My understanding of circular motion is that the resultant vector of the tangential and radial acceleration causes uniform/ non uniform circular motion depending on whether tangential acceleration is in effect with radial acceleration being a prerequisite. So for uniform circular motion where the radial acceleration brings about an inwards velocity causing a curvature in the path of the object, but if the radial acceleration is constant wouldn't the velocity keep increasing thereby eventually sending the object into a spiral and not uniform circular motion? So far I haven't seen anyone mention this other than bringing up the point that it simply curves the trajectory which is fine and well but wouldn't a constant increase in radial velocity dominate the resultant velocity vector and curve the trajectory more and more over time? I apologize if I'm missing something obvious.
2 Answers
The component of velocity towards the center of rotation is always zero. The specific magnitude of the centripetal force causes it to curve only.
If you don't believe me,
$\vec{r}(t) = R cos(\omega t) \hat i + R sin(\omega t) \hat j$
This represents circular motion. Take the second derivative and you will see the absolute magnitude of this expression is the centripetal acceleration. Aka the acceleration needed BY DEFINITION, to cause circular motion.
Remember, the acceleration isn't constant. The magnitude is. There is actually a $-\hat r$ vector attached to it, acceleration points towards the center, Whose direction changes as the mass moves.
Here the yellow line represents the displacement from the straight line path the object WANTS to go in, to the circular path WE want to go in. due to the force, it pulls it back onto the circular path.
The magnitude of the force is chosen such that thus pull, makes it go on a circular path. Any other magnitude wouldn't make it go on a circle path.
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$\begingroup$ I was under the assumption radial velocity was in effect too. In that case, why is the radial velocity zero, I mean other than the "because the radius of the circle remains the same" explanation. As in why does does radial acceleration not bring about a velocity vector in the constantly rotating radial direction. Is it because it's already a component of an acceleration that is the driving force behind an already existing velocity vector? In that case, could you corelate the dynamics of a spiral trajectory since it has a radial velocity present. It would help a lot, thank you! $\endgroup$– RandomCommented May 8, 2022 at 19:13
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$\begingroup$ Starting with a velocity perpendicular to the radial vector) Imagine the force didn't exist, the v object will move in a straight line, as it moves a component in the radial direction increases.(imagine drawing a line connecting the center, to the object moving a straight line) as time increases, the velocity becomes less perpendicular to the radial vector. The force now is turned on, such that it has a specific magnitude ,to curve it away from its initial straight line path, such that the radius is constant. Any force greater than the exact force required to "just curve it," $\endgroup$ Commented May 8, 2022 at 19:18
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$\begingroup$ Will cause it to speed up and come closer $\endgroup$ Commented May 8, 2022 at 19:18
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$\begingroup$ Appreciate the answer but I'll try to explain what I was trying to get at by drawing parallels to projectile motion. In the case of a horizontal projectile we have a velocity vector in the horizontal direction which is completely unaffected by the acceleration of gravity perpendicular to it but it does bring into a reality a velocity vector in the direction of the acceleration so my question was why doesn't the radial acceleration similarly create a radial velocity vector? $\endgroup$– RandomCommented May 8, 2022 at 20:13
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$\begingroup$ The difference with regular projectile motion. Is that you have an acceleration $-g \hat j$, whose direction is invariant to the position of the particle. Radial acceleration changes as a function of the particles position, they are inherently different. The accelerations magnitude in the case of circular motion pulls it towards the center, enough to cancel out the component of velocity wanting to make it keep going jn a straight line. $\endgroup$ Commented May 8, 2022 at 20:31
In circular motion, a predictable centripetal force is required to produce the rate of change in the direction of the velocity vector which is required to keep the mass following a circular path. If the applied force does not match that prediction, then the path will not be circular and there will be a radial component of velocity. The required force does depend on the speed of the mass.
