An almost empty bobbin is pulled along a flat surface by a thread which is wrapped around it. The diameter of the inner reel is 5cm and that of the outer wheels is 10 cm. Assuming no slipping or sliding, how far, in centimeters, has the bobbin moved when the end of the thread has moved 12cm?
I got the answer of 24cm. But the correct answer is 8am. Can anyone try to solve it? Thanks in advance.
On
Note that $12$cm is not the length of the unwrapped thread, but the distance that the end of the thread has moved.
Let the distance the bobbin has moved be $x$ cm, and the number of turns is $n$. Then the length of the thread is $12-x$ Then we have a system of two equations: \begin{align} 12-x&=5\pi n, \\ x&=10\pi n, \end{align} which results in $n=\frac{4}{5\pi}$, $x=10\pi\frac{4}{5\pi}=8$.
When the bobbin rolls one complete turn to the right, it has travelled $10\pi$ cm. So the center of the inner reel has also travelled that distance. The amount of thread in one turn around the inner reel is $5 \pi$ cm. So assuming that the thread is coming off the top of the inner reel, the end of the thread has moved $10 \pi$ cm + $5 \pi$ cm = $15 \pi$ cm. So: for $10\pi$ cm of movement of the bobbin, we get $15 \pi$ cm of movement of the thread-end. That's a ratio of 3 to 2.
When the thread has moved 12cm, the bobbin must have moved $8$cm.
Note: it's also possible for the thread to come off the bottom of the bobbin, and in this case, $10\pi cm$ movement of the bobbin causes only $10 \pi - 5\pi$ = $5 \pi$ cm movement of the thread, i.e., a ratio of 1 to 2. So that when the thread has moved $12cm$, the bobbin must have moved $24$cm (i.e., your answer).
The question is fundamentally ambiguous.