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- The relationship between radius and velocity depends on whether the velocity is linear or angular, and whether the flow is constant or not. Linear velocity is proportional to angular velocity times radius123, while angular velocity is inversely proportional to radius23. Flow rate is proportional to linear velocity times area45, which means that linear velocity is inversely proportional to radius squared at constant flow45.Learn more:✕This summary was generated using AI based on multiple online sources. To view the original source information, use the "Learn more" links.Since the arclength around a circle is given by the radius*angle (l = r*theta), you can convert an angular velocity w into linear velocity v by multiplying it by the radius r, so v = rw.www.khanacademy.org/science/physics/torque-an…Or another way of thinking about it, if you divide both sides by R, the magnitude of our angular velocity is going to be equal to the magnitude of our velocity, or our speed, over R. Or we can say that R is equal to the speed, magnitude of velocity, over the magnitude of our angular velocity.www.khanacademy.org/science/mechanics-essenti…Well, the key thing to realize here, and we've seen this in multiple videos, is the relationship between the magnitude of angular velocity and the magnitude of velocity, linear velocity. The magnitude of angular velocity times your radius is going to give you the magnitude of your linear velocity.www.khanacademy.org/science/mechanics-essenti…This relationship shows that, at a constant vessel radius, changes in flow are proportionate to changes in velocity, and vice versa. Another important relationship is that velocity, at constant flow, is inversely related to the radius squared (V ∝ 1 / r2 at constant flow).cvphysiology.com/hemodynamics/h013The flow rate is the average velocity times the area. If the velocity was constant, you would get a flow rate that scaled with r2 r 2 (the area). But the velocity goes up for larger pipes - in fact, velocity scales with the square of the radius. And the product of these two squares gives us the 4th power relationship.physics.stackexchange.com/questions/168069/ho…
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