motion in a circle

[CaptainBlood]

An object is moving counterclockwise in a circle of radius r at constant speed v. The center of the circle is at the origin of rectangular coordinates (x, y), and at t = 0 the particle is at (r, 0). If the angular frequency is given by ω = v/r, show that x'' + ω^2 r = 0 and y'' + ω^2 r = 0

Attempted Solution:If particle is at (r,0) then r = x we know: ar = v2/r = ω^2 since velocity is constant ar = so x'' has to equal 0 thus x'' + ω2 r = 0 the same argument can be applied to y'' + ω^2 r = 0 proving the second statement

Is this correct? Is there a better way of showing the two statements are true?

Edited April 4, 2011 by CaptainBlood

[DrRocket]

An object is moving counterclockwise in a circle of

radius r at constant speed v. The center of the cir-

cle is at the origin of rectangular coordinates (x, y),

and at t = 0 the particle is at (r, 0). If the "angular

frequency" is given by ω = v/r, show that

 

x'' + ω2 r = 0 and y'' + ω2 r = 0

 

Attempted Solution:

 

If particle is at (r,0) then r = x

 

This is true for t=0 and ωt = 2n pi

 

we know: ar = v2/r = ω2 r

since velocity is constant ar = 0

 

I assume ar is radial acceleration and not a x r. In any case velocity is not constant. Speed is constant.

 

If ar = v2/r and v2 is constant then ar is not zero unless v is also zero. If ar were 0 there would be no such thing as centripetal acceleration.

 

so x'' has to equal 0 thus x'' + ω2 r = 0

 

nope

 

I would approach this by noting that for uniform circular motion x = r cos( ωt) and y = r sin(ωt) and then differentiate to get velocity and acceleration.

 

Your equations are not true in general, but x'' + ω2 r = 0 is true for t=0 while y'' + ω2 r = 0 is not.

[CaptainBlood]

I'm sorry, I don't understand what is t and what is n in ωt = 2n pi. Also why are you using ωt in x = r cos (ωt) instead of theta?

Edited April 4, 2011 by CaptainBlood

[DrRocket]

I'm sorry, I don't understand what is t and what is n in ωt = 2n pi. Also why are you using ωt in x = r cos (ωt) instead of theta?

 

t is time. n is an integer. [math]\omega t = 2n \pi [/math] is required in order that [math] cos( \omega t) = 1 [/math]

 

[math]\omega t = \theta[/math] in uniform circular motion.

 

It would help if you indicated your level of education and where you are encountering this problem.

Edited April 5, 2011 by DrRocket

[CaptainBlood]

Great, thank you very much DrRocket, I gotta run now but I'll try to solve this a little later.

This problem is given in the first semester Physics course and unfortunately before we learned about rotational motion besides the ar = v^2/r , I'm undergrad.

Edited April 5, 2011 by CaptainBlood

[CaptainBlood]

Ok, I did it. Thanks a lot for your help I really appreciate it.

Edited April 5, 2011 by CaptainBlood