You say A stirring blade is attached at the level of the center of the wheel.
sounds like the distance is zero. The blade is attached to what? Does the blade move in some manner? If not, its distance must be constant.
Is this mechanism submerged in a tank of some shape?
I'd like to help, but I still can't visualize the setup. Maybe just describe all the lines, points and circles involved.
A stirring blade is attached at the level of the center of the wheel. Find the distance from the center of the wheel to the stirrer blade for 30° and 225° angles of rotation. The radius is 2ft and the rod of the blade attached to the blade is 5ft long.
Describe the distance of point P from the left or right of the center of the wheel as a function of the angle of the rotation. Assume that the wheel is rotating counterclockwise and the angle of rotation, θ , is the angle between the horizontal ray through the center of the wheel directed to the right and the radius CP. Find x. (There's a circle. Inside the circle is a right triangle. The radius is CP. The horizontal distance of that triangle is x. y is adjacent to the 90 degrees from point P to the horizontal line. There's also another right triangle next to it, it's much longer and the horizontal distance is z)
Describe the horizontal distanc from point P on the wheel to the stirrer blade as a function of the angle of rotation.
How far is the stirrer blade from the center of the wheel for each angle of rotation? Write a function for the distance in terms of the angle.
I can't seem to figure it out, please help.
5 answers
-----------P.*. C............*...........
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( I tried to draw it out..)
The blade is in a box and the circle moves counterclockwise. As the circle moves more to the left, the blade also moves to the left end of the box, same thing with the right. One line starts from the center and goes up about 45 degrees to P, that's theta. The Then it comes down so that it's the same level as the center. That's the first right triangle. The second right triangle is right next to the first one, except it's more narrow and longer. It's opposite side also has to be on the same level as the center. The distance of the first triangle is x and the second triangle is z. y is perpendicular line from P to the center.
Sorry I'm really bad at explaining this. If you search up "circular motion is converted to linear motion" and go to images, it looks sort of like the 6th and 7th images.
-------------*
------------*
----------*
( I tried to draw it out..)
The blade is in a box and the circle moves counterclockwise. As the circle moves more to the left, the blade also moves to the left end of the box, same thing with the right. One line starts from the center and goes up about 45 degrees to P, that's theta. The Then it comes down so that it's the same level as the center. That's the first right triangle. The second right triangle is right next to the first one, except it's more narrow and longer. It's opposite side also has to be on the same level as the center. The distance of the first triangle is x and the second triangle is z. y is perpendicular line from P to the center.
Sorry I'm really bad at explaining this. If you search up "circular motion is converted to linear motion" and go to images, it looks sort of like the 6th and 7th images.
AH! Finally. I have to run now, but if some other genius like Reiny or Damon doesn't get to it, I'll check in when I get back in a while.
OK. Using the figure at
http://4.bp.blogspot.com/-grakz9URobc/Up9-PNgEd4I/AAAAAAAAAtA/_hhPBXCaelY/s1600/reciprocating.jpg
I shall try to give you a hand. I call point Q the other end of the rod from P. And let R be the foot of the altitude from P to the horizontal axis (CQ).
At angle θ,
PR = y = 2sinθ
CR = x = 2cosθ
RQ = z = √(25-y^2) = √(25-4sin^2θ)
CQ = CR+RQ = 2cosθ + √(25-4sin^2θ)
If I have interpreted the wrong drawing, maybe this will get you started.
http://4.bp.blogspot.com/-grakz9URobc/Up9-PNgEd4I/AAAAAAAAAtA/_hhPBXCaelY/s1600/reciprocating.jpg
I shall try to give you a hand. I call point Q the other end of the rod from P. And let R be the foot of the altitude from P to the horizontal axis (CQ).
At angle θ,
PR = y = 2sinθ
CR = x = 2cosθ
RQ = z = √(25-y^2) = √(25-4sin^2θ)
CQ = CR+RQ = 2cosθ + √(25-4sin^2θ)
If I have interpreted the wrong drawing, maybe this will get you started.
Thank you so much!
1 answer