Non-uniform circular motion vectors

In summary, Δv/v = Δr/r is a geometric interpretation of rotational motion, where v and r represent velocity and acceleration, respectively. The limit Δt→0 shows that Δv and the acceleration Δv/Δt are perpendicular. This relation is derived from the conservation of angular momentum, where v is the instantaneous tangential velocity and r is the instantaneous distance from the center of rotation. Angular displacement, velocity, and acceleration are analogous to linear displacement, velocity, and acceleration. The equations for arclength, tangential velocity, and tangential acceleration can be used to simplify the concepts of angular motion. The differentiation of angular displacement and velocity is simply a definition of terms, not a derivation. Expressing concepts in mathematical
  • #1
negation
818
0
Δv/v = Δr/r

what is the v and r denominator? I know they both refers to velocity and acceleration respectively but exactly which vector?
Base on geometric interpretation, →vector 1 + Δv = →vector 2. v is a magnitude but exactly which magnitude are we referring to?
Also it states that the limit Δt→0,Δv and the acceleration Δv/Δt becomes exactly perpendicular, how so?
 
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  • #2
I think that relation is derived from the conservation of angular momentum ... in which case v is the instantaneous tangential velocity and r is the instantaneous distance from the center of rotation.
 
  • #3
Simon Bridge said:
I think that relation is derived from the conservation of angular momentum ... in which case v is the instantaneous tangential velocity and r is the instantaneous distance from the center of rotation.

Capture.JPG


How do I simplify it?
 
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  • #4
I would not have made those definitions...

Angular displacement is ##\theta## unit = radian.
(strictly - angular displacement is the change in angles.)

Angular velocity is ##\omega=\frac{d}{dt} \theta##
Angular acceleration is ##\alpha = \frac{d}{dt}\omega##
... all those are instantaneous definitions.
The average angular velocity is ##\bar\omega = \Delta\theta /\Delta t##

##\theta##, ##\omega##, and ##\alpha##, are to rotational motion what ##s##, ##v## and ##a## are to linear motion (i.e. suvat equations).

Now for the equations on your attachment:
The arclength subtended by an angular displacement of ##\theta## is ##S=r\theta##
Similarly the arclength between ##\theta_i## and ##\theta_f## is ##S=r(\theta_f-\theta_i)##
The tangential velocity is ##v_\perp = r\omega##
The tangential acceleration is ##a_\perp = r\alpha##

Related:
Radial velocity is ##v_r=\frac{d}{dt}r##
Centripetal acceleration is ##a_c = v_\perp^2/r=r\omega^2##

Total linear velocity is the vector sum of the radial and tangential veocities.
Total linear acceleration is the vector sum of the centripetal and tangential accelerations.

That make sense now?
 
  • #5
Simon Bridge said:
I would not have made those definitions...

Angular displacement is ##\theta## unit = radian.
(strictly - angular displacement is the change in angles.)

Angular velocity is ##\omega=\frac{d}{dt} \theta##
Angular acceleration is ##\alpha = \frac{d}{dt}\omega##
... all those are instantaneous definitions.
The average angular velocity is ##\bar\omega = \Delta\theta /\Delta t##

##\theta##, ##\omega##, and ##\alpha##, are to rotational motion what ##s##, ##v## and ##a## are to linear motion (i.e. suvat equations).

Now for the equations on your attachment:
The arclength subtended by an angular displacement of ##\theta## is ##S=r\theta##
Similarly the arclength between ##\theta_i## and ##\theta_f## is ##S=r(\theta_f-\theta_i)##
The tangential velocity is ##v_\perp = r\omega##
The tangential acceleration is ##a_\perp = r\alpha##

Related:
Radial velocity is ##v_r=\frac{d}{dt}r##
Centripetal acceleration is ##a_c = v_\perp^2/r=r\omega^2##

Total linear velocity is the vector sum of the radial and tangential veocities.
Total linear acceleration is the vector sum of the centripetal and tangential accelerations.

That make sense now?

No, can you give me a leg-up in deriving angular acceleration using calculus.
 
  • #6
deriving angular acceleration using calculus.
Well, I'm not sure what that means.
Can you illustrate the question by deriving linear acceleration using calculus?
 
  • #7
Simon Bridge said:
Well, I'm not sure what that means.
Can you illustrate the question by deriving linear acceleration using calculus?

I typed too fast and didn't know what was I fast-typing.

How do I derive angular acceleration using differentiation.
 
  • #8
It's just really strange.
I'm starting to feel really frustrated with the tautological terms revolving around circular motion.

If angular displacement is the change in angle as you put it and angular velocity is the change in angle with respect to time, aren't they tautological, and if they are suppose to mean 2 different concept, isn't there a contradiction?
 
  • #9
If angular displacement is the change in angle as you put it and angular velocity is the change in angle with respect to time, aren't they tautological, and if they are suppose to mean 2 different concept, isn't there a contradiction?
Compare with the case for linear motion.

I define the displacement s to be the change in position.
That's a definition of what the word "displacement" means.
If an object starts at position ##\vec p_1## and ends in position ##\vec p_2## then the displacement is given by $$\vec s = \vec p_2 - \vec p_1$$ ... which is just the same definition written out in the language of mathematics.

The velocity is the rate of change of displacement.
That's the definition of what the word "velocity" means.
In the language of maths I write that definition out as $$\vec v=\frac{d\vec s}{dt}$$
... I trust you do not think that ##\vec v## and ##\vec s## are somehow the same thing?

Now for circular motion.
The rotational concept closest to position is "orientation".
The object's orientation is just the angle that it makes with respect to some reference.
So - angle is like position.

From here, the change in angle gives the angular displacement.
That's just the definition of displacement.

The rate of change of displacement is the definition of velocity, so the rate of change of angular displacement must be the definition of "angular velocity". In math we write that down as: $$\vec \omega = \frac{d}{dt}\vec\theta$$ ... but I usually leave the vector signs off to save typing.

So that's the differentiation - it is just a definition of terms not a derivation.
 
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  • #10
Simon Bridge said:
Compare with the case for linear motion.

I define the displacement s to be the change in position.
That's a definition of what the word "displacement" means.
If an object starts at position ##\vec p_1## and ends in position ##\vec p_2## then the displacement is given by $$\vec s = \vec p_2 - \vec p_1$$ ... which is just the same definition written out in the language of mathematics.

The velocity is the rate of change of displacement.
That's the definition of what the word "velocity" means.
In the language of maths I write that definition out as $$\vec v=\frac{d\vec s}{dt}$$
... I trust you do not think that ##\vec v## and ##\vec s## are somehow the same thing?

Now for circular motion.
The rotational concept closest to position is "orientation".
The object's orientation is just the angle that it makes with respect to some reference.
So - angle is like position.

From here, the change in angle gives the angular displacement.
That's just the definition of displacement.

The rate of change of displacement is the definition of velocity, so the rate of change of angular displacement must be the definition of "angular velocity". In math we write that down as: $$\vec \omega = \frac{d}{dt}\vec\theta$$ ... but I usually leave the vector signs off to save typing.

So that's the differentiation - it is just a definition of terms not a derivation.

Expressing everything in mathematical language is much useful and of greater clarity than expressing it in natural language.
Thanks.
Edit: if Θ is the displacement in circular motion then what is radius . Θ? Isn't it displacement too?

Now, given that ω=ΔΘ/Δt;
can I produce the angular acceleration?
 
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  • #11
if Θ is the displacement in circular motion then what is radius . Θ? Isn't it displacement too?
It is possible to have more than one displacement.

##\theta## is the angular displacement, ##\vec r## is the displacement from the center of the circle that the object is moving in, and the center itself can also be moving ... making for another displacement.

Now, given that ω=ΔΘ/Δt;
can I produce the angular acceleration?
... of course, just use the definition from post #4.
 

1. What is non-uniform circular motion?

Non-uniform circular motion refers to the movement of an object along a circular path at varying speeds. This means that the object is not moving at a constant rate, but instead, its speed is changing as it moves around the circle.

2. How are vectors related to non-uniform circular motion?

In non-uniform circular motion, the velocity and acceleration of the object are represented by vectors. These vectors indicate the direction and magnitude of the object's movement and change in speed, respectively.

3. How do you calculate the velocity vector in non-uniform circular motion?

The velocity vector in non-uniform circular motion can be calculated by finding the rate of change of the object's displacement along the circular path. This can be done using the formula v = rω, where v is the velocity vector, r is the radius of the circle, and ω is the angular velocity.

4. What is the relationship between acceleration and non-uniform circular motion?

In non-uniform circular motion, the acceleration of the object is constantly changing in both magnitude and direction. This is because the object's speed is changing, which causes a change in its velocity and therefore, its acceleration. The acceleration vector is always directed towards the center of the circle.

5. How does centripetal force come into play in non-uniform circular motion?

Centripetal force is the force that keeps an object moving in a circular path. In non-uniform circular motion, the centripetal force is constantly changing to match the changes in the object's acceleration. This force is always directed towards the center of the circle and is responsible for keeping the object moving in a circular path.

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