Suppose you have an object undergoing uniform circular motion, with force vector pointing towards the center and another force vector tangential. Can it be said that the net force pointing in the direction towards the center of the circle is equal to the centripetal force; or, as I seem to have mistakenly assumed, the net force on the object is equal to the centripetal force?
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asked Nov 23, 2014 at 18:33
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$\begingroup$ Try and think about how the friction force functions normally on earth, does it have any magnitude if the person's velocity is 0 with respect to the ground? $\endgroup$– KSabCommented Nov 23, 2014 at 18:44
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$\begingroup$ If there is no friction acting on the guy, he will remain in the same place relative to some observer of the spaceship watching on Earth. With friction, the guy will rotate with the spaceship. Right? $\endgroup$– math_loverCommented Nov 23, 2014 at 19:35
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$\begingroup$ I think I've made multiple attempts in the comments to understand the problem, as well as having generalized the problem, so please unlock. $\endgroup$– math_loverCommented Nov 24, 2014 at 0:00
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$\begingroup$ Hi Joshua, I edited your question to reflect the more general statement of yours since I think that's the conceptual part that is likely to be reopened. $\endgroup$– Brandon EnrightCommented Nov 24, 2014 at 0:21
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$\begingroup$ Thank you, as that is exactly what I meant. Perhaps the below answers should be removed, as they no longer make sense without the rest of the question. $\endgroup$– math_loverCommented Nov 24, 2014 at 0:24
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5 Answers
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Can it be said that the net force pointing in the direction towards the center of the circle is equal to the centripetal force; or, as I seem to have mistakenly assumed, the net force on the object is equal to the centripetal force?
Read the above sentences twice. I'll explain with respect to them.
First, let us get the concept of centripetal force clear. It's definition. It simply means 'force towards the center in circular motion'. It is much like saying 'upward force' or 'downward force'. It is not a special type of force. It just a name given to a force that already exists. So, if you are rotating a stone attached to a string, the tension force is the centripetal force. Suppose some earth-like planet revolves around it's sun-like star in a perfect circle (let's not go into ellipses right now), the gravitational force is the centripetal force.
So, to answer your question, only the force towards the center is the centripetal force. And the force that is tangential to the radius vector is the tangential force. The net force has two components: the centripetal component and the tangential component. Or if I was to explain it to my little sister, "The net force has two components: the one that points towards the center and the one that points in the direction the object moves."
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Friction between the person's shoes and the floor/outer wall of the station would only exist if the person were attempting to start walking around the circumference of the station and thus participate in non-uniform circular motion. Think about how his bowling ball would behave if he just put it down at his side.
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$\begingroup$ But if the person is standing still on the space station, theres static friction making him move with the space station. The bowling ball would move with the space station too. $\endgroup$ Commented Nov 23, 2014 at 19:21
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$\begingroup$ @JoshuaBenabou: This is incorrect; no static friction is needed to make him continue to move with the space station $\endgroup$– DJohnMCommented Nov 26, 2014 at 15:39
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If the person starts with an initial velocity equivalent to the tangent velocity at the edge of the station, he will not slide across the surface on the inside regardless of friction (friction is meaningless in this case, as it cannot act if the person has 0 velocity in respect to the floor beneath him).
Imagine we are looking at this space station as a circle from above and he is at the bottom moving clockwise. His inertia attempts to move him to the left, but the curve of the station is in the way and so exerts an up and slightly right force on him (towards the center of the station). This is, of course, the normal force of the station which is also acting as the centripetal force.
If he stepped onto the station from space, and had 0 velocity relative to our observer, then (assuming the station was frictionless) he would of course not move. The station would endlessly rotate under him (or from his perspective he would be endlessly moving along the frictionless floor). It may seem counterintuitive but he would still be essentially in zero gravity. It would be somewhat analogous to being in orbit around Earth, which is essentially free fall (experiencing zero gravity) with a large horizontal movement.
If it did have friction, as soon as he stepped on, the forces of friction would begin to act on him, as he has a non-zero velocity in respect to the floor of the station, and would continue to do so until he was moving at the same speed, at which point the first case takes over.
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$\begingroup$ Sorry, you are wrong. Consider two books, stacked on top of eachother with friction acting between them. The lower book rests on a frictionless surface. You pull the lower book with some force $F$, and if $F$ is sufficiently small, friction from the lower book acting on the top book will pull the top book so that the top book remains stationary above the lower book. Thus you may have 0 velocity with resepct to the station, but friction can also be acting on you. $\endgroup$ Commented Nov 23, 2014 at 23:59
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$\begingroup$ Also this is question about vectors. $\endgroup$ Commented Nov 24, 2014 at 0:13
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$\begingroup$ @JoshuaBenabou Your point is a valid one, static friction is the type of friction you are talking about, which basically perfectly counteracts any forces working parallel to the ground, however in the case above, where the only force present is a normal force perfectly perpendicular to the ground, friction is still 0. My point was that in the scenario friction is 0 and a non-factor. $\endgroup$– KSabCommented Nov 24, 2014 at 0:33
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Uniform circular motion, almost by definition, involves only a force component in the radial direction, ie the net force only ever points inwards, towards the centre of the described circle, with magnitude $v^2/r$. In this way, the direction of the velocity vector changes at the exact rate required to keep the body "on the circle".
As soon as the net force has another component, in your case a pure tangential component, this will cause the body to accelerate "off the path", moving it either further or closer to the centre of rotation until the extra force ceases to act.
The "centripetal force" (whose popular-culture perception is much disliked by physicists) is simply the name given to the radial net force; it is not a force in of itself, as we have not in any way specified what was generating the acceleration in the first place. It could be gravity, or a piece of string, or electrostatic attraction, but none of that would change the physical analysis of the forces.
Again, in true uniform circular motion, the "centripetal force", however it is caused, should be the only component of the net force.
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Only the component of net force perpendicular to the path contributes to the object following the path (centripetal force) and its reaction (centrifugal force) on the body.
A body moving on a prescribed path with tangent direction unit vector $\hat{e}$ and normal unit vector $\hat{n}$ has its velocity and acceleration decomposed as
$$\begin{align} \vec{v} &= v \hat{e} \\ \vec{a} & = \dot{v} \hat{e} + \frac{v^2}{r} \hat{n} \end{align} $$
where $r$ is the instant radius of curvature of the path. Since net force is $\vec{F} = m \vec{a}$ you can state that
$$\boxed{ \vec{F} = (m\dot{v}) \hat{e} + (m \frac{v^2}{r}) \hat{n} }$$
The first part goes to accelerate the object in the tangential direction, and the second part to force the motion about the curvature $r$.
answered Dec 29, 2015 at 16:31