Solving Orbital Speed with Energy & Angular Momentum Conservation

In summary: So the centripetal force on the particle is now different. But since the force is still directed towards the center of the circle, the energy and momentum are still conserved.
  • #1
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Homework Statement
A particle of mass m moves under the action of a harmonic oscillator force with potential energy 1
2kr2. Initially, it is moving in a circle of radius a. Find the orbital speed v. It is then given a blow of impulse mv in a direction making an angle α with its original velocity. Use the conservation laws to determine the minimum and maximum distances from the origin during the subsequent motion. Explain your results physically for the two limiting cases α = 0 and α = π.
Relevant Equations
Conservation of angular momentum and Conservation of Energy
$$J= const$$ and $$E = const$$
I've already solved the orbital speed by equating the kinetic and potential energy in the circle orbit case.

$$\frac{1}{2}mv^2 = \frac{1}{2}ka^2.$$And so $$v^2 = \frac{k}{m}a^2$$Now when the impulse is added, the particle will obviously change course. If we set our reference point in time just after the impulse is added. We get back our Angular momentum and Energy Conservation.So,

$$Ei = Ef$$

$$\frac{1}{2}mv^2 + \frac{p^2}{2m} + \frac{1}{2}ka^2 = \frac{1}{2}mv'^2 + \frac{1}{2}kr^2$$where v' is the final velocity, and r is new extension of spring.
$$Ji = Jf$$ also gives
$$mva + mvacos(\alpha) = mv'r$$Now we can solve for v' then plug that in the Energy conservation. I will also substitute in the value for v.$$\frac{1}{2}ka^2 + \frac{1}{2}ka^2 + \frac{1}{2}ka^2 =\frac{ma^2v^2(1+cos(\alpha))^2}{2r^2} + \frac{1}{2}kr^2$$Dividing out the $\frac{1}{2}$s and Ks. Then multiplying both sides by $r^2$ we get something like:$$r^4 - 3a^2r^2 + a^4(1+cos(\alpha))^2=0$$which finally gives$$r^2 = \frac{3a^2\pm\sqrt{9a^4-4a^4(1+cos(\alpha))^2}}{2}$$
This is problematic since the term in the root becomes negative, at least for $(\alpha) = 0.$The answer in the book is:$$r^2 = \frac{1}{2}a^2(3+2cos(\alpha) \pm \sqrt{5+4cos(\alpha)}$$I tried solving the problem using the ##\frac{J^2}{2mr^2}## term instead of the ##\frac{p^2}{2m}## term in energy conservation but to no avail.Any help would be appreciated.

Edit. This is my first time using latex. Idk why the last two terms are weird here, they look fine on overleaf. Please excuse an strange looking things.
 
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  • #2
Kyuubi said:
$$\frac{1}{2}mv^2 = \frac{1}{2}ka^2.$$
Why is this true? How does it ensure that the mass is going around in a circle? All you know for the circular motion is that $$\frac{1}{2}mv^2 + \frac{1}{2}ka^2=E_0=\text{constant}.$$ Why do you say that energy and angular momentum are conserved before and after the impulse is delivered? Suppose the mass is initially at rest and has zero mechanical energy. When you give it an impulse ##J## it starts moving while stretching the spring. That says to me that the mechanical energy is no longer zero. So in what sense is energy conserved? Same with angular momentum.

You need to rethink this.
 
  • #3
kuruman said:
Why is this true? How does it ensure that the mass is going around in a circle? All you know for the circular motion is that $$\frac{1}{2}mv^2 + \frac{1}{2}ka^2=E_0=\text{constant}.$$ Why do you say that energy and angular momentum are conserved before and after the impulse is delivered? Suppose the mass is initially at rest and has zero mechanical energy. When you give it an impulse ##J## it starts moving while stretching the spring. That says to me that the mechanical energy is no longer zero. So in what sense is energy conserved? Same with angular momentum.

You need to rethink this.
Ok tbf I had two methods to reach the orbital speed. This was one of them, but like you are saying it feels weird. My other reason for my velocity term is that the potential energy function corresponds to the centripetal force ##F(r) = -kr## which can be equated to ##\frac{mv^2}{r}##.
Plugging in r = a, you'll get the orbital speed expression. But again, this is not where I'm hung up on.
 
  • #4
kuruman said:
Why is this true? How does it ensure that the mass is going around in a circle? All you know for the circular motion is that $$\frac{1}{2}mv^2 + \frac{1}{2}ka^2=E_0=\text{constant}.$$ Why do you say that energy and angular momentum are conserved before and after the impulse is delivered? Suppose the mass is initially at rest and has zero mechanical energy. When you give it an impulse ##J## it starts moving while stretching the spring. That says to me that the mechanical energy is no longer zero. So in what sense is energy conserved? Same with angular momentum.

You need to rethink this.
Sorry for the double reply, but I didn't address your other point.
I'm not saying that Energy and momentum are conserved before and after the impulse is delivered. I'm saying if we think of the system right *after* the impulse is delivered, then conservation holds. Say if the impulse is delivered at t = 0, then at t=0.00000000...01 and at t = inf. Conservation holds. It's as if I'm looking at a new system entirely.
 
  • #5
Kyuubi said:
Ok tbf I had two methods to reach the orbital speed. This was one of them, but like you are saying it feels weird. My other reason for my velocity term is that the potential energy function corresponds to the centripetal force ##F(r) = -kr## which can be equated to ##\frac{mv^2}{r}##.
Plugging in r = a, you'll get the orbital speed expression. But again, this is not where I'm hung up on.
The correct argument is the centripetal force argument because it works for circular orbits regardless of the dependence of the central force on ##r##.

I'll get you started here. From the impulse equation ##\Delta\vec v=\dfrac{\vec J}{m}.## Now consider that the impulse is delivered when the mass is at position ##\vec r=\{a,0\}##. The velocity before the impulse at that moment is ##\vec v=\{0,a\sqrt{k/m}\}=\{0,a\omega\}.##

Let ##t=0## be the time immediately after the impulse is delivered.

Question 1: What are the initial conditions (velocity and position) at time ##t=0##?
Question 2: What is the general expression for the position vector ##\vec r(t)## after the impulse is delivered?
Question 3: What is the particular expression for ##\vec r(t)## that satisfies the initial conditions?

Once you answer Question 3 correctly, you will have everything you need.
 
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  • #6
Kyuubi said:
So,

$$Ei = Ef$$

$$\frac{1}{2}mv^2 + \frac{p^2}{2m} + \frac{1}{2}ka^2 = \frac{1}{2}mv'^2 + \frac{1}{2}kr^2$$
From the second term on the left side, it looks like you assumed that the impulse adds the amount ##\frac{p^2}{2m}## to the kinetic energy. This isn't correct.

Think about how much the impulse will increase the tangential and radial components of velocity. Then you can work out an expression for the total kinetic energy just after the impulse.

Otherwise, your approach to the problem looks good to me.
 
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  • #7
TSny said:
From the second term on the left side, it looks like you assumed that the impulse adds the amount ##\frac{p^2}{2m}## to the kinetic energy. This isn't correct.

Think about how much the impulse will increase the tangential and radial components of velocity. Then you can work out an expression for the total kinetic energy just after the impulse.

Otherwise, your approach to the problem looks good to me.
Thank you so much! This is exactly what I needed, and I got the right answer. I see now that I shouldn't plug in the momentum into the Energy conservation, but rather I should see the resultant change in energy from the applied momentum and use that instead.
 
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  • #8
dobirama said:
Please, can you show the final resolution of your system step by step?
That's not how this forum works. Read the guidelines.
Please start a new thread, filling in the homework template and showing your own attempt.
 

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