3
$\begingroup$

The problem is as follows:

The following machine consists of five interconnected wheels. The machine is let to work until the first wheel completes six turns. How many turns more than the third wheel will give the fifth?. Consider the order goes from left to the right.

See the figure from below:

$\hspace{2cm}$Sketch of the problem

In my attempt to solve this problem I though to use this equation:

$$\textrm{1 turn =}\,2\pi r$$

The first wheel would become into:

$$\textrm{turns}_{\textrm{1st}}=\,2\pi \left( 8\, \textrm{inches} \right)$$

The third wheel would become into

$$\textrm{turns}_{\textrm{3rd}}=\,2\pi \left( 6\, \textrm{inches} \right)$$

Therefore the ratio between the two would be:

$$\frac{\textrm{turns}_{1st}}{\textrm{turns}_{\textrm{3rd}}}=\frac{8}{6}$$

So from this can be established that:

$$\textrm{turns}_{1st}=\frac{8}{6} \textrm{turns}_{\textrm{3rd}}$$

So the third wheel will give $\frac{8}{6}$ times of what the first wheel will do.

Then I did performed the same logic between the third wheel and the fifth wheel.

The third wheel is:

$$\textrm{turns}_{\textrm{3rd}}=\,2\pi \left( 6\, \textrm{inches} \right)$$

while the fifth wheel is:

$$\textrm{turns}_{\textrm{3rd}}=\,2\pi \left( 3\, \textrm{inches} \right)$$

The ratio between the two would be:

$$\frac{\textrm{turns}_{3rd}}{\textrm{turns}_{\textrm{5th}}}=\frac{6}{3}$$

Therefore we can establish a relationship between the third and the fifth wheel:

$$\textrm{turns}_{\textrm{3rd}}=\frac{6}{3}\textrm{turns}_{\textrm{5th}}$$

or

$$\textrm{turns}_{\textrm{3rd}}=2\times\textrm{turns}_{\textrm{5th}}$$

Then I assumed that the number of turns will give the fifth will be two times of the first.

Then If it has been said that the first wheel gives $6$ turns then:

$$6\,\textrm{turns}_{\textrm{1st wheel}}\times\frac{8\,\textrm{turns}_{\textrm{3rd wheel}}}{6\,\textrm{turns}_{\textrm{1st wheel}}}\times \frac{2\,\textrm{turns}_{\textrm{5th wheel}}}{1\,\textrm{turns}_{\textrm{3rd wheel}}}=\,16\,\textrm{turns}_{\textrm{5th wheel}}$$

But because what it is being asked it is how many turns will give the fifth than the third, this can be obtained by subtracting the number of turns between the fifth with that of the third:

$$16\,\textrm{turns}-8\,\textrm{turns}=8\,\textrm{turns}$$

So $8$ must be the answer.

However I'm confused at the way how I did made some interpretations during this problem:

Let's look at the equation I wrote lines above:

$$\textrm{turns}_{1st}=\frac{8}{6} \textrm{turns}_{\textrm{3rd}}$$

Is it okay to say that third wheel will be $\frac{8}{6}$ times of the first wheel?.

I got the same confusion when trying to make an interpretation of the second equation:

$$\textrm{turns}_{\textrm{3rd}}=2\times\textrm{turns}_{\textrm{5th}}$$

From looking at the equation it seems to say that the third wheel is $2$ times the turns of the fifth wheel.

Can somebody help me to clear these doubts?

There is another aspect which I'm also confused and it is, what happens with the wheels in middle. During my attempt to solve this problem I just ignored them and considered the effects as constant. Is it okay to assume this?

Then, other than turns. Is the number of turns related with the speed?. I'm aware of the tangential speed which is $\frac{2 \pi r}{t}$. So would the speed remain constant or be higher for the small wheel?.

$\endgroup$
3
  • $\begingroup$ It is related to the circumferences -- but they are in direct proportion to the size (radius or diameter). So $π$ can be ignored as well as in between wheels. $\endgroup$ – dan post Apr 13 '18 at 2:09
  • $\begingroup$ @Chris Steinbeck you mean gear wheels? $\endgroup$ – Narasimham Apr 13 '18 at 19:43
  • $\begingroup$ @Narasimham I think this logic can be applied to gear wheels problem. But I meant on how to avoid getting confused when establishing a ratio between the two as for matters of reading I want to avoid assigning the incorrect value to some of the wheels. $\endgroup$ – Chris Steinbeck Bell Apr 14 '18 at 2:15
3
$\begingroup$

Much simpler is to say the first wheel turns $96\pi$ inches, so all the wheels turn $96\pi$ inches. The third wheel then makes $8$ turns and the fifth makes $16$, so the fifth does $8$ more turns than the third.

The linear speeds of the circumference of the wheels is the same. The angular speeds are in inverse proportion to the radii.

$\endgroup$
3
  • $\begingroup$ Sorry for the delay, I'm some slow at understanding these things. Perhaps can you explain me where does the $96\pi$ comes from?.Is it from the $6$ turns so all those are like a big wheel of that length?. My question arises on how do I notice that the third is $8$ turns the first?. It is like reversing the radius from the bigger to the one which is in the middle. But why?. It's radius is $6$ not $8$, this is the part where I'm stuck at. As I mentioned, can these relationships be better expressed into some mathematical equation that can avoid getting any confusion?. $\endgroup$ – Chris Steinbeck Bell Apr 13 '18 at 2:21
  • $\begingroup$ @ChrisSteinbeckBell it is only my guess but 96 could be served as the least common multiple. Just to illustrate full rotations of the smaller wheels and to avoid fractions. $\endgroup$ – Przemyslaw Remin Sep 9 '19 at 10:28
  • $\begingroup$ @ChrisSteinbeckBell: the $96\pi$ comes because the circumference of the first wheel is $2\pi\cdot 8=16\pi$ inches and it is specified that the first wheel makes $6$ turns. $\endgroup$ – Ross Millikan Sep 9 '19 at 13:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.