Explain which measurements you will should take and also extra information that you might need in stimulate to calculation the thickness of a soda can. You might assume the the deserve to is do of aluminum. Because of the hazard of injury, cutting the can and also directly measuring its thickness is not permitted.

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This is a variation of ""How thick is a soda can? variation I"" which allows students to work-related independently and think about how they can determine exactly how thick a soda can is. The teacher should explain clearly that the goal of this task is to come up v an ""indirect"" means of assessing just how thick the can is, that is straight measuring that is thickness is no allowed. Students must be provided with cans to aid in this process. After permitting the students to think around this because that a while, it would be good to have a discussion about what tactics students have developed and also what dimensions they think the taking. The many relevant measurements for this problem are surface ar area, weight, and also density so the teacher may wish to discuss these if they are not lugged up through the students.

If students have access to tape measures and relatively carefully calibrated scales, then these can be used and would allow for a discussion around precision in this measurements. As much as the thickness of aluminum goes, the instructor may wish to provide this information to students ($2.70$ grams every cubic centimeter). So the key pieces of details which the students may identify as being essential to calculation the thickness the the can are:

the surface ar area of the can, the weight of the can, the thickness of the can.

Assuming that the thickness that the can is uniform, the students would then usage the formula: $$ m density approx frac m surface,,, m area imes m thickness m weight $$ to find the almost right thickness. The factor why the $approx$ is suitable in this vault formula, rather of $=$, is the we perform not know that the thickness that the can is uniform and the ""edges"" that the can will additionally influence this relationship. Moreover, soda cans are not in reality cylindrical as they space slightly tapered both at the top and bottom.

A second method for estimating the thickness the the deserve to uses Archimedes" principle, presented in ""Archimedes and the King"s crown,"" and also this does not require the density. Archimedes" method calculates the volume the aluminum in the can by submerging the in water in a calibrated beaker and also measuring just how much water is displaced. This is perhaps much more appropriate and feasible in a chemistry class yet the concepts can be questioned even if that is not feasible to execute the dimensions in math class. Most importantly, Archimedes" technique avoids knowing the thickness of aluminum and additionally the load of the can. So using this method, the students would certainly then usage the formula $$ m Volume approx m surface,,, m area imes m thickness $$ The $approx$ sign is suitable both due to the fact that the surface area and volume of the deserve to are both estimates and because the wherein the top and bottom of the have the right to are joined with the cylindrical shame piece, closer modeling would certainly be vital to get an accurate formula.


Solutions

Solution:1 Use thickness of aluminum

A photo of the soda deserve to is offered here. This can be mutual with the studentsif they perform not have accessibility to soda can be ~ or come the suitable measuring devices that castle will need to estimate that is thickness.

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The only part of one aluminum have the right to which provides us a visual feeling for how thick the can could be is the opening from which us drink the soda. Feeling and also observing this,it is fairly sharp and plainly very thin, certainly too thin to be measured v a regular tape measure. The thickness of the deserve to will, however, be proportional to just how much aluminum is in the can and also this, in turn, deserve to be captured by weighing the can. Inaddition come the load of the can, the college student will require to know the thickness of aluminum (that is, just how much walk a provided amount that aluminum weigh) and they willneed to approximate the surface ar area the the can.

The surface ar of the deserve to is comprised of the top and bottom together with thecylindrical component of the can. In practice, the top and bottom are not flatbut room well approximated by circles. According to the provided information,the radius of these circles is around $1 frac316$ inches. For this reason the area isabout $pi imes left(1 frac316 ight)^2$ square inch or about $4.4$ squareinches. If the cylindrical part of the deserve to were do flat, it would certainly be a rectangle v dimensions $4 frac34$ customs by $pi imes 2 imes1 frac316$ inches or about $35.4$ square inches. For this reason the complete surfacearea of the deserve to is about $44.2$ square inches.

If there are about $15$ grams of aluminum in the soda can and the density ofaluminum is about $2.70$ grams every cubic centimeters climate there room about$$frac15 mbox grams2.70 mbox grams every cubic centimeter approx 5.6 mbox cubic centimeters$$of aluminum in the soda can.

Since the amount of aluminum is offered in cubic centimeters and the area the thesoda deserve to in square inches we need to make a conversion in bespeak to estimate the almost right thickness the the can. There are about 2.54centimeters every inch and so there are around $(2.54)^2 approx 6.5$ squarecentimeters every square inch. So$$44.2 mbox square inches approx 44.2 imes 6.5 mbox square centimeters.$$This is about $287.3$ square centimeters. To uncover the approximate thicknessof the can we recognize that$$ m thickness imes 287.3 mbox cm^2 approx 5.6 mbox cm^3.$$This way that the almost right thickness the the have the right to is around $0.02$ centimetersor $0.2$ millimeters.

If the teacher wishes to interact students in a discussion of precision that measurements and also how results should be recorded, there room three measurementswhich are subject to error and the reported thickness of aluminum which isonly precise the nearest one hundredth the a gram every cubic centimeter. Themeasurements that the can, if made v a ice cream measure marked to the nearestsixteenth of an inch, deserve to be assumed come be specific to within $frac132$ ofan inch. This is about $2.5$ percent the the measurement of the radius and also seventenths of one percent that the measurement for the height. Detect the approximatearea that the top and bottom the the have the right to requires squaring the radius i m sorry willroughly twin the error come $5$ percent. We have that $5$ percent of $8.8$square inches is almost fifty percent a square inch. The error because that the cylindrical part ofthe can will depend on the error for the height and also the error for the circumferenceand will certainly be a little an ext than $3$ percentor a little more than one square inch. So the total possible error hereis close come one and a fifty percent square inches. One appropriate method to document this wouldbe $44 pm frac32$ square centimeters and then this error continues throughthe final estimation that the thickness that the aluminum can. The canis no perfectly cylindrical and also it has a tab to open the can: these will certainly contributeto the error too but the in its entirety estimate that $0.2$ mm for thickness have to be an excellent provided this thickness is uniform.

Solution:Solution 2 usage Archimedes" principle

In order come estimate just how thick the soda have the right to is, we can estimate the surfacearea that the can and the volume of the aluminum in the can and then use the reality that the volume that aluminum in the can is approximately equal to the surface area of the have the right to times the thickness. Because that the surfacearea, the method of the an initial solution can be used and we repeat this here.

The surface ar of the have the right to is made up of the top and bottom along with thecylindrical part of the can. In practice, the top and bottom are not flatbut are well approximated by circles. Follow to the offered information,the radius of this circles is about $1 frac316$ inches. For this reason the area isabout $pi imes left(1 frac316 ight)^2$ square inch or around $4.4$ squareinches. If the cylindrical part of the deserve to were made flat, it would be a rectangle v dimensions $4 frac34$ customs by $pi imes 2 imes1 frac316$ customs or around $35.4$ square inches. Therefore the total surfacearea of the deserve to is about $44.2$ square inches. There are about 2.54centimeters per inch and also so there are around $(2.54)^2 approx 6.5$ squarecentimeters per square inch. So$$44.2 mbox square inches approx 44.2 imes 6.5 mbox square centimeters.$$This is around $287.3$ square centimeters.

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The volume the the can is not given and so this needs to it is in assessed by part indirect means. Making use of a method dating earlier to Archimedes, if we submerge the have the right to in water, climate the quantity of water displaced by the deserve to will be equalto that is volume. One means to perform this would certainly be to crush the deserve to (although some care is needed below to make sure that in so doing we do not trap too much air in ~ the can). Another an approach would be to very closely fill the have the right to to the optimal with water and then the filled have the right to will sink. The lot of water displaced by the aluminum in the deserve to should be close come $5.6$ cubic centimeters as wasfound in the an initial solution. Speculative values might differ, however, so mean we find that the aluminum in the soda deserve to displaces $x$ cubic centimeters the water. Then making use of the equation$$ m Volume approx m surface,,, m area imes m thickness$$we deserve to plug in $x$ because that the volume and $287.3 m cm^2$ for the surfacearea and we find$$ m thickness approx fracx287.3 m, cm^2.$$

Notice that this technique is much faster than the first and requires less information. ~ above the other hand, it does need some lab equipment.