circular motion conditions

Nov 2019
15
1
Athens
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)???
 
Last edited:
Jun 2016
1,142
514
England
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
 
Nov 2019
15
1
Athens
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.
 
Nov 2019
15
1
Athens
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
Apr 2008
2,926
608
On the dance floor, baby!
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
 
Nov 2019
15
1
Athens
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
\(\displaystyle \vec{r(t)}-\vec{r_0}=\lambda \vec{v(t)}\)
 
Nov 2019
15
1
Athens
or for every t, it should be \(\displaystyle \frac{y(t)}{x(t)}= \frac{ \pi}{6}\)
 

topsquark

Forum Staff
Apr 2008
2,926
608
On the dance floor, baby!
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
 
Nov 2019
15
1
Athens
Do you mean that those equations for (a) (mine and yours) are equivalent ?
 
Nov 2019
15
1
Athens
as for the (c) question , it doesn't say that the angle is \(\displaystyle \frac{\pi}{6}\). It says the slope is \(\displaystyle \frac{\pi}{6}\)