Motion in a Plane: Ellipse, K.E, P.E, and Constant Energy

[abramsay]

Homework Statement

A particle of mass m is in the xy plane so that it's position vector is r = acoswti + bsinwtj, where a, b and w are positive constants a>b.
(a)Show that,
i) The particle moves in an ellipse
ii) the force acting on the particle is always directed towards the origin

(b) Find the,
i) Kinetic energy of the particle at A(a,0) and B(0,b)
ii) Potential energy at A and B
iii) The total energy of the particle and show that it is always constant.

Homework Equations

The Attempt at a Solution

(a)
i) I know this is the equation of an ellipse: (x - h)² / a² + (y - k)² / b² = 1 but I find it hard to show the particle moves in an ellipse.
ii) I differentiated twice to get a = -w²r so the minus sign shows that it is always directed to the origin.
iii) For a force to be conservative, the curl must be equal to zero. I tried doing that and got zero.

(b) I'm somehow stuck. I tried using 1/2mv² but I don't know where to plug in the A and B.
iii) I know I should add the potential and kinetic energy to get the total energy and differentiate it to get something without t, but since I would not get the P.E and K.E, I couldn't go any further.

 

[gabbagabbahey]

2[/SUP] / a² + (y - k)² / b² = 1 but I find it hard to show the particle moves in an ellipse.

Well, if \textbf{r}(t)=a\cos(\omega t)\textbf{i}+b\sin(\omega t)\textbf{j}, what are x(t) and y(t)? What are x^2 and y^2? Can you multiply each by some number, add together the results and get one?

ii) I differentiated twice to get a = -w²r so the minus sign shows that it is always directed to the origin.
iii) For a force to be conservative, the curl must be equal to zero. I tried doing that and got zero.

Good...

(b) I'm somehow stuck. I tried using 1/2mv² but I don't know where to plug in the A and B.

Again, what are x(t) and y(t) if \textbf{r}(t)=a\cos(\omega t)\textbf{i}+b\sin(\omega t)\textbf{j}?

iii) I know I should add the potential and kinetic energy to get the total energy and differentiate it to get something without t, but since I would not get the P.E and K.E, I couldn't go any further.

You should be able to answer this without plugging in any points (like A and B). What is \textbf{v}(t)? What does that make the kinetic energy (as a function of time)? If \textbf{F}(t)=-m\omega^2\textbf{r}(t), what is the potential that gives rise to this force? What do you get when you add the two together?

 

[abramsay]

Do you think what I did is right?

Well, if \textbf{r}(t)=a\cos(\omega t)\textbf{i}+b\sin(\omega t)\textbf{j}, what are x(t) and y(t)? What are x^2 and y^2? Can you multiply each by some number, add together the results and get one?

i) I squared both each component of i and j and divided by a² and b² respectively to get 1.

Again, what are x(t) and y(t) if \textbf{r}(t)=a\cos(\omega t)\textbf{i}+b\sin(\omega t)\textbf{j}?

I assumed at A, the y component is zero, differentiated the x component, squared it and multiply by half the mass. Did the same at point B.
[/QUOTE]

 

[gabbagabbahey]

I assumed at A, the y component is zero, differentiated the x component, squared it and multiply by half the mass. Did the same at point B.

You seem to be missing the point. If \textbf{r}(t)=a\cos(\omega t)\textbf{i}+b\sin(\omega t)\textbf{j}, then you can immediately say x(t)=a\cos(\omega t) and y(t)=b\sin(\omega t).

 

[abramsay]

Ok, I got that side.

I assumed at A, the y component is zero, differentiated the x component, squared it and multiply by half the mass. Did the same at point B.

This is for getting the Kinetic energy at points A and B.

 

[gabbagabbahey]

Ok, I got that side.

This is for getting the Kinetic energy at points A and B.

Okay, and what were your results?

 

[abramsay]

Okay, and what were your results?

I got it to be: 0.5m(aw²coswt)² at A and 0.5m(aw²sinwt)² at B.

 

[gabbagabbahey]

I got it to be: 0.5m(aw²coswt)² at A and 0.5m(aw²sinwt)² at B.

But at each point (A and B), t will have a definite value...what value of t makes \textbf{r}(t)=a\cos(\omega t)\textbf{i}+a\cos(\omega t)\textbf{j}
equal to point A (a,0)=a\textbf{i}+0\textbf{j}?

 

[abramsay]

But no t was given or told to be found in the question...I think I'm confused.

 

[gabbagabbahey]

But no t was given or told to be found in the question...I think I'm confused.

I think you're confused too

At a particular time the particle will be at point "A" (and at some other time it will be at point "B"). Since you know the position of the particle at any time is \textbf{r}(t)=a\cos(\omega t)\textbf{i}+b\sin(\omega t)\textbf{j}, you can calculate what time the particle reaches "A" and what time it reaches "B".