Constant Force: Projectile Motion

Transcription

1 Contant Force: Projectile Motion Abtract In thi lab, you will launch an object with a pecific initial velocity (magnitude and direction) and determine the angle at which the range i a maximum. Other tak, uch a determining the initial velocity, comparing the meaurement to model calculation and calculating the angle at which to launch to trike a wall at the highet point may be performed. Introduction We have done everal lab in which a contant force wa applied to an object, reulting in a contant acceleration (a indicated by Newton' Second Law of motion AKA, the energy principle): F NET =m a (1) (Recall that ince m i a calar, the acceleration i in the ame direction a the new force). In thoe experiment, we implified matter by tarting at zero initial velocity. Then, the velocity wa alway in the ame direction a the acceleration (and o the ame direction a the force): v f = at (2) Again, a calar multiplying a vector give a new vector in the ame direction. But thi will not be true for the cae we're doing today, where the initial velocity i NOT zero. Although the CHANGE in velocity will be in the ame direction a the acceleration: v= v f v 0 = at (3) the FINAL velocity will NOT be in the ame direction a the acceleration: v f = v 0 + at (4) Note that the final velocity i the um of two vector (the initial velocity and the change in the velocity), and will NOT in general be in the ame direction a either of thoe two. The figure how the parabolic path of a v 0 projectile, with the a t velocity hown at two point. At the ide i hown the vector v addition given in f Equation (4): Note that all three Drawing 1: Parabolic projectile path with velocity at two point. vector ( v 0, v f, at ) are in different direction.(you hould draw a velocity vector with correct direction AND magnitude at the mall cro. Draw the analogou vector addition diagram. Note well that there' no guarantee that the triangle you

2 make will be a right triangle like mine i.) Neglect Air Drag When conidering projectile, you're told invariably to neglect air drag. So much that it almot eem like phyicit are a little defenive about air drag. We aren't. It jut really i the cae that mot of the time (not ALL of the time) air drag i a mall effect compared to weight. You can etimate the likely effect of fluid drag (notice how I changed air to fluid? Tricky.) in the following way. Uing hi Second and Third Law of Motion, Iaac Newton developed an expreion for the magnitude of the force exerted by a moving fluid upon an object: = 1 2 ρ A v2 (5) (You hould check that the unit work OK.) To be fair, thi expreion wa already known. Leonarda DaVinci had already argued that drag force were proportional to cro ectional area (A) and Galileo had claimed that drag wa proportional to the denity of the fluid ( ρ ). And a cientit named Marriotte had uppoedly hown that the drag force wa proportional to the quare of the peed (but I haven't been able to track thi down for ure). But Newton deerve ome credit for howing that thi expreion follow from the fundamental law of motion (on the other hand, it' alway eaier to get the right anwer if you already know what the right anwer i :). A couple of hundred year later, in the early 20 th century, the development of fluid dynamic wa ued to how that the equation for drag look like: = 1 2 C Dρ Av 2 (6) See the difference? It' not eay. The only difference i that there i a drag coefficient, C D. And the drag coefficient only varie from about 0.1 (for omething very treamlined) to about 1 (for omething blocky, like a truck). So the baic phyic i in Equation (5). To undertand omething about thi expreion, imagine that I releae an object from high up. Initially, the only force acting will the weight, acting down. So the object will begin to accelerate downward. A the peed increae, the drag force, acting oppoite the motion (or up) will increae. So there will be two force, the larger one (weight) acting down and the weaker one (drag) acting up. So the net force will be maller. I can write Newton' Second Law of motion a: F NET = W + F DRAG =W down+ 1 2 C DρA v 2 up=m a (7) Now, noting that up and down are oppoite direction (that i, down= up ), I can rewrite (7) a: (W 1 2 C Dρ A v 2 ) down=m a (8) So we ee that the a long a the weight i greater than the drag (in thi cae, it will be), the acceleration will alway be down. That i, the object will continually peed up. But a the peed increae, the NET force will get maller and maller, o the acceleration (while till pointing down) will get maller and maller. Eventually, (if I drop the object from high enough), the acceleration will get o mall that

3 it might a well be zero (i.e., too mall to meaure). And that mean that the peed will top increaing. The object won't jut STOP. Zero acceleration doen't mean mean zero velocity. Zero acceleration jut mean UNCHANGING velocity. So the velocity of a falling object will approach a contant magnitude obtained from etting the force in Eq. (8) to zero: v TERM= 2W C D ρ A (9) Thi give a way of calculating the terminal velocity of variou object. Shortly, we'll ee how thi will allow u to ae whether we have to worry about air drag for a projectile. A an example, I'll calculate the terminal velocity of a kydiver. The weight will be in the ball park of 700 N. The denity of air at room temperature i about 1.2 kg/m 3 (you can get thi from the ideal ga law). The area of a peron falling face firt i about A (2 m)(0.3m)=0.6 m 2. When in doubt, I ue a value of 0.5 for the drag coefficient. Putting it all together, I get a terminal velocity of 60 m/ (about 140 mph, which i what the Interweb ay it i...). But the real goal i to be able to etimate in advance whether air drag will be an iue or not. To do that, go back to Equation (6). I can ue the expreion for terminal velocity (Eq. (9)) to rewrite the magnitude of the drag force a: v =W( v TERM ) 2 There' nothing really new here (thi i really the ame a Eq. (6)) but it' in a form that tell u omething important. The drag force will be mall, COMPARED TO THE WEIGHT, whenever the peed i le than the terminal velocity. And it goe a the quare, o that if the peed i one-tenth of the terminal velocity, the drag force will be one percent of the weight (right?). So we can now ee a way to tell pretty eaily whether we have to worry about a drag force or not. Etimate the verminal velocity (from Eq. (9)) and ue it in Eq. (10). For example, when Uain Bolt et the world record in the 100 m race, wa air drag a ignificant factor? Uing the reult (100m in 9.58 econd), the drag force i about: =W( 10.44m/ 2 60m/ ) =0.03W (11) So, no, air drag i not a ignificant factor, at leat compared to hi weight. (On the other hand, 3% can be a big deal in world record conider that the record before Uain Bolt wa 9.74 econd, by Aafa Powell). In today' lab, you'll launch a phere made of teel or wood or hard platic. Should you be worried about air drag in interpreting your reult? Contant Acceleration If we DO ignore air drag, then the kinematic i eaier. The acceleration i contant: a= g 9.8 m/ down=9.8 m/ (up)=9.8 m/ (10) up (12) o it make ene to orient one axi of our two-dimenional problem along that acceleration. Conventionally, people orient the y-axi along g (but with the poitive direction up intead of down weird). So that' what we'll do. Then, the kinematic i eay. The generic expreion for poition

4 when the acceleration i contant: r=r 0 +v 0 t+ 1 2 a t 2 (13) become two equation (one for x and one for y): y= y 0 +v 0y t 1 2 g t 2 x= x 0 +v 0x t (14) I want to find out how far the object will fly until it hit the ground ( y=0 ), o I call the horizontal ditance travelled the range, R= xx 0, and the initial height, y 0 =H. After a whole bunch of algebra (I lay it out for you later), the equation for the range become: R= v 2 0 g inθ coθ [ 1+ 2gH θ] 1+ (15) v 2 0 in 2 where v 0 i the initial peed and θ i the angle at which it i launched. There' no eay way to ue thi formula to find the maximum range, but it i traightforward to find it experimentally. Jut load up and fire. And ue the equation to compare the meaured range to the calculated range. But note that you need everything on the right ide. So ee if you can figure out how to meaure each of thoe parameter. The Experiment In hort, the experiment conit of launching a projectile at varying angle (but keeping a many other parameter the ame a poible) to find the maximum range and the angle at which it occur. Note that you will need to figure out how to find the initial peed and initial height. You may ue g=9.8n /kg. Meauring the ditance to the landing point i eaier if you lay a piece of paper down with a piece of carbon paper on top the projectile will make a mark on the paper that you can eaily ee. Concluion A you may have gueed, the maximum range will NOT occur at a launch angle of 45º. Why do you uppoe that i? Your lecture teacher didn't lie, o apparently there i ome difference between what we're doing here and what your lecture teacher howed you. Do you uppoe you could figure out how to launch a projectile at a wall a fixed ditance away o that it hit the wall at the highet point? If your projectile hit the highet, you win.

5 APPENDIX To ee how the range equation come about, tart with the kinematic equation for 2D motion: y= y 0 +v 0y t 1 2 g t 2 x= x 0 +v 0x t (16) Make the ame definition a before: the landing point i the ground ( y=0 ), the horizontal ditance travelled the range, R= xx 0, and the initial height, y 0 =H. Then the kinematic equation become: 0=H+(v 0 inθ)t 1 2 gt 2 R=(v 0 coθ)t (17) (Note that I've ued the initial velocity, v 0 =v 0x î+v 0y ĵ=v 0 (coθ î+inθ ĵ) ). I can olve the x- equation for time ( t=r/(v 0 coθ) ), ubtitute thi into the equation for y, and get an equation that tell me how far the object goe (but not when it get there): g 0=H +tanθ R 2v 2 0 co 2 θ R2 (18) So far, the algebra i traightforward and well within your capabilitie. But the next tep i kind of tricky (it involve the quadratic formula), o THIS you aren't reponible for. If you DO go through it, you get the final expreion for the range: R= v 2 0 g inθcoθ [ 1+ 2gH θ] 1+ (19) v 2 0 in 2 Thi i Equation (15). Note that the maximum range will not be when the launch angle i 45 degree. A the figure how, an angle that give maller range AT THE SAME HEIGHT will give a longer range if the landing point i below the launch point. (Eentially, the lower height give more time for the larger x-velocity to give a larger x-ditance). Two different angle 1.5 Height [m] Range [m]