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The problem states the following

The figure shows a small iron sphere hanging from a spoon-like structure by a flexible wire labeled as $\textrm{QH}$. The sphere is pushed in the indicated direction in such a way that the wire always stays taut so that the sphere reaches $\textrm{MN}$. Calculate the length traveled by the sphere. Consider these lengths $\textrm{MN=NP=PQ=9 inch}$.

The figure is attached below:

Sketch of the problem

What I did to tackle this problem was to separate the distance traveled by the hanging sphere into three different arc segments to which I labeled as $\textrm{L}_{1}$, $\textrm{L}_{2}$, $\textrm{L}_{3}$. See figure below of the proposed solution

Sketch of the proposed solution

Taken into consideration I noticed that the angle of all circle segments were $60^\circ$, $30^\circ$ and again $60^\circ$.

However for this to be consistent I took the fact that the gravity acting on the iron sphere would make that part such as a leveling instrument and therefore the angle on $\angle HQ\textrm{wall on the right side}=90^\circ$. From then on I calculated them as this way:

$$\textrm{I define the point where the structure is fixed on the right wall=R}$$

$$\angle PQR = 150^\circ$$ $$\angle PQH= 150^\circ-90^\circ=60^\circ$$

$$\angle HPN= 180^\circ-150^\circ=30^\circ$$ $$\angle HNM= 180^\circ-120^\circ=60^\circ$$

With these angles in mind I proceeded to calculate the lengths by using the formula $L=\theta\times R$

The first radius is calculated directly as is the same as the length of the wire thus is $21\,\textrm{inches}$.

$$R_{1}=21$$

The second radius is calculated as the difference between the 9 inches of the right wall which is on the back of the spoon and the total 21, hence the acting radius will be:

$$R_{2}=21-9=12$$

The third radius same as the previous one, will be the difference between those acting 12 inches and the other 9 inches from the base of the spoon becoming:

$$R_{3}=12-9=3$$

The the last step will be just replacing those ones with the angles already found as (taken into consideration transforming the sexagesimal angles in radians):

$$L_{1}=\theta \times R=(60^\circ\times\frac{\pi}{180^{\circ}})\times 21=7\pi$$

$$L_{2}=\theta \times R=(30^\circ\times\frac{\pi}{180^{\circ}})\times 12=2\pi$$

$$L_{3}=\theta \times R=(60^\circ\times\frac{\pi}{180^{\circ}})\times 3=1\pi$$

Hence the total distance would be the sum of each distance:

$$\textrm{Distance traveled}=7\pi + 2\pi + \pi= 10 \pi$$

Would this approach be okay?. I'm not very sure if my assumptions are correct as I'm just assuming that the structure is leveled with the ground and forming parallel lines.

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  • $\begingroup$ Wasn't there anything said about the sphere's size? $\endgroup$ – J. M. is a poor mathematician Mar 2 '18 at 2:24
  • $\begingroup$ @J.M.isnotamathematician No, sorry. The problem omitted that information. $\endgroup$ – Chris Steinbeck Bell Mar 2 '18 at 2:26
  • $\begingroup$ You might surmise why I asked about it. If the sphere was sufficiently sizable, your last arc will be a bit short of $60^\circ$. If the sphere is negligibly small, that $60^\circ$ would be fine. $\endgroup$ – J. M. is a poor mathematician Mar 2 '18 at 2:28
  • $\begingroup$ @J.M.isnotamathematician Why would it be shorter?. $\endgroup$ – Chris Steinbeck Bell Mar 2 '18 at 2:30
  • $\begingroup$ If the sphere bumps into the last wall, its center point would not be touching the wall, but would be slightly offset from it. If all you had was a string with a heavy end, you'd be able to cover the full $60^\circ$ $\endgroup$ – J. M. is a poor mathematician Mar 2 '18 at 2:34

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