Dynamics Polar Coordinates question

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Discussion Overview

The discussion revolves around the application of radial and tangential forces in dynamics, particularly in the context of polar coordinates. Participants explore when to consider these forces in various scenarios, including circular motion and changes in angular position. The conversation includes technical explanations and examples related to forces acting on objects in motion.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Homework-related

Main Points Raised

  • Some participants express confusion about when to use radial versus tangential forces, particularly in relation to the orientation of motion (horizontal or vertical).
  • There is a discussion about the equations of motion (F = ma) for radial and tangential components, with some suggesting that one can only use the radial equation when the radial force is known.
  • Participants propose that a change in angle indicates the presence of an angular force, but there is uncertainty about whether this applies uniformly or only under certain conditions.
  • Clarifications are made regarding centripetal acceleration and radial acceleration, with some participants noting the need to differentiate between them in calculations.
  • There is a question about the existence of radial force in certain scenarios, with some arguing that not all situations involve a distinct radial force, while others emphasize the radial component of the net force.
  • Participants discuss the representation of forces in free body diagrams, particularly the distinction between radial and transverse components.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and agreement on the concepts discussed. Some points remain contested, particularly regarding the definitions and applications of radial and tangential forces, as well as the conditions under which they apply. The discussion does not reach a consensus on several technical aspects.

Contextual Notes

Limitations include potential misunderstandings of the definitions of radial and tangential forces, the conditions under which they are applicable, and the representation of these forces in diagrams. There is also ambiguity regarding the relationship between centripetal acceleration and radial acceleration.

Marchese_alex
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Hi everyone. I am a little desperated cause my exam is on monday and still much stuff to do.
I don't get when I am supposed to use/consider radial and tranversal forces. Most excercises say "it rotates on the horizontal or vertical" I guess this is the info that tells me if there is radial/tranversal forces but i don't get it. Some time they consider the forces and sometimes they dont. I attach an example

Ps. Please keep it simple cause the profesor isn't that good and basically I am learning by my self
Thanks!
 

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Welcome to PF!

Hi Marchese_alex! Welcome to PF! :smile:
Marchese_alex said:
I don't get when I am supposed to use/consider radial and tranversal forces.

Usually, you have an F = ma equation for each,

but one of them involves an unknown force …

in that case, you have to use the other one!​

in this case, the only radial force is the spring (because there's no friction), and you know what that is, so you can use the radial equation

you can't use the tangential equation because you don't know what the tangential force is (until you work it out from the radial equation!) :wink:
 
tiny-tim said:
Hi Marchese_alex! Welcome to PF! :smile:


Usually, you have an F = ma equation for each,

but one of them involves an unknown force …

in that case, you have to use the other one!​

in this case, the only radial force is the spring (because there's no friction), and you know what that is, so you can use the radial equation

you can't use the tangential equation because you don't know what the tangential force is (until you work it out from the radial equation!) :wink:

What I've gather from internet(because my profesor never teach us this) is like you say... i will have two forces radial (Fr=ma) in the direction of "r" (according to book) and angular force, that will be tangential to the path or just perpendicular to Fr (Fbeta=ma) correct me if wrong. If i see that the object angel changes I say that there's an angular force. For example, if the rod goes from horizontal (flat on floor) to vertical, i can say that a angular force is present. Correct?

And since there's a change in angle it will be going in a circular motion, and if there's a circular motion I have a centripetal force. Is that is the same thing as radial force or is the samething as radial aceleration(Ar)--> Fr=mAr
 
Marchese_alex said:
What I've gather from internet(because my profesor never teach us this) is like you say... i will have two forces radial (Fr=ma) in the direction of "r" (according to book) and angular force, that will be tangential to the path or just perpendicular to Fr (Fbeta=ma) correct me if wrong.

correct, but confusing :confused:

there are two F = ma equations, one radial and one tangential

(essentially, there's only one F = ma vector equation, of which you can take the components in any two independent directions, which you usually take to be perpendicular)​

there may be two forces, or more than two forces, it's the number of equations that matters
If i see that the object angel changes I say that there's an angular force. For example, if the rod goes from horizontal (flat on floor) to vertical, i can say that a angular force is present. Correct?

if the angel o:) accelerates then yes, there must be a force in the tangential direction

(but if the angle changes uniformly, no)
And since there's a change in angle it will be going in a circular motion, and if there's a circular motion I have a centripetal force.

correct :smile:

(even if the angle changes uniformly)

except you really ought to call it centripetal acceleration, and not mention centripetal force unless you're using a rotating frame of reference
Is that is the same thing as radial force or is the samething as radial aceleration(Ar)--> Fr=mAr

if there is angular motion, then there is always a centripetal acceleration

this is separate from the radial acceleration, which is d2r/dt2…

you must subtract the centripetal acceleration (because it's always negative :wink:) from the radial acceleration to get the total "a" to put in the radial F = ma :smile:​
 
to illustrate the last point, try this problem …

a smooth rod rotates uniformly and horizontally with angular velocity ω

a bead is free to slide (without friction) on the rod … obviously, it will move away from the centre!

but the only force on it (there's no spring or anything else) is the normal force from the rod, which is tangential …

so why does it have a radial motion, and what is the equation? :smile:​
 
if the angel o:) accelerates then yes, there must be a force in the tangential direction

(but if the angle changes uniformly, no)


if there is angular motion, then there is always a centripetal acceleration

this is separate from the radial acceleration, which is d2r/dt2…

you must subtract the centripetal acceleration (because it's always negative :wink:) from the radial acceleration to get the total "a" to put in the radial F = ma :smile:​
[/QUOTE]


When you say substract centripetal acceleration you mean ~~> see part III picture

Is my diagram correct? Is radial force(Fr) always in that direction?

http://i174.photobucket.com/albums/w105/marchese_alexander/cd953eb0b2ce1c91d11cce442007042f.jpg
 
Marchese_alex said:
When you say substract centripetal acceleration you mean ~~> see part III picture

yes, the radial component of acceleration is always ar = r'' - r(θ')2 :smile:

r'' is what i call the radial acceleration (i don't know whether other people call it that, or whether they call the whole of ar the radial acceleration)

and r(θ')2 is the centripetal acceleration
Is my diagram correct? Is radial force(Fr) always in that direction?

there isn't necessarily a radial force

for example, if a weight on a string attached to the ceiling moves in a horizontal circle, the only forces are vertical (the weight) and along the string (the tension) …

but the radial direction is horizontal :wink:​
 
What exactly make this excercises different that they say Fr exist?
 

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no, they don't call it the radial force, they call it the radial component of the force :wink:

the (net) force always has a radial component,

but there isn't always an individual force along the radial direction :smile:​
 
  • #10
tiny-tim said:
for example, if a weight on a string attached to the ceiling moves in a horizontal circle, the only forces are vertical (the weight) and along the string (the tension) …

but the radial direction is horizontal :wink:​

and since there angular motion I will have centripetal acceleration. Right?
 
  • #11
Ok i get the centri acceleration. Now i just don't get
1.the radial and tranversal component part of a force... Is it radial for "x" component and tranversal for y

2.what exactly is F and Fr representing in the free body diagram
 
  • #12
Marchese_alex said:
and since there angular motion I will have centripetal acceleration. Right?

Right! :smile:
Marchese_alex said:
Ok i get the centri acceleration. Now i just don't get
1.the radial and tranversal component part of a force... Is it radial for "x" component and tranversal for y

2.what exactly is F and Fr representing in the free body diagram

In the diagram, "F" seems to be a mistake for "Fθ".

The problem asks for the "radial and transverse components of the force exerted on A", which are Fr and Fθ.

Radial is the r component, and transverse is the θ component.

(tangential isn't often used, it's the component of the force along the line of motion, which won't be transverse unless r is constant)
 
  • #13
tiny-tim said:
Right! :smile:


In the diagram, "F" seems to be a mistake for "Fθ".

The problem asks for the "radial and transverse components of the force exerted on A", which are Fr and Fθ.

Radial is the r component, and transverse is the θ component.

(tangential isn't often used, it's the component of the force along the line of motion, which won't be transverse unless r is constant)

Got it!
 
  • #14
How about this exercise.

The only part I don't get is why I can't say that the first derivative of the angle is 0 and they put that is w(angular velocity). I know that the first derivative is angular velocity and the second derivative is angular acceleration. Also I know if I say is 0 then I would not be able to find what the exercise is asking. I am just trying to understand the "why".


Heres my data http://i174.photobucket.com/albums/w105/marchese_alexander/5ce37c38be157e984f3f113fe1a3a65f.jpg
 

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  • #15
Marchese_alex said:
… why I can't say that the first derivative of the angle is 0 and they put that is w(angular velocity).

θ' is the angular velocity, ω (same as x' is the ordinary velocity, v)

i don't understand why you think ω = 0 :confused:

if it's moving, then its speed isn't 0, and its angular speed isn't 0
 
  • #16
tiny-tim said:
θ' is the angular velocity, ω (same as x' is the ordinary velocity, v)

i don't understand why you think ω = 0 :confused:

if it's moving, then its speed isn't 0, and its angular speed isn't 0

Ok! Thanks
 
  • #17
  • #18
Bump... Need to know please
 
  • #19
Marchese_alex said:
Why in excercice ( 14.94) its wrong to say Fr-mgsin(angle)=m(bla bla) and rigth to say Fr=m(bla bla). Or symply put why the weight is not being considered?

Whats the difference in this excercice ( 14.96) that they consider "mg"

14.94 is in the x-y plane (ie horizontal) :wink:
 
  • #20
tiny-tim said:
14.94 is in the x-y plane (ie horizontal) :wink:

What do you mean? Still dumb lol

The #95 #97 (can see in pictures) they don't say anything about x,y plane and they put it the same way
 
  • #21
14.96 obviously includes gravity, from the main diagram

the others don't mention gravity (and 14.95 makes it clear that the plane is horizontal, and asks for the horizontal force)
 
  • #22
thats what I don't understand, when you say horizontal. What does horizontal mean? What information it let's me know?
 
  • #23
tiny-tim said:
14.96 obviously includes gravity, from the main diagram

the others don't mention gravity (and 14.95 makes it clear that the plane is horizontal, and asks for the horizontal force)

do you mean that in a plane the W is not considered and at horizontal (no plane) it is concidered
 
  • #24
no, they're the same … if the problem is purely horizontal, then gravity isn't considered either

(i wouldn't worry … exam questions are very clearly worded, even if book examples sometimes aren't :smile:)
 

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