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Non-uniform circular motion

by negation
Tags: circular, motion, nonuniform
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negation
#1
Jan2-14, 06:58 PM
P: 819
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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Simon Bridge
#2
Jan2-14, 08:18 PM
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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.
negation
#3
Jan3-14, 12:55 AM
P: 819
Quote Quote by Simon Bridge View Post
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.
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How do I simplify it?

Simon Bridge
#4
Jan3-14, 02:09 AM
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Non-uniform circular motion

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?
negation
#5
Jan3-14, 02:22 AM
P: 819
Quote Quote by Simon Bridge View Post
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.
Simon Bridge
#6
Jan3-14, 02:28 AM
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deriving angular acceleration using calculus.
Well, I'm not sure what that means.
Can you illustrate the question by deriving linear acceleration using calculus?
negation
#7
Jan3-14, 02:29 AM
P: 819
Quote Quote by Simon Bridge View Post
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.
negation
#8
Jan3-14, 02:38 AM
P: 819
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?
Simon Bridge
#9
Jan3-14, 02:55 AM
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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.
negation
#10
Jan3-14, 03:28 AM
P: 819
Quote Quote by Simon Bridge View Post
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?
Simon Bridge
#11
Jan3-14, 05:36 PM
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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.


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