EPT2 objects, same shape, surface, and size. different masses.
both are dropped from an airplane.
before either object reaches terminal velocity, will they accelerate at the same rate?
hey physics geniuses
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Deleted User 3420
24492 Posts
2 objects, same shape, surface, and size. different masses. both are dropped from an airplane. before either object reaches terminal velocity, will they accelerate at the same rate? | ||
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XCetron
5226 Posts
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Deleted User 3420
24492 Posts
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L
Canada4732 Posts
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Crisis_
United States165 Posts
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Clutch3
United States1344 Posts
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TheYango
United States47024 Posts
Drag force is given by -0.5ACdv^2 Where A is the surface area, C is the drag coefficient, d is the density of the fluid, and v is the velocity. Since they both feel the same drag force initially, they experience different accelerations, because the drag force is not proportional to the mass of the object. Since their acceleration is different initially, their acceleration functions have to be different. Another way to look at it is that since terminal velocity is proportional to the square root of mass, they must have different terminal velocities, and therefore must reach 0 acceleration at different rates. | ||
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iloahz
United States964 Posts
cancel the mass you have a=g | ||
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TheYango
United States47024 Posts
On October 23 2008 08:18 Crisis_ wrote: Both objects will accelerate at the same rate if they are dropped at the exact same time. Gravity is the only acceleration acting upon both objects in this free-fall simulation, so right before one object reaches terminal velocity, their accelerations will be exactly the same, regardless of mass. Um, I assume the problem implies air resistance is present. He only ruled out air pressure variability and wind GUSTS (where the fluid itself as a result of forces other than the motion of the objects). | ||
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micronesia
United States24753 Posts
edit: but yeah, they will have different acceleration curves edit2: because frictional force will be the same for both but one has less inertia edit3: they should also have different terminal velocities since Fnet=0 where Fnet = W + Ff, and W1!=W2.... and Ff is a function of velocity edit4: Fnet = net force W = weight Ff = force of friction | ||
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Deleted User 3420
24492 Posts
we are assuming normal atmospheric conditions. so this means there is air resistance. this is not a vacuum! | ||
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TheYango
United States47024 Posts
Micronesia will probably provide a better, clearer explanation if he has time. Just as a side note, showing formulas would be much easier if the forums had [math][/math] tags to hold Latex code.  | ||
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Deleted User 3420
24492 Posts
On October 23 2008 08:20 TheYango wrote: No. Drag force is given by -0.5ACdv^2 Where A is the surface area, C is the drag coefficient, d is the density of the fluid, and v is the velocity. Since they both feel the same drag force initially, they experience different accelerations, because the drag force is not proportional to the mass of the object. ok wait what? if in a vacuum they would accelerate at the same rate why would their rates of acceleration change when both subjected to an equal drag force ? Another way to look at it is that since terminal velocity is proportional to the square root of mass, they must have different terminal velocities, and therefore must reach 0 acceleration at different rates. terminal velocity determines acceleration? | ||
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micronesia
United States24753 Posts
edit: TheYango I think you are over-complicating this for the purpose of Travis' question haha. | ||
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TheYango
United States47024 Posts
On October 23 2008 08:27 travis wrote: ok wait what? if in a vacuum they would accelerate at the same rate why would their rates of acceleration change when both subjected to an equal drag force ? Gah, I'll try to be clearer. Suppose at some time t, the two objects have the same velocity, and therefore, the same drag force. Since the drag FORCE is the same, we'll call it f. The equations for Newton's 2nd law would be m1a1 = m1g - f a1 = g - f/m1 m2a2 = m2g - f a2 = g - f/m2 Where a1 and a2 are the accelerations for objects 1 and 2, g is acceleration due to gravity, and m1 and m2 are their masses.Unless m1 and m2 are equal, those accelerations CANNOT be equal, which means the acceleration functions must be different. An equal change in force does not mean an equal change in acceleration. It only is so for cases in which the force is proportional to the mass (as with the force of gravity). terminal velocity determines acceleration? The time when you reach terminal velocity corresponds to a point on your acceleration function where a = 0. If a = 0 at different times for the two acceleration curves, doesn't that imply that the two curves must be different? | ||
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Deleted User 3420
24492 Posts
On October 23 2008 08:32 micronesia wrote: What prompted this question? BTW if you didn't see my edits I put them in the earlier post. yes I saw them but I didn't understand them. can you give me an equasion, with all variables explained? pretty please? this question was prompted by a prop bet I made with someone, me getting 10 to 1. though I will admit I really didn't know, so I expected to lose this bet | ||
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Mooga
United States575 Posts
On October 23 2008 08:27 travis wrote: ok wait what? if in a vacuum they would accelerate at the same rate why would their rates of acceleration change when both subjected to an equal drag force ? terminal velocity determines acceleration? His answer assumes that you have a drag force due to air resistance. But in a vacuum, the accelerations are equal. I wasn't really sure whether you wanted to include air resistance either. | ||
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micronesia
United States24753 Posts
On October 23 2008 08:34 travis wrote: yes I saw them but I didn't understand them. can you give me an equasion, with all variables explained? pretty please? this question was prompted by a prop bet I made with someone, me getting 10 to 1. though I will admit I really didn't know, so I expected to lose this bet Okay, wasn't sure of what physics background you have or remember. When you apply a force to an object, it accelerates. The more mass it has, the more it resists accelerating (inertia). The formula is Fnet = m*a The force of friction will be the same on both objects. But since one object has a bigger m, it will have a smaller a because of the above equation. Therefore, the heavier object will accelerate towards its terminal velocity faster (higher acceleration at any given time prior to nearing terminal velocity). The tricky thing about air resistance is that it changes depending on speed. For high speeds such as falling objects, you say that the force of friction due to air resistance is proportional to speed squared. So if I double the speed of an object, it experiences four times the force. This is why objects will quickly arrive at a terminal velocity. The heavier object will have a higher terminal velocity, everything else being equal. This is because the object has no acceleration (terminal velocity means constant speed). Let's show this. a=0 so from Fnet=ma we know Fnet=0. Fnet is really the sum of the forces acting on the object. The two forces are weight downward, and friction upward. These forces are equal in magnitude and opposite in direction. When the object is moving fast enough that the air resistance gets large, the object can arrive at terminal velocity (specifically when the force of friction becomes equal to the weight). With a smaller weight, you can arrive at terminal velocity at a lower speed. An interesting question is to see which object arrives at specific percentages of its terminal velocity faster, but that's beyond the scope of the original question. edit: minor corrections | ||
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Elvin_vn
Vietnam2038 Posts
On October 23 2008 08:06 travis wrote: here is my question. 2 objects, same shape, surface, and size. different masses. both are dropped from an airplane. before either object reaches terminal velocity, will they accelerate at the same rate? same shape + surface + size ==> same resistance ==> they will accelerate at the same rate | ||
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Mooga
United States575 Posts
On October 23 2008 08:41 Elvin_vn wrote: same shape + surface + size ==> same resistance ==> they will accelerate at the same rate Except the shape of the object affects the drag coefficient, so this is not true unless you assume there is no air resistance. | ||
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