Tangential and Centripetal Acceleration in Circular Motion

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  • #26
Nathanael
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Right. But I was talking about the velocity being tangential (that definitely ensures that it moves in a circle).
Ah, yeah you can look at it that way too, but the fact velocity is tangent to the path is implied for any path ##\vec R(t)## because the definition of the unit tangent vector is ##\hat T=\frac{\vec V}{|\vec V|}## so it is enough to show that ##\vec R(t)## traces out a circular path.
 
  • #27
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Yup. This post has finally come to a conclusion. I hereby declare that I won't post any more messages on this thread. :)
 
  • #28
sophiecentaur
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No. That how does radial component of acceleration change the particle's direction so as to keep it moving in a circle.
Reading all your questions and replies to the answers that have been posted it strikes me that you find it hard to 'believe' a proof that involves Calculus. It you are not prepared to get involved with the Calculus then it doesn't surprise me that you are still having difficulty. Why do you think that Calculus was 'invented' if not to deal with these types of problems. It was found, of course, that arm waving and arithmetic are not enough.

Calculus approaches problems by looking at how things change when the steps are taken to be smaller and smaller. Approaching your OP as 'verbally' as I can, I would look at what happens after a very small time interval, when there is a radial force applied. The average longitudinal component of the force will approach zero as the time interval approaches zero (simple trigonometry and vector addition of a tiny force forwards and a tiny force backwards). That means there will be no tangential acceleration. The acceleration will all be centripetal - also by simple trig, adding two forces in the same direction and dividing by 2. (See what a drag it is to use words and not symbols?)
 
  • #29
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Reading all your questions and replies to the answers that have been posted it strikes me that you find it hard to 'believe' a proof that involves Calculus. It you are not prepared to get involved with the Calculus then it doesn't surprise me that you are still having difficulty. Why do you think that Calculus was 'invented' if not to deal with these types of problems. It was found, of course, that arm waving and arithmetic are not enough.

Calculus approaches problems by looking at how things change when the steps are taken to be smaller and smaller. Approaching your OP as 'verbally' as I can, I would look at what happens after a very small time interval, when there is a radial force applied. The average longitudinal component of the force will approach zero as the time interval approaches zero (simple trigonometry and vector addition of a tiny force forwards and a tiny force backwards). That means there will be no tangential acceleration. The acceleration will all be centripetal - also by simple trig, adding two forces in the same direction and dividing by 2. (See what a drag it is to use words and not symbols?)
You don't understand what I am asking. I never avoided calculus. The explanation that Nathaniel gave was that the particle moves in a tangent because from the very beginning R(t) traces out a circle. But what I was asking was, that if it moves in a circle because of the nature of R(t), then how does ar help to keep the particle on the circular path? You see? You say the particle moves in a circle because of the way R(t) is defined. And you also say that ar helps to keep the particle in a circle. But you never explained why ar keeps the particle in a circular path. And the closest anyone came to giving an answer was this: "The centripetal acceleration can't change the speed of the particle (the reason for which I provided in the beginning and nobody else did) and pulls on the particle so as to keep it in a circular trajectory at all times." That "pull" word is not sufficient enough to explain why ar keeps the particle in a circular path.
 
  • #30
A.T.
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But what I was asking was, that if it moves in a circle because of the nature of R(t), then how does ar help to keep the particle on the circular path?
The acceleration a is the second derivative, and thus a key part of "the nature" of R.
 
  • #31
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The acceleration a is the second derivative, and thus a key part of "the nature" of R.
Sorry, I didn't understand a thing.
 
  • #33
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Yeah, I know that. This is maths and not physics. But you still didn't explain that how ar changes the direction of the particle to keep it moving in a circle. My question is so simple, yet no one is able to answer it. I wonder why.
 
  • #34
A.T.
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My question is so simple, yet no one is able to answer it. I wonder why.
Maybe It's so simple that there is nothing to explain.
 
  • #35
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Maybe, I am not a physicist. But I don't think it won't have any explanation.
 
  • #36
A.T.
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But I don't think it won't have any explanation.
No explanation beyond the math already posted
 
  • #37
sophiecentaur
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I never avoided calculus.
But you seem to be avoiding what it is telling you. The point of describing Physics (your sporting prfeormance and your Bank Balance) with maths is that Maths describes the relationships between Physical (and other) quantities very well and it predicts outcomes. What more is required? You seem to want to have a verbal narrative running in parallel with what the Maths is telling you. Why should this be necessary?
Verbal speech is just not adequate for dealing with the more sophisticated relationships that constitute modern Science.
 
  • #38
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Velocity is a vector, so it has a magnitude and a direction. If something changes the vector but does not change the magnitude then it must change the direction.

It seems that you already understand how the component parallel to the velocity is the component that changes the magnitude. Since the acceleration component perpendicular to the velocity changes the velocity but does not change its magnitude then it must change its direction.
 
  • #39
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You can also show this as follows. Without loss of generality, we consider a coordinate system where the velocity vector and the acceleration vector are both in the x-y plane and the velocity vector is aligned with the positive x axis. The magnitude of the velocity is:
##s=\sqrt{v_x^2 + v_y^2 + v_z^2} = v_x ##
##\dot{s}= (v_x \dot{v_x} +v_y \dot{v_y} +v_z \dot{v_z} )/(\sqrt{v_x^2+v_y^2+v_z^2}) = \dot{v_x} = a_x ##

The direction of the velocity is:
##\theta = \arctan(v_y/v_x) ##
##\dot{\theta} = (v_x \dot{v_y} - v_y \dot{v_x})/(v_x^2+v_y^2) = \dot{v_y}/v_x = a_y/v_x##

So the change in speed is due to the acceleration in the x-direction, which is parallel to the velocity, and the change in direction is due to the acceleration in the y-direction, which is perpendicular to the velocity.
 
  • #40
Drakkith
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Yeah, I know that. This is maths and not physics. But you still didn't explain that how ar changes the direction of the particle to keep it moving in a circle. My question is so simple, yet no one is able to answer it. I wonder why.
You are aware that it is the centripetal force that causes the change in direction, which is what centripetal acceleration is, right? The particle is under a net force which changes its velocity, and that change in velocity is acceleration. The acceleration doesn't cause the particle to move in a circle, it is the result of the particle being forced to move in a circular path by a force.
 
  • #41
In case of a uniform circular motion, let us add two constant velocities in a chosen x and y directions. In other words, we shift the frame of reference to another inertial frame of reference.The acceleration components would obviously 0, since the velocity components of new frame of reference are constants. Thus the uniform circular motion shall remain a uniform circular motion under transformation into inertial frames of reference and there would be no chance in radial and tangential velocities under such transformation.
 
  • #42
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@andyrk I don't want to hijack your thread, it's interesting. But I also have a simple question on centripetal force or acceleration.
I decided not to start a new thread for such small info.

Can anyone prove mathematically F=(mv2)/R ? or acceleration a=v2/R ?
All proofs I have seen so far in books etc are Graphical, using vector addition method. Again prove by using simple Mathematics not Graphs.
Thanks.
 
  • #43
DrGreg
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@andyrkCan anyone prove mathematically F=(mv2)/R ? or acceleration a=v2/R ?
All proofs I have seen so far in books etc are Graphical, using vector addition method. Again prove by using simple Mathematics not Graphs.
Thanks.
See post #20 with [itex]\ddot \theta = 0[/itex].
 
  • #44
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See post #20 with [itex]\ddot \theta = 0[/itex].
Yes, it does show a=v2/R, Thanks.
I wonder why no book uses such simple derivation instead of geometric hoopla?
Doesn't it also tell us acceleration of a rotating mass tied to string length R is the same as acceleration of
a rotating rod of length R?
 

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