Finding x in Radians: A Puzzling Problem for Pavadrin

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The discussion revolves around solving a problem involving the motion of a piston and its relation to radians and percentages. The user, Pavadrin, initially calculated the movement of the piston as 68% but found a discrepancy with the book's answer of 17%. Key insights revealed that the piston’s movement should be evaluated using trigonometric functions, specifically the cosine of the angle, to determine the vertical displacement accurately. Additional details about the geometry of the problem were provided, leading to a refined calculation that resulted in approximately 17% for x in terms of h. The importance of including all relevant information in problem statements was emphasized to facilitate better assistance.
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Hey
I have recently been studying for my upcoming exams. I am stuck on the following problem involving radians. Here is the diagram for it.
Geometrya nd trig problem.JPG

Where PQ = the radius of the circle
I need to find x in as a percentage in terms of h
My working for this problem is as follows
in a circle there is 2pi radians, therefore in half a cirlce there is pi radians, which means for that for pi radians the pistion moves for the high position to the low posotion.
I then calculated the remainder of the angle in half a circle to determine how much the piston had moved, which resulted in being 2.1416 radians.
I then calculated this as a percentage by dividing it by pi and multiply it by 100, which gave me an answer of 68% (correct to one decimal place)
Form here I took that value, 68% from 100% which gave me an answer as being 32% which I assumed as being the value for x.
However I checked the answer in the back of the book and it said 17%.
Could someone please tell me where I have gone wrong? Or if the answer in the book is wrong?
Thanks in advance,
Pavadrin
 
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I feel like there is some information missing. Could you please post the complete problem statement.
 
Integral said:
I feel like there is some information missing. Could you please post the complete problem statement.
the complete problem statement is as follows:
The diagram shows how the vertical motion of a piston can be used to produce rotational motion. As the piston travels from the low position to the high position and back again the wheel will rotate. If the minor arc PQ is eqaul to the radius in length to r, the radius of the wheel, express x as a percentage of h correct to the nearest percentage
thanks
 
Last edited:
pavadrin said:
Hey
I have recently been studying for my upcoming exams. I am stuck on the following problem involving radians. Here is the diagram for it.
View attachment 6901
Where PQ = the radius of the circle
I need to find x in as a percentage in terms of h
My working for this problem is as follows
in a circle there is 2pi radians, therefore in half a cirlce there is pi radians, which means for that for pi radians the pistion moves for the high position to the low posotion.
I then calculated the remainder of the angle in half a circle to determine how much the piston had moved, which resulted in being 2.1416 radians.
I then calculated this as a percentage by dividing it by pi and multiply it by 100, which gave me an answer of 68% (correct to one decimal place)
Form here I took that value, 68% from 100% which gave me an answer as being 32% which I assumed as being the value for x.
However I checked the answer in the back of the book and it said 17%.
Could someone please tell me where I have gone wrong? Or if the answer in the book is wrong?
Thanks in advance,
Pavadrin
You're on the right track, but forgot two things. The piston didn't move 2.1416 radians straight down. Sine and cosine aren't linear functions.

You need the cosine of 2.1416 radians to find how far down you moved. That will be in proportion to the radius. The maximum up and down range varies by twice the radius (i.e. - by the diameter).
 
BobG said:
You need the cosine of 2.1416 radians to find how far down you moved. That will be in proportion to the radius. The maximum up and down range varies by twice the radius (i.e. - by the diameter).
BobG, I am not quite sure what you mean.
 
Hey
I’m really stuck on this problem. There more time I spend on it the more I get confused. I have tried another approach, where I drew triangles and substituted values for the variable r (the radius of the wheel). I looked at the diagram I had posted beforehand and noticed I had left out some detail (the distance between the top of the wheel and the low position is also equal to the radius of the wheel). I am sorry if this has caused an inconvenience in anyone’s calculation if this was also significant. The length of the arm from PR = 3r and the length of height of h = 2r. From this point onwards I have absolutely no idea where to go.
Thank you to anyone who replies
Pavadrin
 
pavadrin said:
Hey
I’m really stuck on this problem. There more time I spend on it the more I get confused. I have tried another approach, where I drew triangles and substituted values for the variable r (the radius of the wheel). I looked at the diagram I had posted beforehand and noticed I had left out some detail (the distance between the top of the wheel and the low position is also equal to the radius of the wheel). I am sorry if this has caused an inconvenience in anyone’s calculation if this was also significant. The length of the arm from PR = 3r and the length of height of h = 2r. From this point onwards I have absolutely no idea where to go.
Thank you to anyone who replies
Pavadrin

That was actually a critical piece of info. I just tried it before but since you hadn't provided that info it wasn't doable. I'll post up the solution in a bit.

EDIT : Please remember to include every piece of info in the question, no matter how insignificant it seems to you. It would help us to help you faster and with less hassle. :smile:
 
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See the attachment and all the labels.

l\cos\theta = r\cos\alpha + r + r + x --(1)

Also,

l = 3r --(2)

and l\sin\theta = r\sin\alpha --(3)

From (3),

l\cos\theta = \sqrt{l^2 - r^2\sin^2\alpha} --(4)

Put (4) and (2) into (1) and rearrange,

x = r(\sqrt{9 - \sin^2\alpha} - \cos\alpha - 2)

Since h = 2r,

\frac{x}{h} = \frac{1}{2}(\sqrt{9 - \sin^2\alpha} - \cos\alpha - 2)

Now we're basically given that alpha = 1 radian (chord length PQ = radius). Therefore, x/h works out to about 16.96 % as required.
 

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    Geometrya nd trig problem.JPG
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thank you so much Curious3141~!
 
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pavadrin said:
thank you so much Curious3141~!

You're welcome.:smile:
 
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