# circular motion conditions

#### Maidenas

An object with mass m, is moving in the xy-plane and is subject to a force $$\displaystyle \vec{F}=-k \vec{r}$$ where $$\displaystyle k$$ is a positive constant and $$\displaystyle \vec{r}$$ the position vector of the object.

a) Solve the differential equations of the motion of the object, with initial values: for $$\displaystyle t=0$$ we have $$\displaystyle \vec{r}(t=0)=(x_0,y_0)$$ and $$\displaystyle \vec{v}(t=0)=(v_{0x},v_{oy})$$

b) Which is the condition in order the object to make circular motion

c) Which is the condition in order the object to make rectilinear motion with slope $$\displaystyle \frac{\pi}{6}$$ with respect to x axis.
The a) question i managed to solve it !
I solved the differential equations and i found that

$$\displaystyle x(t)=x_0 \cdot cos\left ( \sqrt{\frac{k}{m}}\cdot t \right)+\sqrt{\frac{m}{k}} \cdot v_{0x} \cdot sin\left ( \sqrt{\frac{k}{m}}\cdot t \right)$$
$$\displaystyle y(t)=y_0 \cdot cos\left ( \sqrt{\frac{k}{m}}\cdot t \right)+\sqrt{\frac{m}{k}} \cdot v_{0y} \cdot sin\left ( \sqrt{\frac{k}{m}}\cdot t \right)$$

For the b) question teacher told us to write equations...

I thought that if i take $$\displaystyle x(t)^2+y(t)^2$$ that must always equal to something constant. And in $$\displaystyle t=0$$ the object was in the position$$\displaystyle (x_0,y_0)$$. That means that the magnitude of the position vector is equal to $$\displaystyle \sqrt{x_0^2+y_0^2}$$

Thus, for every t, it has to be $$\displaystyle x(t)^2+y(t)^2=x_0^2+y_0^2$$

But adding the two equations i found in the a) question i can't find a condition in order x(t)^2+y(t)^2 to be indepedent from t.

For the c) question i know that the centripetal acceleration vector has to be equal το 0... only this...

Any help for b) and c)???

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#### Woody

Are you sure you have question 1 correct?
I haven't tried to work this out myself,
but I would expect to see the opposite trigonometric functions for $$\displaystyle x$$ and $$\displaystyle y$$ components.
i.e. where $$\displaystyle x$$ has cos, $$\displaystyle y$$ would have sin and visa versa.

Then when you combine the two equations you would end up with the $$\displaystyle sin^2+cos^2=1$$ arrangement,
which would then allow you to cancel large parts of the equation

#### Maidenas

Yes, they are correct both of the equations 100%. I checked it my calculations many times.

We need to put conditions in order
$$\displaystyle x(t)^2+y(t)^2=x_0^2+y_0^2$$ for every t.... That is what i think has to be done, but i don't know how.

#### Maidenas

Ι finally did the b) question !

the conditions in order that object to make circular motion is:

$$\displaystyle v_0 = \sqrt{ \frac{k}{m}}r_0$$ and
$$\displaystyle x_0v_{0x}+y_0 v_{oy}=0$$
or $$\displaystyle <\vec{r_0}, \vec{v_0}>=0$$

Now only c) has left (!)

#### topsquark

Forum Staff
Your solution holds true (I didn't derive the whole thing but it seems to be correct.) However as Woody suggests there is a different form that is simpler:
x(t) = r_0 cos ( sqrt{ k/m } t + phi )
y(t) = r_0 sin ( sqrt{ k/m } t + phi )

(LaTeX is giving me problems again.)

-Dan

#### Maidenas

Your solution holds true (I didn't derive the whole thing but it seems to be correct.) However as Woody suggests there is a different form that is simpler:
$$\displaystyle x(t) = r_0 ~ cos \left ( \sqrt{ \dfrac{k}{m} } ~ t + \phi \right )$$
$$\displaystyle y(t) = r_0 ~ sin \left ( \sqrt{ \dfrac{k}{m} } ~ t + \phi \right )$$

-Dan

Sorry, i didn't understand. Are you talking about the answer of (b) or (c).

Now i am trying to handle the (c) question.

My first thought is that for every t, it should be
$$\displaystyle \vec{r(t)}-\vec{r_0}=\lambda \vec{v(t)}$$

#### Maidenas

or for every t, it should be $$\displaystyle \frac{y(t)}{x(t)}= \frac{ \pi}{6}$$

#### topsquark

Forum Staff
Ι finally did the b) question !

the conditions in order that object to make circular motion is:

$$\displaystyle v_0 = \sqrt{ \frac{k}{m}}r_0$$ and
$$\displaystyle x_0v_{0x}+y_0 v_{oy}=0$$
or $$\displaystyle <\vec{r_0}, \vec{v_0}>=0$$

Now only c) has left (!)
That was for the solution to the differential equation for a).

For c): Start over with the equation F= kr, where r has components $$\displaystyle x(t) = r~cos \left ( \dfrac{\pi}{6} \right )$$ and $$\displaystyle y(t) = r~sin \left ( \dfrac{ \pi}{ 6} \right )$$

-Dan

#### Maidenas

Do you mean that those equations for (a) (mine and yours) are equivalent ?

#### Maidenas

as for the (c) question , it doesn't say that the angle is $$\displaystyle \frac{\pi}{6}$$. It says the slope is $$\displaystyle \frac{\pi}{6}$$