Centripetal Force
Discussion
seeking the center
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A proof somewhat like this one was first laid out by the English scientist, mathematician, inventor, and theologian Isaac Newton (1642–1727). Here's an image from one of his notebooks showing how he thought of it.
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equations
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a_{c} = | v^{2} |
r |
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a_{c} = − ω^{2}ra (m/s^{2}) | device, event, phenomenon, process |
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0 | moving at a constant speed in a straight line |
2.32 × 10^{−10} | galactic acceleration at the Sun |
8.8 | International Space Station in orbit |
0–150 | human training centrifuge |
10^{4}–10^{6} | medical centrifuge |
derivation using euclidean geometry and algebra
The left side of the diagram below shows a dashed line representing the path of an object following a circular path at a constant speed. (This is called uniform circular motion for people who like technical terms.) The red arrows point to the position of the object at some moment (r_{0}) and at another moment a little bit later (r). The origin of this coordinate system is at the center of the circular path. That means the position vectors are also radii. The blue arrows show the velocity of the of the object at those two moments (v_{0}, v). Velocity vectors are always tangent to the path the object is following. As you were probably taught in a geometry class somewhere, a radius at a point and a tangent at the same point on a circle are always perpendicular.
Let's move the vectors off to the middle part of the diagram so we can better see what's going on. The two position vectors are coming out of the same point — and we'll just keep them that way. The two velocity vectors are coming out of different points. Let's move them so they come out of the same point. The definition of a vector is a quantity with both magnitude and direction. It doesn't say anything about location. Relocating vectors is not a problem. Add an arrow from the tip of the initial vectors (r_{0}, v_{0}) to the tip of the final vectors (r, v). Those are the changes in those vectors (Δr, Δv).
We now have two similar triangles: one with sides r_{0}, r, Δr and another with corresponding sides v_{0}, v, Δv. Both triangles are isosceles (have two equal length sides) and have the same vertex angle (θ). The position vectors r_{0} and r are equal because they're both radii, and all radii are the same length in any one circle. The velocity vectors v_{0} and v are equal because I said the speed was constant. The angle θ in both triangles are the same because the position and velocity vectors are always perpendicular. When you rotate the position vector by some amount you also rotate the velocity vector by the same amount.
For similar triangles, the ratio of the corresponding sides is constant. I know how I want this to end up looking when I'm done, so I'm going to write in in this order. (The last ratio is kind of redundant, since it's sort of a copy of the middle one. We're not going to use it for anything again.)
Δv | = | v | = | v_{0} |
Δr | r | r_{0} |
When I set up the description of this derivation, I intentionally used the phrase "a little bit later" to describe the change in position. I did that for a reason. The two position vectors, r_{0} and r, are also the sides of a sector of a circle. (A sector of a circle is like a slice of a pizza — as long as your pizza is round and "diagonal cut".) The arc length of that sector is the distance traveled by the object. If the time interval is "small", then that distance (Δs) is almost the same as the change in position (Δr) — and the smaller it is, the closer they get.
Δs → Δr | as | Δt → 0 |
Since the speed was constant, there's a simple equation for the distance.
Δr ≈ Δs = vΔt
Substitute back into the similar triangles ratio.
Δv | = | v |
vΔt | r |
Collect like quantities by cross multiplying just the speed (v).
Δv | = | v^{2} |
Δt | r |
Recognize the quantity on the left side? It's acceleration — in this case, centripetal acceleration (a_{c}).
a_{c} = | v^{2} |
r |
derivation using analytical geometry and calculus
Here's one method that can be used to derive the equations the equation for centripetal acceleration. Imagine an object in uniform circular motion — something like a person standing still on a rotating platform. They stay at the same distance from the point of rotation (r = constant) but travel around it at a constant angular velocity (ω = constant).
Represent their position vector (r) a components in a rectangular coordinate system (x, y). Take the derivative of these coordinates to get the velocity components (v_{x}, v_{y}) of the person's velocity vector (v). Then take the derivative of the velocity components to get the acceleration components (a_{x}, a_{y}) and acceleration vector (a). Do it something like this…
r | ⇒ | x = | +r cos(ωt) | y = | +r sin(ωt) | ||
v = | dr | ⇒ | v_{x} = | −rω sin(ωt) | v_{y} = | +rω cos(ωt) | |
dt | |||||||
a = | dv | ⇒ | a_{x} = | −rω^{2} cos(ωt) | a_{y} = | −rω^{2} sin(ωt) | |
dt |
We will now analyze these results to determine the magnitudes and relative directions of the three kinematic vectors. Use the pythagorean theorem to get the magnitudes of position, velocity, and acceleration.
Position first.
r^{2} = x^{2} + y^{2} r^{2} = [+r cos(ωt)]^{2} + [+r sin(ωt)]^{2} ^{0}r = r |
The person stays a constant distance from the center of rotation. That's what it means to follow a circular path. A circle is the locus of points equidistant from a point.
Velocity second.
v^{2} = v_{x}^{2} + v_{y}^{2} v^{2} = [−rω sin(ωt)]^{2} + [+rω cos(ωt)]^{2} v_{t} = rω |
The translational and angular velocities are related by the expected equation.
Acceleration third.
a^{2} = a_{x}^{2} + a_{y}^{2} a^{2} = [−rω^{2} cos(ωt)]^{2} + [−rω^{2} sin(ωt)]^{2} a_{c} = rω^{2} |
Speed has a constant value, but direction is changing. There is an acceleration and that's its equation. If you don't like angular quantities, you can use algebra to state centripetal acceleration in terms of tangential velocity. Do that and you get…
a_{c} = | v^{2} |
r |
Use trig identities to get the relative directions of position, velocity, and acceleration. Start with position and acceleration, since that's the easier pair to compare. The trig functions match, but the signs are opposite. This means that whatever direction the position vector points, the acceleration vector points the opposite way. Since the position vector always points out and away from the center of rotation, the acceleration vector always points in and towards the center. In other words, the acceleration is centripetal.
x = | +r cos(ωt) | y = | +r sin(ωt) | |
a_{x} = | −rω^{2} cos(ωt) | a_{y} = | −rω^{2} sin(ωt) | |
a_{x} = | −xω^{2} | a_{y} = | −yω^{2} | |
a = −r ω^{2} |
Less obvious is what to do with the velocity. The sine and cosine functions are full of symmetries by themselves and with each other. There are all kinds of ways they can be flipped and shifted so that one function becomes the other. By flip I mean a change in sign or a reflection and by shift I mean the addition of a phase angle or a translation. For example…
These identities can be used to show that the velocity vector is 90° ahead of the position vector during uniform circular motion; that is to say, tangent to the circular path in the direction of motion. Maybe you learned it somewhere, but if not I'm telling you now, a tangent and a radius are perpendicular.
x = | +r cos(ωt) | y = | +r sin(ωt) | |
v_{x} = | −rω sin(ωt) | v_{y} = | +rω cos(ωt) | |
v_{x} = | +rω cos(ωt + 90°) | v_{y} = | +rω sin(ωt + 90°) | |
There is no nice way to write this mathematically^{} |
Since centripetal force and radius are 180° apart, centripetal force and velocity are 90° apart. This is another proof that can be done analytically, but I don't want to do it. Just refer back to the graphic near the start of this part of the discussion.
centrifugal
discussion
rotating reference frame