who can answer this the fastest?
there is a cylindrical bobbin which is 1 centimeter in diameter, and a strip of tape which is 10 meters in length and 0.5 milimeters in thickness.
what is the maximum number of times you could wrap the tape around the bobbin if the thickness of the tape remains the same…?
Are we to assume that the bobbin and the tape are the same width?
With slight approximations, I get 242.5 times.
Damn! Dropped a factor of ten. You change units too many times! It’s really 79.5 times.
Geez, I keep making errors. Now I saw it’s 23.3 times.
Final answer, I hope.
My guess:
The surface of the resulting circle (if you look at the thing from the side) must equal 10 meters * 0.5 millimeters = 5,000 mm².
r² * pi = 5,000 mm² —> 5,000 mm²/pi = 1,591.5 mm²
—> r = SQR 1,591.5 mm² = 39.9 mm
That’s about 80 times wrapped around.
I’m not sure if this calculation is waterproof, but it sounds reasonable to me.
[sub]Hell, I wrote a 180 mins physics exam today - I should finally forget stuff like this.[/sub]
but let’s say you wrap it around the cylinder…
the first wrap would be let’s say 3cm around. the next wrap would be 3cm + 0.5 mm.
the next wrap would be 3cm +0.5mm +0.5mm… and so on…
True, but I suppose this is all taken into account if you just look at the surface the wraps form.
The radius of the first turn is 5 mm + .25 mm = 5.25 mm and each succesive turn has a radius .5 mm larger
this yields 70 whole turns (and a fraction)
Pretty close. You forgot to account for the bobbin, which doesn’t matter too much in the area, but gives you an extra ten wraps.
I got r = 40.2 mm for the outside radius. Subtract 5 mm for the bobbin, and divide by 0.5mm to get about 70 wraps, which agrees with Sailor’s result.
The answer is 70.4. . . There are several easy ways of doing this. One would be to consider it a spiral and integrate. But there is an even easier one: observed from the side you will see the area of the side of the tape plus the area of the bobbin: 5000 plus 78.54 mm2 = 5078.54 mm2. The radius of this circle is 40.206 mm and we subtract the radius of the bobbin and we have 30.206 as the total thickness of the tape. Divide by .5 mm and you get about 70.4 turns.
>> who can answer this the fastest?
If you mean “correctly” that would be me and it took me a couple of minutes to work it out.
About ten years ago I invented a way of determining how much tape was left in a VCR tape just by measuring the speed of the counter because the speed of the counter varies depending on how much tape is already wrapped on the spool. No need to rewind and set the counter to zero: the speed of the counter (regardless of the number on it) will tell you exactly how much time you have left on the tape. It was very useful then but I haven’t recorded anything in years now.
Sailor’s solution neglects the fact that the radius of each layer is increasing. His method can be used to determine the outer radius, given the inner radius, the overall length, and the thickness. And, in fact, I used that calculation just recently. Calculating the number of times around is more complex because of that detail.
The circumference of a section of radius r is 2pir. That of radius r + t is 2pi(r+t), so each subsequent wrapping (ignoring the fact that it’s really a spiral) is as long as the previous one plus 2pit. If there are M wrappings, the total length is M2piR + 2pitM(M+1)/2. (Here we’ve used the fact that the sum of 0 to M is M(M+1)/2). We set the total length L equal to L = M2piR + pitM(M+1) and solve the resulting quadratic equation. Plug in your initial values and you get 23 wrappings and change, as I noted above.
dang …
maybe someone should actually try it with physical materials…
haha…
Cal’s method sounds reasonable… but only 23 wraps?.. sounds too little to me…
i used excel and i came up with 147 wraps about…
not sure though…
I wish I was that good at math…was never my strongest subject
oh and i used quadratic equation and came up with 160 wraps…
** CalMeacham**:
Shouldn’t the part in bold be M(M-1)/2?
On the first wrap, the radius is R;
on the second wrap, radius R + t;
.
.
.
on the Mth wrap, radius R + (M-1)t.
So we’re summing from 0 to M - 1. With that change, I believe your answer is very close to sailor’s.
Actually, after writing that and thinking about it a bit, I don’t think the answer can be entirely clear. When the tape is wrapped around, will the inner surface be 10 meters long (and so the outer surface must stretch to accomodate this), or will the outer surface be 10 meters long (and so the inner surface is compressed to accomodate this)?
?? I don’t see where the increasing radius matters at all in sailor’s method. The side surface area (length x thickness) of the tape gives the surface area of the top of the resulting cylinder, and the radius of the cylinder minus the radius of the bobbin gives the total thickness of the tape as it wraps around. Divide this by the thickness of each layer (which is .5 mm) and you get the number of layers, which is the number of times the tape had to be wrapped. Each layer will have a greater length, of course-but the thickness of each layer will be exactly the same.
Plug in M = 70, R = 5, t = 0.5, total length = 2199.1 + 7806.86 = 10,006 mm, which is about the right length. How’d you get M = 23?