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My physics textbook says that at any moment, displacement in uniform circular motion is tangent to the centripetal force, so their dot product is always 0 and there is no work done.

What is the displacement being taken relative to? If the displacement of the object is taken relative to the center of the circle, then the displacement vector is always in the same direction of the force. Why then does the textbook say that displacement is tangent to the centripetal force?

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  • $\begingroup$ "displacement in uniform circular motion is tangent to the centripetal force" If your book says that, it's a pretty bad typo: the displacement is tangent to the circle, and orthogonal to the force. "the displacement vector is always in the same direction of the force" Why do you say that? If the force is centripetal it points radially to the center, so a displacement in the same direction would be a straight line, not a circular motion. The displacement is, by definition, along a circle: and it's pure geometry that every point of this curve (circle) is perpendicular to its radius. $\endgroup$ – stafusa Oct 9 '17 at 20:34
  • $\begingroup$ @stafusa I feel like I am missing something obvious here... What is the difference between displacement and position? $\endgroup$ – u8y7541 Oct 9 '17 at 20:37
  • $\begingroup$ In physics jargon, displacement is usually a change in position. So an initial position might be $x_0=50$ and a second position, assumed some time later, $x_1=65$: the displacement is $\Delta x = x_1-x_0 = 15$. $\endgroup$ – stafusa Oct 9 '17 at 20:39
  • $\begingroup$ @stafusa So what you mean is that the instantaneous displacement is always perpendicular to the force? $\endgroup$ – u8y7541 Oct 9 '17 at 23:52
  • $\begingroup$ In uniform circular motion, yes. $\endgroup$ – stafusa Oct 10 '17 at 7:08
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You may read "displacement" in this context as similar to "velocity". It doesn't mean the absolute displacement from the center, but the relative displacement over time.

Over a time period $\Delta t$, the object is displaced by a small amount. In circular motion, this displacement will be oriented along the circle in the direction of motion.

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Displacement in this context is relative to the (centripetal) force. The force is always at right angles to the velocity - so you can do no work.

Work done per unit time is $\mathbf{F}\cdot\mathbf{v} = Fv\cos\theta$, and when the velocity is at right angles to the force, the angle is $\pi/2$, $\cos\theta=0$ and therefore no work is done.

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  • $\begingroup$ "Displacement in this context is relative to the (centripetal) force." What do you mean by that? $\endgroup$ – u8y7541 Oct 9 '17 at 19:59
  • $\begingroup$ I mean that when you talk about displacement because you want to figure out work, then what matters is the direction of the force relative to the displacement. In circular motion these two are perpendicular so no work is done. $\endgroup$ – Floris Oct 11 '17 at 20:10
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The definition of the work done on a mass is the integral of the force acting on the mass and the infinitesimal displacements $ds$: $E=\int\vec{F}ds$ (this integral is equal to the increase or decrease of the kinetic energy of the mass, depending on the relative direction of the force). By the way, if the force is constant along the way the integral can be written as $E=\vec{F}{\Delta}s $. Now because the distance of the object to the center of the circle on which the object moves with constant velocity doesn't change, the gravitational force doesn't exert work, which means the kinetic energy of the object stays the same. So it's the constancy of the object's distance to the circle's center which ensures that no work is done, and the velocity of the object stays the same.

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