the static friction is replaced by kinetic friction. There is a net force F net = F push f k in the direction of F push.

Transcription

1 the static friction is replaced by kinetic friction. There is a net force F net = F push f k in the direction of F push. Exaple of kinetic friction. Force diagra for kinetic friction. Again, we find that f k N, or Figure 42: Kinetic friction. where µ k is the coefficient of kinetic friction. f k = µ k N, (4.29) Rolling friction f r : the frictional force that appears when an object rolls on a surface. As in equations (4.28) and (4.29), we find that f r N, or f r = µ r N, (4.30) which defines the coefficient of rolling friction, µ r. Typical values for various object/surface coefficients of the various frictional forces are shown in table 1. Notice, in particular, that µ k < µ s ; i.e. the kinetic frictional force is less than the axiu static friction. 65

2 Materials µ s µ k µ r Rubber/Concrete Steel/Steel (dry) Steel/Steel (lubricated) Wood/Wood Wood/Snow Ice/Ice Table 1: Typical friction coefficients. µ r µ k : i.e. it is easier to roll an object on wheels than to slide it. Figure shows how the frictional force changes as the applied horizontal force F push on an object increases. F push = 0 : object is at rest. F push < f s ax : static frictional force increases proportionally with F push. F push = f s ax : applied force is equal to the axiu static frictional force. The object slips, and begins to accelerate. The frictional force on it is now kinetic. F push > f s ax : object is oving. Kinetic frictional force reains constant Drag Force Recall that the drag force is a resistive force acting on an object due to its otion through a fluid. e.g. objects oving through air experience a drag force due to air resistance. 32 Knight, Figure 5.13, page

3 Figure 43: Variation in frictional force with F push. 67

4 Object Drag coefficient, C D Strealined body 1.00 Sports car 0.80 Typical car 0.10 Racing cyclist 0.50 Table 2: Typical drag coefficients in air. To quantify the drag force, D, we suppose that it is in soe way proportional to the following: the speed of the object, v s. the cross-sectional area of the object, A. the density of the fluid, ρ. Fro diensional analysis (see Additional Materials 33 ), we derive the for D = C D A vs 2 ρ (4.31) for D, where C D is the drag coefficient. Therefore, D depends upon the size and shape of the object, but not on its ass, as well as on the density of the fluid. Note: in general, C D is not constant it depends upon factors such as the object s speed and shape, roughness of the object s surface and viscosity of the fluid. Soe typical values of C D for objects in air are shown in table 2. Consider the otion of a ball being thrown vertically upwards (figure ). If D = 0 (i.e. no air resistance), a s = g throughout its flight. If D 0, a s changes. 33 Derivation of the Drag Force via Diensional Analysis, ne131/ 34 Knight, Figure 5.21, page

5 Figure 44: Vertical otion of a ball. (1) Ball rises with a net force (F net ) y = W + D y = g D acting on it, so by Newton s 2nd Law, its acceleration is a s = (F net) y = g D ( = g + D ) i.e. ball decelerates, and agnitude of this acceleration is > g. (2) Fro equation (4.31), D decreases as ball s speed decreases., (4.32) (3) At the axiu height of ball, its velocity v s = 0, so that D = 0, and ball s acceleration is siply a s = a free fall = g. (4) Fro equation (4.31), D increases as ball s speed increases. (5) Net force on ball is (F net ) y = W D y = g + D. By Newton s 2nd Law, its acceleration is a s = (F net) y = g + D ( = g D ). (4.33) Equations (4.32) and (4.33) show that since D is independent of the object s ass: 69

6 Upwards otion: less-assive objects subject to larger acceleration than ore-assive objects of sae size. Downwards otion: less-assive objects subject to saller acceleration than ore-assive objects of sae size. Galileo s experient: although haer > feather, the cross section of the feather is greater then that of the haer, and so the feather experiences a greater drag. Terinal speed Fro (4.31), D varies quadratically with v s. Furtherore, in the exaple shown in figure 44, we observe that the net vertical force on a falling object is (F net ) y = g + D (4.34) i.e. the weight and drag forces act in opposite directions. If the object falls far enough, the agnitude D will increase until it becoes equal to its weight. The net force (F net ) y in equation (4.34) will thus disappear: There will be no further acceleration of the object - it will reach a constant speed, and so by equation (4.31), D will also reain constant. This constant speed, at which D = g, is called the terinal speed v ter. Once the object reaches this terinal speed, it will continue falling until it reaches the ground. Using equation (4.31) in the condition D = g, i.e. C D A v 2 ter ρ = g, 70

7 yields the expression v ter = g C D Aρ (4.35) for v ter, i.e. V ter (thus, a ore assive object has a larger terinal speed than a less assive object). Consider the speed-tie graphs for free-fall otion with and without drag (figure 45); the effect of the forer can clearly be seen. Without drag. With drag Figure 45: Speed-tie graphs for free fall including and not including the drag force due to air resistance. Note: although the above refers to objects in free fall, D can also act to slow down objects oving horizontally in the sae way as those oving vertically. e.g. an airplane reaches a axiu speed when the drag force is equal and opposite to the thrust (e.g. for a passenger jet, this axiu speed is 550 ph) Relative Motion Recall our exaple involving the aircraft containing the following observers 71

8 Man fixed to Earth observes an airplane accelerating along the runway. Stewardess standing still in aircraft. Let us change the exaple such that The aircraft now oves along the runway such that its speedoeter reads a constant speed of 20 s 1 (i.e. inertial reference frae). The stewardess walks along the aisle, towards the cockpit, with a constant speed of 2 s 1 (i.e. inertial reference frae). Fixing a reference frae to both people, what does the an and stewardess see? Man: Observes airplane oving with a constant velocity of 20 s 1. Observes stewardess oving with a constant velocity of 22 s 1. Stewardess: Observes airplane oving with a constant speed of -2 s 1. Observes an oving with a constant speed of -22 s 1. So each person disagrees on what the speed of the airplane is. Which one is correct? Each observer s co-ordinate syste is an inertial frae - so both are correct. Each of the clais to be at rest - what they are really doing is reporting the relative otion of the aircraft, i.e. relative to their own reference frae. So how do we analyse the dynaics of an object in differing inertial fraes? 72

9 Let an object have position vectors r and r in inertial reference fraes S and S, respectively. Velocity of object in S and S are v and v, respectively. Frae S oves at a constant speed V with respect to frae S. It can be easily shown (see Additional Materials 35 ) that the velocities are related by v = v + V, v = v V. (4.36) Equation (4.36) constitute the Galilean Transforations (intuitively obvious?). Thus, if we know an object s velocity in one inertial frae, we can transfor it into the velocity easured by an observer in a different inertial frae. Exaple: 36 Police are chasing a bank robber. While driving at 50 s 1, they fire a bullet to shoot out a tire of his car. The police gun shoots bullets at 300 s 1. What is the bullet s speed relative to a TV caera crew parked beside the road? Solution: confining all otion to 1-diension, let frae S be fixed to the TV crew, and frae S fixed to the police car. We therefore have the following inforation: v x = 300 s 1, V x = 50 s 1. Using the x-coponent equation of (4.36), the speed v x of the bullet in the TV crew s frae is v x = v x + V x = = 350 s The Galilean Transforations, ne131/ 36 Knight, Exaple 6.8, page