[Lab #5 - AIR RESISTANCE LAB Purpose To investigate the effects of air resistance on falling objects. Equipment Small coffee lters meter-stick or two Introduction Question: Why does a hammer fall faster than a feather? Answer: Because air resistance has a greater affect on the motion of the feather than on the hammer. Why? Air resistance (R) is a frictional force that opposes the motion of objects moving through air. The degree of air resistance that a falling object experiences is proportional to its surface area (i.e., the greater the surface area, the greater the air resistance). On the other hand the force of gravity (i.e., weight (or W)) on an object is proportional to its mass. Ihus the effect of air resistance is greater on low weight, high surface area objects (like feathers) than on high weight lower surface area objects (like hammers). As it happens, both these factors are embodied in an obj ect's density (Density = massfvolume). High density objects like hammers have relatively high mass (and therefore high weight) and low surface areas compared to low density objects (like feathers) which have relatively low mass and weight and high surface areas. Recall that 2F = ma. Ihmihm an object\" s acceleration must decrease if the net force on the object decreases. [The net force (FNET) on an object equals the sum of the forces (SF) on the object (i.e., FNET = EF).] Now the net force on a falling object is equal to its weight minus air resistance (i.e., 2F = W R). And so, the net force on a falling object decreases if R increases. Thus acceleration drops with increasing R. As it also happens, the air resistance experienced by a falling object depends on its velocity relative to the air. Thus, the faster an object falls, the greater the air resistance it experiences. As a result, the acceleration of a falling object decreases as its velocity increases (because the net force on the object decreases as its velocity increases). When the air resistance has increased to the point at which it equals the object weight, the net force drops to zero (and therefore so does its acceleration). At this point the falling object has reached its maximum possible velocity. This velocity is known as terminal velocity (because it's the top end velocity for a falling object). Because of its low density, a feather has a large surface area-to-weight ratio while, because of its high density, a hammer has much smaller surface area-to-weight ratio. 4.1 As a result air resistance on a falling feather almost immediately equals its weight while the air resistance on a falling hammer takes a much longer time (by comparison) to reach terminal velocity. Elms a dropped feather accelerates for only an instant before reaching its terminal velocity (yx) while the hammer continues to accelerate to higher and higher velocities for a much longer time. Because they have a relatively large surface area-to- weight ratio, the common coffee lter nicely demonstratethe effect air resistance. If we drop two same sized coffee filters side-by-side, they fall at the same rate and both hit at the ground at the same time. On the other hand, if drop at the same time a single coffee lter and a double weight lter (two single weights stuck together), the double weight lter will hit the ground rst because of its higher density. Question: Just how does the air resistance experienced by an object falling through air affect its velocity? In today's lab we shall test two hypotheses of how air resistance affects the velocity of a falling object. To do so will make use of the facts that at terminal velocity (m), air resistance equals the weight of the falling object and that distance fallen after reaching terminal velocity is equal to the product of terminal velocity and the time it spends falling at terminal velocity (or: d = yxx t.) | Eote: It is assumed that the coffee lters reach terminal velocity almost instantly after being released] First Hypothesis Air Resistance is directly propoliv'tt'ortoir to velocity (i.e., R CC V). If air resistance actually is proportional to velocity, then the terminal velocity must also be proportional to the object's weight (W), because the air resistance equals the weight at terminal velocity (i.e., because R = W at terminal velocity). So, if hypothesis 1 (i.e., R CC V) is correct, then 3;; 0c W. So (since d = m): t) we can also say: d 0c W x t. Thus (if hypothesis 1 is correct), a double lter (of weight 2W) will fall twice the distance (2d) and a triple tilter (3W) will fall three times the distance (3d), etc., etc., etc., than a single lter (1W) in the some amount of time We can test hypothesis 1 by seeing if experimental results match the above predictions. 4 17. Second Hypothesis Air resistance is proportional to the square of the velocity (R cc 122). If R 0c v2, then W 0c y___ because R = W at terminal velocity). So: vT 0c M , and therefore: (1 0c 3 W t (because d = 339): t) g the distance fallen is proportional to the square root of the weight. Under these conditions, if we drop a single lter from one meter above the oor and simultaneously drop a double lter from 1.41 meters (i.e., the '5 ), they will both hit the ground at the same time. We can test hypothesis 2 by seeing if the experimental results match the predicted behavior. Procedure 1 Test hypothesis 1 (R ~ v) In this case, it ~ Wt. If air resistance actually is proportional to velocity, then a double lter, when dropped from twice the height of a single lter, will hit the ground at the same time as the single lter when the single and double lters are released at the same time. To test hypothesis 1, drop the lters from the heights given on the procedure lworksheet and see if they all hit the floor at the same time. Do three trials from each height. If all the lters do hit the ground at same time, then your results strongly support hypothesis 1. But, if all (or even just some) don't hit very nearly at the same time, then your results (if accurate) disprove the hypothesis. Procedure 2 Test hypothesis 2 (R ~ 122). In this case, a ~ W t If air resistance actually is proportional to velocity squared, then a double lter, when dropped from the square root of W the height of a single filter, will hit the ground at the same time as the single lter when the lters are released at the same time. To test hypothesis 2, drop the coffee lters from the heights given on the procedure 2 worksheet and see if they hit the floor at the same time. Do three trials from each height. If all the lters hit the ground at very nearly same time, then your results strongly support hypothesis 2. But, if all (or even just some) don't hit at very nearly at the same time, then your results (if accurate) disprove the hypothesis. {Eotez The care and precision with which you carry out your tests will strongly affect your test results.) Diagram for start of 2 meters Procedure 1 1 meter Diagram for start of Procedure 2 2 meters 1.41 metersWORKSHEET NAMES CLASS DAY AND TIME Procedure 1 Test Results HEIGHTS ( In METERS) HIT AT THE SAME TIME ? (Y/N) SINGLE FILTER DOUBLE FILTER Trial 1 Trial 2 Trial 3 0.500 1.000 0.600 1.200 0.700 1.400 0.800 1.600 0.900 1.800 1.000 2.000 Procedure 2 Test Results HEIGHTS (In METERS) HIT AT THE SAME TIME ? (Y/N) SINGLE FILTER DOUBLE FILTER Trial 1 Trial 2 Trial 3 0.500 0.707 0.600 0.846 0.700 0.990 0.800 1.131 0.900 1.273 1.000 1.410