# circular motion conditions

#### topsquark

Forum Staff
Do you mean that those equations for (a) (mine and yours) are equivalent ?
I think they are. They both seem to solve the same differential equation but looking at the Physics they do the same things. (I'll work out your solution but I don't expect that I'll find anything wrong.)

-Dan

#### topsquark

Forum Staff
as for the (c) question , it doesn't say that the angle is $$\displaystyle \frac{\pi}{6}$$. It says the slope is $$\displaystyle \frac{\pi}{6}$$
That's almost certainly a typo, but in case it's not then my angle would be $$\displaystyle \theta = tan^{-1} \left ( \dfrac{ \pi }{6} \right )$$.

-Dan

#### Maidenas

I think they are. They both seem to solve the same differential equation but looking at the Physics they do the same things. (I'll work out your solution but I don't expect that I'll find anything wrong.)

-Dan
Can you tell me how did you find these equations?

The way I followed was that:

$$\displaystyle \vec{F}=m\vec{a}$$
$$\displaystyle -k\vec{r}=m\frac{d^2 \vec{r}}{dt^2}$$

Which means for the x axis:

$$\displaystyle x''(t)+\frac{k}{m}x(t)=0$$

And for y axis:
$$\displaystyle y''(t)+\frac{k}{m}y(t)=0$$

We find complex roots ect...

#### topsquark

Forum Staff
Yup. Both of our solutions follow from that. For some reason a Physicist would say $$\displaystyle x(t) = A~cos( \omega t + \phi )$$ instead of using the sine function. But you could do it with the sine function also. That just means the $$\displaystyle \phi$$ would be different.

For the sake of simplicity I would use my suggestion for x(t) and y(t) but there's no reason you can't do it the way you did.

-Dan

#### Woody

I am not a skilled mathematician,
I can usually grind my way through to the answer after great effort,
which is why I didn't check your initial answer in the initial post.

However I was surprised to see that you got identical answers for X and Y
For circular motion (in general) I would expect to see very similar solutions for X and Y,
But I would expect one to be 90degrees out of phase with the other.

Is this one of those situations where there are multiple possible solutions,
possibly at 90 degree phase intervals (pure guess)

My suspicion is that your solutions are correct, just not the appropriate ones.
(like the multiple roots of a polynomial, which do you choose).

#### Maidenas

I just asked the teacher, she told me she made wrong in the (c) question. She wanted to use the word angle instead of slope. So we want the angle to be $$\displaystyle \theta = \frac{\pi}{6}$$.

Afterwards, i also made mistake above in my thought that $$\displaystyle \frac{y(t)}{x(t)}= \frac{\pi}{6}$$ for every t.
Because the angle between $$\displaystyle \vec{r}$$ and the x axis is not the same with the angle between the line and the x axis.

So we want $$\displaystyle \frac{\Delta y(t)}{\Delta x(t)}= \frac{\pi}{6}$$ for every t.

#### Maidenas

If the images are allowed i can show you with details how i conclude to (a) and (b). For using the equations of x(t) and y(t) as topsquark says, i have to prove it before using it :/
How can i conclude there. Because if i can use it then the (c) question is very easy.

Either way, thank you both for your time, i appriciate it very much)

#### Maidenas

I think i solved it, but i am not sure about one argument i used:

We want $$\displaystyle \frac{\Delta y(t)}{\Delta x(t)}= \frac{\pi}{6}$$ for every t . So if $$\displaystyle \Delta y(t)\rightarrow 0$$ and $$\displaystyle \Delta x(t)\rightarrow 0$$ then we will have: (this argument i am not sure if it's correct)

$$\displaystyle \frac{d y(t)}{dx(t)}= \frac{\pi}{6}$$ which is equivalent to

$$\displaystyle \frac{y'(t)}{x'(t)}=\frac{\pi}{6}$$

$$\displaystyle \frac{-y_0 p sin(pt)+v_{0y} cos(pt)}{-x_0 p sin(pt)+v_{0x} cos(pt)} =\frac{\pi}{6}$$ where $$\displaystyle p=\sqrt{ \frac{k}{m} }$$

and from this i conclude to the asked conditions:

$$\displaystyle 6y_0=\pi x_0$$
and
$$\displaystyle \pi v_{0x}=6 v_{0y}$$

#### Maidenas

CORRECTION:

We want $$\displaystyle \frac{\Delta y(t)}{\Delta x(t)}= tan\left ( \frac{\pi}{6}\right )$$ for every t . So if $$\displaystyle \Delta y(t)\rightarrow 0$$ and $$\displaystyle \Delta x(t)\rightarrow 0$$ then we will have: (this argument i am not sure if it's correct)

$$\displaystyle \frac{d y(t)}{dx(t)}= \frac{\sqrt{3}}{3}$$ which is equivalent to

$$\displaystyle \frac{y'(t)}{x'(t)}=\frac{\sqrt{3}}{3}$$

$$\displaystyle \frac{-y_0 p sin(pt)+v_{0y} cos(pt)}{-x_0 p sin(pt)+v_{0x} cos(pt)} =\frac{\sqrt{3}}{3}$$ where $$\displaystyle p=\sqrt{ \frac{k}{m} }$$

and from this i conclude to the asked conditions:

$$\displaystyle 3y_0=\sqrt{3} x_0$$

and

$$\displaystyle \sqrt{3} v_{0x}=3 v_{0y}$$

topsquark