Solve Centripetal Force Equation for Pivot Point

[avolaster]

as you all know the equation for centripetal force is (m((2 pie r)/t)^2)/r). ok so imagine this. a weight attached to a string attached to a pivot point. the weight is rotating around the pivot point. the pivot point can move back and forth along the x axis. if looking at a x, y graph. the pivot point starts at the origin (0,0). obviously when the weight reaches the y-axis all of its force is facing the x-axis in one direction or another. i want to know how much force would be acted upon the pivot point at any given point in time with an equation. i realize this is possibly difficult to visualize so if you need clarifications please ask

 

[HallsofIvy]

In order to answer that, you would have to know how the force affects the motion of the pivot point. Is there friction acting on it?

 

[avolaster]

yes there is. and what do you mean by how the force affects the motion?

 

[K^2]

Assuming no friction and a massless arm/string, we need to know the mass of the pivot. If the mass of the pivot is zero, the thing won't spin around it at all due to unbalanced forces.

If the pivot has a mass, then in direction that the pivot is allowed to slide, the center of mass of the system will not accelerate. You can use this fact to deduce the motion of the pivot, and from that, find forces acting on it.

 

[avolaster]

the pivot point does have a mass. k^2 you sound spot on. yet i fail to fully understand what you are saying clearly. would you please mind explianing it more?

 

[K^2]

We had a discussion about this, maybe even on this forum. The guy I was arguing with sent me a video of the exact setup we are talking about, and I superimposed predicted motion over that video.

Here is the resulting video.

The red dot on the arm is the center of mass between the slide and the rotating mass. The green line is predicted motion of center of mass based on assumption of no friction. As you can see, red dot deviates a bit in the end, most likely due to friction in the slide.

Here is what's going on. While the rotating mass is on the left, the tension pulls the slide to the left, and it's prevented from moving that way. So it works as a stationary pivot. As soon as the arm passes the low point, the tension starts to pull to the right. From that moment on, the slide is free to move.

Now, the slide + rotating mass can be viewed as a closed system. There are no sources of horizontal force, so horizontal velocity of center of mass must remain constant, while vertical velocity is just given by rotation of the arm around the pivot. That's the green line, and that's the motion of the red dot.

The angular momentum of the arm around the pivot point is conserved, since there is no torque about the pivot. That means angular velocity of the arm is constant. Knowing these facts, you should be able to project the motion of both the slide and the rotating mass. Knowing velocities at each instant of time, you can derive accelerations, and therefore, forces acting on the slide and the rotating mass. Naturally, you'll find these to be equal and opposite in the horizontal direction. In vertical direction, you also have to account for the normal force provided by the rail.