# an almost empty bobbin is pulled along a flat surface by a thread which is wrapped around it. The diameter of inner wheel is 5 cm and outer wheel is 10cm. Assuming no slippage how far has the...

an almost empty bobbin is pulled along a flat surface by a thread which is wrapped around it. The diameter of inner wheel is 5 cm and outer wheel is 10cm. Assuming no slippage how far has the bobbin moved when the end of the thread has moved 12cm?

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### 1 Answer

So we know that a bobbin has an inside wheel that has a diameter of 5 cm and an outer wheel with a diameter of 10 cm.

In order to determine how far the end of the thread has moved, we have to calculate how far the bobbin is moving as well as the thread from the bobbin. Lets start by figuring out how far the bobbin moves for every 1 cm of thread we pull away from the bobbin.

To do this we must find the circumference of the two wheels first.

Circumference of Inner Wheel: `C=5pi=15.71 `

Circumference of Outer Wheel: `C=10pi=31.42 `

So if we pull 1 cm of string from the bobbin, we can find what percentage of the wheel spun by the following equation:

`(1cm)/(15.71cm)=.064`

This shows us that to pull one cm of thread from the bobbin, the whole wheel spins 6.4% of the whole circle. We can find how far the outer wheel will roll by setting it equal to this percentage:

`(xcm)/(31.42cm)=.064 `

If we know that the proportion of the distance the outer wheel moved should be equal to the distance that the inner wheel moved, then we can set this proportion above. By solving for "x", we should get the following:

`x=2cm `

So whenever we pull 1 cm of thread from the bobbin, the wheel will move 2 cm in the same direction. So it has a proportion of **1:2**

With this information we can figure out how to calculate how far the bobbin must move in order to get the end of the thread to move **12 cm**.

Let's let **x** equal the length of the thread that we pull **from the bobbin**

If that is so, then **2x** should be the length that the whole **bobbin moves**.

Therefore, if we have `x+2x=12 `

Then we should be able to solve for "x" and then calculate the length the bobbin moved.

`3x=12 cm `

`x=4 cm `

So if x is 4 cm, then we know that "**2x**" was how far the bobbin moved. **2*4=8cm**

So the bobbin moves 8 cm, and the thread moves 4 cm. Then if we combine those, the end of the string should have moved **4 cm +8 cm** or **12 cm. **So our answer is correct!