Mass describing an elliptical motion connected to the center by a spring

[obarquero]

Since $$\vec{r}\bot\vec{v}$$ the module of the angular momentum is

$$L = rmv\sin{90} = rmv$$

The centripetal force is exerted by the spring Force, so

$$F_{k} = m \frac{v^2}{r} = m\omega^2 r\ \text{(i)}$$

$$kr = m\omega^2r \Rightarrow$$ $$\omega^2 = \frac{k}{m}.$$

Therefore, the angular momentum

$$L = rm\omega r = r^2\sqrt{km}$$

My doubt here is how to write r as a function of "a" and "b". Evenmore, was the derivation of angular momentum well done?.


Regarding b), according (i), and

$$\omega = \frac{2\pi}{T}$$


We could conclude

$$kr = m\omega^2r$$

and

$$k = \frac{m4\pi^2}{T^2}$$

finally

$$T^2 = \frac{m4\pi^2}{k} \Rightarrow T = 2\pi\sqrt{\frac{m}{k}}$$

Once more there must be a way to write r as a function of a and b, and I am not completely sure if the calculations are correct.

I will appreciate help.

 

[tiny-tim]

welcome to pf!

hi obarquero! welcome to pf!

(have a pi: π and an omega: ω and a square-root: √ )

why is r perpendicular to v?

this is elliptical motion, not circular

(and the spring force is not proportional to the total length, only to the extension)

 

[gneill]

For elliptical motion, the velocity vector is only perpendicular to the radius vector at certain points on the trajectory. Otherwise it's circular motion, not elliptical!

More importantly, ,what precisely is the question you're trying to answer?

 

[obarquero]

Sorry, for some reason I lost my statement, here it is:

A particle of mass m is connected by a spring, constant k, to a fixed point O. Knowing that the trajectory of m is an ellipse centered at O, with semiaxes b and a, determine: a) The angular momentum of the particle with respect to O, b) the period of the particle.

 

[gneill]

Sorry, for some reason I lost my statement, here it is:

A particle of mass m is connected by a spring, constant k, to a fixed point O. Knowing that the trajectory of m is an ellipse centered at O, with semiaxes b and a, determine: a) The angular momentum of the particle with respect to O, b) the period of the particle.

Here's a thought; Suppose the motion of the particle describing the ellipse were to be decomposed into separate X and Y components (X parallel to the major axis of the ellipse, Y parallel to the minor axis). What would be the properties of the individual motions?

 

[obarquero]

An ellipse is the combination of two sinusoidal, with different amplitudes, namely:
$$x = a\ \sin{\omega t}$$
$$y = b\ \sin{\omega t}$$

Lets focus on the questions: a) angular momentum respect O.

My main doubts are: the modulus of the angular momentum can be write as: rmv; i.e. is r perpendicular to v in an ellipse, when the origin is the center of the ellipse?

Regarding b) Period

Is it enough the condition
$$F_k = m\frac{v^2}{r} \Rightarrow kr = m\omega^2 r$$

 

[gneill]

An ellipse is the combination of two sinusoidal, with different amplitudes, namely:
$$x = a\ \sin{\omega t}$$
$$y = b\ \sin{\omega t}$$

Lets focus on the questions: a) angular momentum respect O.

My main doubts are: the modulus of the angular momentum can be write as: rmv; i.e. is r perpendicular to v in an ellipse, when the origin is the center of the ellipse?

Sure. The radial distance is r and the motion is perpendicular to r at the ends of the axes. It satisfies the requirements. In general, L = m*(r x v), for vectors r and v, where the vector r doesn't need to be axes-aligned.

Regarding b) Period

Is it enough the condition
$$F_k = m\frac{v^2}{r} \Rightarrow kr = m\omega^2 r$$

Enough for what? This scalar version would apply at the ends of the axes. You'd need to write a vector version for it to be general.

Also, the problem doesn't state what the relaxed length of the spring is. Is there a logical choice?

 

[gneill]

Regarding b) Period

Is it enough the condition
$$F_k = m\frac{v^2}{r} \Rightarrow kr = m\omega^2 r$$

Enough for what? This scalar version would apply at the ends of the axes. You'd need to write a vector version for it to be general.

I take that back. For an ellipse, the centripetal acceleration is not given by m v²/r, even at the ends of the axes. It's not circular motion.

For a relationship that works, try angular momentum which has a nice neat form at the axes ends.

 

[obarquero]

First of all, the motion can be described as the combination of two independent and perpendicular movements, namely

$$x = a\cos{\theta} = a\cos{\omega t}$$
$$y = b\sin{\theta} = b\sin{\omega t}$$

where a is the semimajor axis and b is the semiminor axis.

Taking derivates we can obtain the velocity components

$$v_x = -a\omega\sin{\omega t}$$
$$v_y = b\omega\cos{\omega t}$$

The velocity modul is

$$v = \omega\sqrt{a^2\sin^2{\omega t} + b^2\cos^2{\omega t}}$$

Regarding question a), angular momentum respect O, we start analyzing the forces. The only force is the spring force, F_k, which in this problem is a
central force respect O. Therefore, the torque must be zero

$$\tau_O = rF_k \sin{180} = 0$$

This implies that the angular momentum is conserved and that we can calculate it in any point of the trajectory. So we choose a convinient one, e.g., point
(a,0), where the angle(wt = 0) is zero, or the time t=0.

$$L_O = rmv = am\omega b$$

We can verify this by calculating it in other point, e.g., (0,b), where the angle(wt = pi/2) is pi/2

$$L_O = rmv = bm\omega a$$

Regarding question b). Since both motions are s.h.o, originating by a mass m and spring with constant k, the angular frequency, which in this case is the
same that the angular velocity (is this statement true?), must be

$$\omega^2 = \frac{k}{m}$$

So, the period must be

$$T = 2\pi\sqrt{\frac{m}{k}}$$


I have some doubts in some assumptions, but so far, is the best I do.

 

[tiny-tim]

Yes, your equation L = abmω looks fine.

And don't forget, from your definition of ω, that T = 2π/ω.

From that you should be able to get k (perhaps using some centripetal acceleration formula).

(Although I think ω is √(k/m), I'm not sure that the way you got it is valid.)

(Technically, we don't know that ω is constant, but it seems a sensible working assumption)

 

[obarquero]

I found a, maybe, more elegant way to solve the problem by means of energetic considerations.

The idea came to me looking at Kittel's book, when I saw the an expression of the total energy for a mass m in a circular motion. There was in a problem
in chapter 6, page 207 (in my spanish version of the book):

The total energy is:

$$E = U(r) + \frac{1}{2}m\dot{r} + \frac{L^2}{2mr^2}$$

when the last term is only true in locations when r and are perpendicular. In those points r must be a maximum or a minimum, so dr/dt must be 0. Since
energy must be conserved, the energy in (0,b) is equal to the energy in (a,0), so:
$$E_a = E_b \Rightarrow \frac{1}{2}ka^2 + \frac{L^2}{2ma^2} = \frac{1}{2}kb^2 + \frac{L^2}{2mb^2}$$

After some careful manipulations we can obtain L

$$L = ab\sqrt{mk}$$

Which is the same result that we obtained by the previous method.

In order to obtain the period we can use expression that relates the areolar velocity and the angular momentum:
$$\frac{dS}{dt} = \frac{1}{2m}L$$

Integrating this expresion for a complete cicle, that implies the area of the ellipse divided by the period

$$\frac{\piab}{T} = \frac{1}{2m}ab\sqrt{mk} \Rightarrow T = 2\pi\sqrt{\frac{m}{k}}$$

As we expected.