# How to calculate the centripetal force along a cartesian axis?

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.

• 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. Commented May 1, 2023 at 20:47
• @DKNguyen There is one continuously increasing centrifugal force as the centrifuge is accelerating.
– Dale
Commented May 1, 2023 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. Commented May 1, 2023 at 21:08
• As the centrifuge accelerates the centripetal force increases the way a spiral does.
– Dale
Commented May 1, 2023 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 Commented May 1, 2023 at 22:53

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 $$X_1= x cos(ωt) + y sin(ωt)= x cos(ωt) +0=x cos(ωt)$$ $$Y_1= -x sin(ωt) + y cos(ωt) = -x sin(ωt) +0= -x sin(ωt)$$
• @Dale, plugin initial values and $cos(\alpha t^2)$ and integrate. Commented May 2, 2023 at 5:48