## Bicycle Sprockets and Angular Velocity

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
GreenLantern
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### Bicycle Sprockets and Angular Velocity

The Full Question (the hint was included so I though I'd put it here too):
An expert cyclist can attain a speed of 40mi/hr if the sprocket assembly has r1=5in, r2=2in, and the wheel is 28in in diameter.

Approximately how many revolutions per minute of the front sprocket wheel will produce a speed of 40mi/hr? (Hint: First change 40mi/hr into in/sec)

Clarifications:
The front sprocket is referring to the larger sprocket (5in)(it has the pedals on it). The smaller sprocket (2in) is attached to the wheel (28in diameter). This is all shown in a diagram in the book that you, unfortunately cannot see.

My post:
I've worked this problem a number of different ways, and have proved that s/t=r(theta)/t, but I haven't come up with much else.

The answer in the back of my book is telling me 192.08 revolutions/minutes. I have yet to connect the two through a mathematical equation.

I'm mostly just frustrated by it.

stapel_eliz
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To learn how to convert the units, try here.You should get 704 inches per second.

Given that value, and the radius of the wheel that the bike rides on, you can find the number of rotations needed per second to get the required linear velocity: Divide the distance per second by the circumference, to find the number of circumferences (that is, the number of rotations) needed to cover that distance in a second.

I think you're supposed to assume that the small gear revolves in synch with the wheel, so you have that many revolutions per second for the small gear. Use the radius of the small gear to find the "distance" that the gear moves in that number of revolutions.

Then use the radius and circumference of the large gear to find the number of revolutions of the large gear. This will give you the revolutions per second.

Multiply by 60 to get the revolutions per minute. (I get the same answer as what you've been given.)

GreenLantern
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### Re: Bicycle Sprockets and Angular Velocity

That's one of the things I tried! Maybe I just put down some numbers wrong...

40mi/hr = 704in/sec I had that before, but it was the only thing that seemed right.

Unless I'm misinterpreting all this, which is probably the case since I got the wrong answer anyway...

The second link is telling me that I should do 704/28(pi)= 78.988... revs/sec
Now since the wheel and small sprocket rotate as one they both need the same revs/sec, so I can assume that the smaller sprocket will also need 78.988... revs/sec.

The distance the small sprocket spins is: 78.988... x 4(pi)= 992.600... in/sec

992.600... / 10(pi)= 311.834... revs/sec And should equal how many times the large sprocket needs to rotate for the small sprocket to rotate 78.988... but the numbers don't make sense. I'm not even going to bother multiplying by 60 yet because I'm starting with a number larger than my needed answer anyway. Where did this go wrong?

stapel_eliz
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The second link is telling me that I should do 704/28(pi)= 78.988... revs/sec
How did you get this value?

You're dividing approximately 700 by approximately 30*3 = 90; you should not be getting a value that is larger than 700/10!

GreenLantern
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### Re: Bicycle Sprockets and Angular Velocity

You're totally right! That makes no sense. I put them both in my calculator to find that 700/90=7.777 and 704/28(pi)=78.988 BUT! 704/(28(pi))=8.00321! The new set of parenthesis changed the whole thing! I was sitting here with 3 different calculators in front of me with on my face.

Now, following your instructions took me straight there and made perfect sense. I was bamboozled by my calculator.

Thanks Liz!