
#1
Jan214, 06:58 PM

P: 710

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? 



#2
Jan214, 08:18 PM

Homework
Sci Advisor
HW Helper
Thanks ∞
PF Gold
P: 11,114

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.




#4
Jan314, 02:09 AM

Homework
Sci Advisor
HW Helper
Thanks ∞
PF Gold
P: 11,114

nonuniform 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? 



#5
Jan314, 02:22 AM

P: 710





#6
Jan314, 02:28 AM

Homework
Sci Advisor
HW Helper
Thanks ∞
PF Gold
P: 11,114

Can you illustrate the question by deriving linear acceleration using calculus? 



#7
Jan314, 02:29 AM

P: 710

How do I derive angular acceleration using differentiation. 



#8
Jan314, 02:38 AM

P: 710

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
Jan314, 02:55 AM

Homework
Sci Advisor
HW Helper
Thanks ∞
PF Gold
P: 11,114

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. 



#10
Jan314, 03:28 AM

P: 710

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? 



#11
Jan314, 05:36 PM

Homework
Sci Advisor
HW Helper
Thanks ∞
PF Gold
P: 11,114

##\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. 


Register to reply 
Related Discussions  
Circular Motion,Uniform motion in Vertical Circles  Introductory Physics Homework  2  
Do uniform circular motion and nonuniform circular motion have same formula for a?  Classical Physics  2  
Vertical circular motion in uniform circular motion  Introductory Physics Homework  3  
Uniform circular motion  plane in a circular arc  Introductory Physics Homework  1  
Uniform Circular Motion, Rotational Motion, Torque, and Inertia  General Physics  1 