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If a centrifuge without a ballast is centered on the origin of an x-y coordinate plane, and starts at the x-axis rotating with increasing velocity counter-clockwise around the origin, how can the centripetal force of this accelerating centrifuge be calculated in the x-direction, and in the y-direction separately? This is to mathematically understand the "knocking" that is happening in an unbalanced centrifuge.

My attempt: Look up a polar function for a spiral as the centripital force is radial, then convert to cartesian coordinates and integrate dy or dx. Attempted this but ran into the problem that for a spiral there are multiple arms of the spiral for each x or y coordinate.

Dale
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  • There is only one centrifugal force so don't you just calculate it, and then along with the RPM project it onto the X and Y-axis? Could also be done in a single step with vectors but same idea. So if you look at it from the side of the X-Y plane, depending on whether you were looking parallel to the X or Y, you would see the unbalanced mass oscillate in 2 dimensions along the other axis. I don't think it would look like anything special though: two sinusoids 90 degrees out of phase with a frequency matching the RPM with peaks equal to the centrigufal force. – DKNguyen May 01 '23 at 20:47
  • @DKNguyen There is one continuously increasing centrifugal force as the centrifuge is accelerating. – Dale May 01 '23 at 21:06
  • Well if you have a time varying RPM then you need to do a vector projection and it sounds like you are already on that road. But what do you mean by "What do you mean by "Attempted this but ran into the problem that for a spiral there are multiple arms of the spiral for each x or y coordinate." Why can't the function for each X and Y axis just track the center of gravity? Otherwise I guess you would need a function for each point along the rotating circle you were tracking. – DKNguyen May 01 '23 at 21:08
  • As the centrifuge accelerates the centripetal force increases the way a spiral does. – Dale May 01 '23 at 21:19
  • The spiral is the problem. At any time t the centripetal force is mrw^2, along the arm. the arm is at an angle theta to the x axis so the x component is x cos(theta) and theta =integral w dt – Greg Locock May 01 '23 at 22:53
  • so the x component is mrw^2* cos(theta) and theta =integral w dt – Greg Locock May 01 '23 at 22:59
  • @GregLocock I get the mrw^2 and shouldn't the the x component should be r cos(theta)? I am trying to understand how theta = integral w dt. Please provide some context for this last equation so I might understand it better. – Dale May 02 '23 at 01:15
  • That's just integrating the angular velocity to get the the angle. I can see no error in my equations in the last comment – Greg Locock May 02 '23 at 20:51

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let's say your centripetal vector is rotating with an angular speed of $$\omega= \alpha t $$

  • $\alpha =$ angular acceleration

Let's say the instantaneous x and y of the vector on a frame aligned with the vector and rotating with it are $$x=\frac{mv^2_{instant}}{r_{instant}}\quad y=0$$

Then the projection of this vector to our stationary cartesian coordinates is $$X_1= x cos(ωt) + y sin(ωt)= x cos(ωt) +0=x cos(ωt)$$ $$Y_1= -x sin(ωt) + y cos(ωt) = -x sin(ωt) +0= -x sin(ωt)$$

kamran
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