Motion in a Circle: Proving the Angular Frequency Equation | Homework Help

[Stochastic13]

Homework Statement

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

\ddot{}x + ω² r = 0 and y'' + ω² r = 0

Homework Equations

ω = v/r

The Attempt at a Solution

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

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

Sorry for using '' instead of like I did on the first x + ωr = 0 since they rarely work

 

[cepheid]

since velocity is constant a_r = 0

Wrong. Velocity is not constant. Speed is constant, but the velocity is changing, because its direction is continuously changing.

The fact that the speed is constant means there is no tangential acceleration (a_t). But there is certainly non-zero radial or centripetal acceleration (a_r). Circular motion requires it.

so x'' has to equal 0 thus x'' + ω² r = 0
the same argument can be applied to y'' + ω² r = 0
proving the second statement.

This argument doesn't make sense at all. Even if ω²r were 0 (it's not by the way), how does that imply that x'' has to be zero? (It's not either, by the way). Think for a second. Does the projected motion in the x-direction ever change speed? Absolutely. In fact, it's simple harmonic oscillation, which is what you are trying to prove by solving this problem.

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

Try expressing x and y in terms of polar coordinates (which describe circular motion more naturally).

Sorry for using '' instead of like I did on the first x + ωr = 0 since they rarely work

Try this (click the equation image to see the LaTeX code for it):

\ddot{x} + \omega^2 x = 0

 

[Stochastic13]

So, should I use x = r cos(\theta) and y = r sin(\theta)?

 

[cepheid]

So, should I use x = r cos(\theta) and y = r sin(\theta)?

Yes. Now, for constant angular speed, what is the expression for theta as a function of time? Note that theta = 0 at t = 0.

 

[Stochastic13]

the expression for \theta as a function of time is ωr
so i take a second derivative of x = r cos(ωr) and y = r sin(ωr)
and get: x'' = ω^2 r cos(ωr) and y'' = - ω^2 r sin(ωr)
so at t = 0 at 0 or 2n*pi: we get x'' = ω^2 which is the same as x'' + ω^2 = 0
but wouldn't y'' = 0 since sin(ωr) = 0 at 0 and 2*pi? If so I didn't prove the second condition.
Am I on the right track? Thanks.

 

[cepheid]

the expression for \theta as a function of time is ωr

No.

ω is the angular speed. It is the rate of rotation in radians/s. Given the (constant) rate of rotation with time, how do you figure out the total amount of rotation in radians in a given time interval?

In other words, given the angular speed, how do you figure out the angular displacement after a certain amount of time?

The units have to work out.

 

[Stochastic13]

Sorry I meant to write ωt instead or ωr, but I still have the same problem when I take a second derivative of x = r cos(ωt) and y = r sin(ωt) and get: x'' = ω^2 r cos(ωt) and y'' = - ω^2 r sin(ωt) so at t = 0 : we get x'' = ω^2 which is the same as x'' + ω^2 = 0 but wouldn't y'' = 0 since sin(ωt) = 0 at t = 0? If so I didn't prove the second condition.

 

[cepheid]

Sorry I meant to write ωt instead or ωr, but I still have the same problem when I take a second derivative of x = r cos(ωt) and y = r sin(ωt) and get: x'' = ω^2 r cos(ωt) and y'' = - ω^2 r sin(ωt) so at t = 0 : we get x'' = ω^2 which is the same as x'' + ω^2 = 0 but wouldn't y'' = 0 since sin(ωt) = 0 at t = 0? If so I didn't prove the second condition.

Why are you evaluating things at t = 0 specifically? Both \ddot{x} and \ddot{y} are functions of time. The point is that the equations you are deriving hold true for ALL t. Sure, at some points in time, the second derivatives happen to be equal to zero. So what? This is a necessary part of simple harmonic motion.

Once you have the result that x'' = -ω²rcos(ωt), the thing you are supposed to do is to notice that on the right hand side of the equation, rcos(ωt) is just equal to the function itself, x(t). Therefore, the equation becomes:

x'' = -ω²x

Similarly, on the right hand of the y equation, the factor rsin(ωt) is just equal to the function y(t) itself. Therefore, this equation becomes:

y'' = -ω²y

Done.

 

[Stochastic13]

Wow thanks for your help, it was very helpful. I just don't get this rotation thing, now I think I do, but before, no. I'm not that slow usually, and actually have had no problems solving equations using Newtons Laws, but this rotation is hard for me to grasp for some reason.