## single piston engine problem - position of piston

1. The problem statement, all variables and given/known data
Consider the simplified single-piston engine in the figure. The wheel rotates at a constant angular speed of 1 rad/s and the radius of the wheel is 2 m. If the piston is fully extended at time t=0, find the position of the piston at 2s. The figure depicts a wheel with a peg that draws a piston in and out of a bore as the wheel rotates (counter clockwise, with the piston horizontal to the right).

2. Relevant equations
x=A cos (omega*t)
where A is the radius of the wheel

3. The attempt at a solution

cos 2 * 2m = x.
since 2 > pi/2, I add 2meters to the resulting x.

The answer I get is reportedly wrong. what's wrong with my approach?
 PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
 Mentor It would help to see a diagram, but are you taking into account the angle of the connecting rod? When the pin is not at 0 or PI, you lose some of the stroke distance since the pin is above or below the axis of the cylinder.
 Mentor Actually it looks like you are trying to include that factor, but the distance x is not cos(2rad*2m). I don't think that's the angle that you should be taking the cos of...

## single piston engine problem - position of piston

after doing some searching, I was able to find another problem with an identical diagram.

http://faculty.ksu.edu.sa/alkurtass/Tut102/Tut102-9.pdf

problem number 7 shows the piston and the wheel. my thinking of the angle being 2 rad stems from the piece that catches on the piston assembly following the path of the wheel. Since the piston is fully extended at time t=0, I believe that puts the initial angle at 0. am I wrong in thinking so?
 Mentor At t=0, the pin is at the far right. At 2s into rotation, the wheel is 2 radians CCW, which is most of the way to PI radians which is the farthest left. The x displacement to the left of the piston is most of the way to max displacement possible, but a little less. To calc the displacement x, draw a triangle. The theta that you take the cos of is most definitely not 2 radians. Draw the triangle on that figure, and show us your calculation...

Mentor
 Quote by berkeman At t=0, the pin is at the far right. At 2s into rotation, the wheel is 2 radians CCW, which is most of the way to PI radians which is the farthest left. The x displacement to the left of the piston is most of the way to max displacement possible, but a little less. To calc the displacement x, draw a triangle. The theta that you take the cos of is most definitely not 2 radians. Draw the triangle on that figure, and show us your calculation...
And actually, I don't think it's a simple cosine. It's a simple angle calculation, though...
 Sorry to intrude, but what is it? I have the same problem, with the same omega = 1 rad/s, but my r = 1.3m. I am trying x = Acos(pi - 2rad) but this is not working. I have drawn a little triangle and concluded that the angle is pi minus omega(t), but no bueno! Any suggestions would be greatly appreciated

Mentor
 Quote by livewire852 Sorry to intrude, but what is it? I have the same problem, with the same omega = 1 rad/s, but my r = 1.3m. I am trying x = Acos(pi - 2rad) but this is not working. I have drawn a little triangle and concluded that the angle is pi minus omega(t), but no bueno! Any suggestions would be greatly appreciated