution by finding the momentum t= and t=.5 sec: mv 1 = 3[kg] 2 = mv 2 = 3 kg v 2 = 3v 2 And the impulse applied to the box by friction with the conveyor belt will be: t 2.5 W 1 2 = F = 1 = 1 t.5 kg m = 5 t 1 s Substituting these values into the governing equation gives: + 5 = 3v 2 v 2 = 5 m 3 s There are several important things we should note. First, we never needed to find acceleration, or know anything about the distance travelled. Second, it is very important to note that momentum and impulse are vectors, and while this problem is very simple, care must be taken to keep track of direction in more complex problems. 7 D y n a m i c s
It might be interesting to see how we can find the momentum-impulse balance starting with Newton s Second Law. F = ma Using the definition of acceleration, we can put this in terms of velocity and time. F = m dv If we assume that mass is constant, we can bring it inside the derivative. F = d mv We can move the to the other side of the equations F = d mv integrate this to get: t 2 v F = mv 2 v 1 = mv 2 mv 1 t 1 Finally, rearrange the above equation to get the balance we expected. mv 1 + t 2 F = mv 2 t 1 Conclusion Relating acceleration velocity time and position allows you to manipulate any knowledge you may have about a problem to either answer a question directly, or put it into a form that is easier to use. It is not a theoretical exercise. The three ways we discussed to relate force and motion, Newton s 2 nd Law, Work and Energy, and Impulse and Momentum, can all be applied to any problem. An important question for the practicing engineer is to determine which is most appropriate for the job at hand. 8 D y n a m i c s