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The following diagram summarizes what, I think, are the forces involved in the movement of a rotating ball at the end of a rope.

Picture

Here, tension and weight are real forces (non fictitious) for an external observer (even for an observer on the ball surface). Weight is compensated by the $T_y$ component of tension.

Hence: what does compensate the $T_x$ component of the real tension force? The fictitious centrifugal force?

Thanks.

EDIT

Many thanks for the answers. I summarize the actual situation in a diagram, including how the centripetal acceleration $\vec{a_c}$ in $T_x$ must be added to speed $\vec{v}$ to change speed's direction but not speed magnitude (the sum of vectors must be an equilateral triangle).

enter image description here

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    $\begingroup$ Nothing compensates for the $T_x$ component. If the $T_x$ component wasn't there, the ball would have a net force of zero acting on it, and hence it would travel in a straight line. $T_x$ is responsible for the circular motion. $\endgroup$
    – ummg
    May 15, 2021 at 20:41

3 Answers 3

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From an inertial observer, the $T_x$ is what provides the center-seeking centripetal force, such that $T_x=\dfrac{mv^2}r$ which is what ensures the motion is circular. If the $T_x$ was balanced (and $\sum \bf \vec F=0$), then there would be no circular motion and as pointed out in the comments, the ball would travel in a straight line.

However, from the non-inertial frame that tracks the conical pendulum bob, the $T_x$ force would in fact be balanced by the fictitious centrifugal force (such that the net force is zero, since the pendulum bob is not accelerating with respect to its own frame).

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    $\begingroup$ Well put. +1 from me. $\endgroup$
    – Gert
    May 15, 2021 at 21:17
  • $\begingroup$ Just one thing. Speed $v$ is tangential to the circumference, centripetal force ($T_x$) is normal to the circumference pointing inwards with acceleration $a_c = v^2/r$. If $v$ is constant, how you possibly can calculate the sum of vectors $v + a_c$ to get a new $v$ vector with modulus unchanged? (and, this is clear, differentially turned to follow the circular path). $\endgroup$ May 15, 2021 at 21:38
  • $\begingroup$ @cibercitizen1 not sure I understand your question. If the speed is constant, then $|\bf\vec v|$ and $|{\bf\vec T}_x|$ remain constant. Furthermore, ${\bf\vec T}_x$ and $\bf\vec v$ are always orthogonal, so $|{\bf\vec v}+{\bf\vec T}_x|$ will remain unchanged. Is this what you meant? $\endgroup$
    – user256872
    May 15, 2021 at 22:25
  • $\begingroup$ @user256872 Yes that's what I meant. But when you add up two orthogonal vectors _| $v$ and $T_x$ so to obtain the "new" $v$, this "new" $v$ is the diagonal which is greater in length than the "older" $v$. I must be missing something. $\endgroup$ May 16, 2021 at 7:04
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    $\begingroup$ @cibercitizen1 That is correct. The ${\bf\vec v}+{\bf\vec T}_x$ vector will have a different (greater) length. Not sure why you would want to add these vectors, but the only particular thing about ${\bf\vec v}+{\bf\vec T}_x$ is that, under the aforementioned conditions, $\|{\bf\vec v}+{\bf\vec T}_x\|=\rm constant$. $\endgroup$
    – user256872
    May 16, 2021 at 7:42
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Centrifugal force

To illustrate @user25's point, see the diagram above.

In the non-inertial frame made up of the two black axis, the centrifugal force $F_c$ is real and prevents $T_x$ from dragging the object along the $x$-axis.

The term 'fictitious' is a misnomer: we experience the centrifugal force every time we're inside a vehicle that changes direction.

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It's not clear why you suggest that the $T_x$ component requires compensation.

In order to sustain circumnavigating motion a centripetal force must be provided.

When it comes to motion there is only one reference: the inertial coordinate system.

Granted: mathematical apparatus exists that allows you to use, say, a rotating coordinate system. Now: the equation of motion that you then use will contain a term for a centrifugal acceleration, commonly referred to as 'the centrifugal force'. That centrifugal term contains the angular velocity of the rotating coordinate system with respect to the inertial coordinate system.

There is no escaping the inertial coordinate system: you always use the inertial coordinate system as the reference of motion.


Since the inertial coordinate system is your reference of motion anyway the straightforward choice is to simply see it that way.

An unbalanced force causes acceleration. Here the $T_x$ component is providing a centripetal force, sustaining circumnavigating motion.




[Later edit] (in response to a comment)

Cognitive psychology

There a psychological dimension here that needs to be recognized. It has to do with how perception of gravity works.

The perception of gravity is different from the perception of other ways force can be exerted upon you. Example: let's say someone is tugging your coat. Some parts of your coat will press against your skin, and you notice that.

Gravity however, acts on all parts of your body the same.

Perception of gravity builds up as follows: when you are standing up your feet have to carry the entire weight of your body, but your pelvis only has to carry your weight from the pelvis up; your neck only has to carry the weight of your head. All parts of your body have sensor cells that report how much compression they are sensing. From your ankle bones up to your neck bones there is a gradient in how much they are compressed. That directional gradient, combined with the sensor input of your equilibrium organ, gives rise to the perception of gravity.

That is: our perception of gravity is not a direct perception. Instead evolution has equipped us with a built in system to infer the presence of a gravitational force.

That system operates automatically. Wenever we sense this directional gradient of compression we infer the presence of a gravitational force.


In a car negotiating a bend

When you are sitting in a car, and the car is taking a corner a force towards the inside of the curve must be provided to sustain the curvilinear motion.

Your arm is against the door (of the car). The car is pushing against your arm, your arm is pushing against your torso. Your arm is the most compressed, followed by your torso, and your free arm is not compressed. So your physical sensation is one of a gradient of compression, with the compression increasing towards the part of your body that presses onto the door of the car.

That physical sensation is identical to being subjected to a centrifugal force because inertial mass is equivalent to gravitational mass.


This, I believe, is why everybody who first starts thinking about cirvilinear motion automatically starts with thinking in terms of a centrifugal force.

What is actually happening is that you are subjected to a centripetal acceleration, and because your body has inertia your body gets compressed a bit.

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  • $\begingroup$ Since the inertial coordinate system is your reference of motion anyway the straightforward choice is to simply see it that way. So when your car goes into a bend you don't experience a force pushing you away from the centre of rotation? Oh well... $\endgroup$
    – Gert
    May 15, 2021 at 21:33
  • $\begingroup$ @Gert I added a discussion of the psychological dimension. $\endgroup$
    – Cleonis
    May 15, 2021 at 22:03
  • $\begingroup$ Good discussion. $\endgroup$ May 16, 2021 at 7:10
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    $\begingroup$ @cibercitizen1 Good to hear you appreciate the discussion. Addtional remark: the name "fictitious force" is the worst possible name; it isn't a force, and it isn't fictitious. Inertia opposes change of velocity. The inertial coordinate system is the reference of motion because inertia is the omnipresent and uniform reference of acceleration. It's just that inertia cannot be categorized as a force. A phenomenon can be categorized as a force if it can act as a third law force pair. That's why gravity falls in the category 'force' and inertia is in a category of its own. $\endgroup$
    – Cleonis
    May 16, 2021 at 7:34
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    $\begingroup$ @cibercitizen1 You will encounter a very, very wide variety of opinions about Newton's third law. The way to learn is to read and think it through. Avoid rote learning. Two moving objects, exerting a force upon each other, will each cause change of velocity of the other object. That type of interaction is is referred to as 'momentum exchange'. The concept of a third law force pair defines what counts as a force. $\endgroup$
    – Cleonis
    May 16, 2021 at 7:56

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