Angular Momentum 10-8-2020

Classical Mechanics Level pending

Two identical particles of mass m m orbit the origin of the x y xy plane in uniform circular motion of radius R R . The particles are diametrically opposed to each other, and their mutual gravitation is solely responsible for maintaining the orbit.

Let P ⃗ 1 \vec{P}_1 and P ⃗ 2 \vec{P}_2 be the positions of the particles, and let P ⃗ \vec{P} be a reference point. Define two displacement vectors:

r ⃗ 1 = P ⃗ 1 − P ⃗ r ⃗ 2 = P ⃗ 2 − P ⃗ \vec{r}_1 = \vec{P}_1 - \vec{P} \\ \vec{r}_2 = \vec{P}_2 - \vec{P}

Let the velocities of the particles be v ⃗ 1 \vec{v}_1 and v ⃗ 2 \vec{v}_2 . The magnitudes of the angular momenta of the particles are:

L 1 = ∣ r ⃗ 1 × m v ⃗ 1 ∣ L 2 = ∣ r ⃗ 2 × m v ⃗ 2 ∣ L_1 = |\vec{r}_1 \times m \, \vec{v}_1| \\ L_2 = |\vec{r}_2 \times m \, \vec{v}_2|

The expressions for L 1 L_1 and L 2 L_2 each consist of a constant term added to an oscillating term. What is the peak value of the oscillating term?

Bonus: Plot both angular momenta on a graph vs. time. How does the total angular momentum behave? Does it depend on the reference point?

Details and Assumptions:
1) m = R = 1 m = R = 1
2) Universal gravitational constant G = 1 G = 1
3) P ⃗ = ( P x , P y ) = ( 0 , 3 ) \vec{P} = (P_x, P_y) = (0,3)
4) ∣ ⋅ ∣ |\cdot| denotes the magnitude of a vector, and × \times denotes the vector cross product


The answer is 1.5.

Steven Chase by Steven Chase , 37, USA

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2 solutions

Karan Chatrath
Oct 8, 2020

Let the positions of each particle be:

P 1 = ( cos ⁡ θ , sin ⁡ θ ) P_1 = (\cos{\theta},\sin{\theta}) P 2 = ( − cos ⁡ θ , − sin ⁡ θ ) P_2 = (-\cos{\theta},-\sin{\theta})

It is given that the particles move in UCM due to mutual gravitation. Then the magnitude of the gravitational force must be equal to m R θ ˙ 2 mR\dot{\theta}^2 . This gives:

θ ˙ = 1 2 \dot{\theta} = \frac{1}{2}

The above value can also be negative, but that is ignored for now.

The velocity of each particle is:

v ⃗ 1 = θ ˙ ( − sin ⁡ θ i ^ + cos ⁡ θ j ^ ) \vec{v}_1 = \dot{\theta}(-\sin{\theta} \ \hat{i} + \cos{\theta} \ \hat{j}) v ⃗ 2 = θ ˙ ( sin ⁡ θ i ^ − cos ⁡ θ j ^ ) \vec{v}_2 = \dot{\theta}(\sin{\theta} \ \hat{i} - \cos{\theta} \ \hat{j})

r ⃗ 1 = cos ⁡ θ i ^ + ( sin ⁡ θ − 3 ) j ^ \vec{r}_1 = \cos{\theta} \ \hat{i} + (\sin{\theta}-3) \ \hat{j} r ⃗ 2 = − cos ⁡ θ i ^ − ( sin ⁡ θ + 3 ) j ^ \vec{r}_2 = -\cos{\theta} \ \hat{i} - (\sin{\theta}+3) \ \hat{j}

Computing each of the angular momenta and simplifying and using the value of θ ˙ \dot{\theta} gives:

L ⃗ 1 = 1 − 3 sin ⁡ θ 2 k ^ \vec{L}_1 = \frac{1-3\sin{\theta}}{2} \ \hat{k} L ⃗ 2 = 1 + 3 sin ⁡ θ 2 k ^ \vec{L}_2 = \frac{1+3\sin{\theta}}{2} \ \hat{k}

The constant term is 0.5 while the amplitude of the oscillating term is 1.5 \boxed{1.5} .

  • It is obvious here that the total angular momentum of the system is a constant, throughout the motion, along the Z Z direction. This is because there is no external force acting on this two particle system.

  • It is also obvious that this total angular momentum of the system is independent of the reference point. But note that the individual angular momenta of each particle is reference point dependent.

  • I am not adding plots, as these expressions are relatively easy to understand. They are just sinusoids offset by 0.5 0.5 with an amplitude of 1.5 1.5 . An interested reader may roughly sketch them by hand.

3 Helpful 3 Interesting 3 Brilliant 0 Confused

@Karan Chatrath I am getting ( d θ d t ) 2 = 1 4 ( cos ⁡ θ i ⃗ + sin ⁡ θ j ⃗ ) (\frac{d \theta}{dt} )^{2}=\frac{1}{4}(\cos \theta \vec{i}+\sin \theta \vec{j})
Is that correct?

Talulah Riley - 8 months ago

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No, it is not correct. How can you equate a scalar to a vector?

Karan Chatrath - 8 months ago

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@Karan Chatrath Yeah exactly that is what I was also thinking.
Now. What should I add in Left hand side to make it correct?

Talulah Riley - 8 months ago

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@Talulah Riley – I would encourage you to think about that. Apply Newton's 2nd law carefully for each particle and you should arrive at the result naturally.

Karan Chatrath - 8 months ago

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@Karan Chatrath – @Karan Chatrath using newtons laws I am getting d v ⃗ d t = 1 4 ( cos ⁡ θ i ⃗ + sin ⁡ θ j ⃗ ) \frac{d \vec{v}}{dt}=\frac{1}{4} (\cos \theta \vec{i} +\sin \theta \vec{j})

Talulah Riley - 8 months ago

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@Talulah Riley – Applied to which mass? What is the next step?

Karan Chatrath - 8 months ago

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@Karan Chatrath – @Karan Chatrath one of the mass
I am. Only not able to understand that how do calculated v 1 ⃗ \vec{v_{1}} , v 2 ⃗ \vec{v_{2}} ?

Talulah Riley - 8 months ago

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@Talulah Riley – The equation you showed above cannot be applied to both masses. It applies to one of them. Which one?

How do you calculate velocities when you have position coordinates?

I will only give hints this time. You should have solved this easily.

Karan Chatrath - 8 months ago

@Karan Chatrath Your solution is upvoted.

Talulah Riley - 8 months ago

Let at an instant of time t t , the position coordinates of the particles be ( R cos ⁡ θ , R sin ⁡ θ ) (R\cos \theta, R\sin \theta) and ( − R cos ⁡ θ , − R sin ⁡ θ ) (-R\cos \theta, -R\sin \theta) respectively. Then

r 1 ⃗ = P 1 ⃗ − P ⃗ = i ^ R cos ⁡ θ + j ^ ( R sin ⁡ θ − 3 ) \vec {r_1}=\vec {P_1}-\vec P=\hat i R\cos \theta +\hat j (R\sin \theta -3)

The magnitudes of velocities of the particles will be the same, obtained from the force balance equation :

v = ∣ v 1 ⃗ ∣ = ∣ v 2 ⃗ ∣ ; G m 2 4 R 2 = m v 2 R ⟹ v = G m 4 R v=|\vec {v_1}|=|\vec {v_2}|; \dfrac {Gm^2}{4R^2}=\dfrac {mv^2}{R}\implies v=\sqrt {\frac{Gm}{4R}}

v 1 ⃗ = − i ^ G m 4 R sin ⁡ θ + j ^ G m 4 R cos ⁡ θ \vec {v_1}=-\hat i \sqrt {\frac{Gm}{4R}}\sin \theta +\hat j \sqrt {\frac{Gm}{4R}}\cos \theta

r 2 ⃗ = P 2 ⃗ − P ⃗ = − i ^ R cos ⁡ θ − j ^ ( R sin ⁡ θ + 3 ) \vec {r_2}=\vec {P_2}-\vec P=-\hat i R\cos \theta -\hat j (R\sin \theta +3)

v 2 ⃗ = i ^ G m 4 R sin ⁡ θ − j ^ G m 4 R cos ⁡ θ \vec {v_2}=\hat i \sqrt {\frac{Gm}{4R}}\sin \theta -\hat j \sqrt {\frac{Gm}{4R}}\cos \theta

Therefore, L 1 ⃗ = m r 1 ⃗ × v 1 ⃗ = k ^ m G m 4 R ( R − 3 sin ⁡ θ ) \vec {L_1}=m\vec {r_1}\times \vec {v_1}=\hat k m\sqrt {\frac{Gm}{4R}}(R-3\sin \theta)

L 2 ⃗ = m r 2 ⃗ × v 2 ⃗ = k ^ m G m 4 R ( R + 3 sin ⁡ θ ) \vec {L_2}=m\vec {r_2}\times \vec {v_2}=\hat k m\sqrt {\frac{Gm}{4R}}(R+3\sin \theta)

So, the amplitude of the oscillating term is

3 m G m 4 R = 1.5 3m\sqrt {\frac{Gm}{4R}}=\boxed {1.5}

Total angular momentum of the system is

L ⃗ = L 1 ⃗ + L 2 ⃗ = k ^ G m 3 R \vec L=\vec {L_1}+\vec {L_2}=\hat k \sqrt {Gm^3R} , independent of θ \theta , that is, the positions of the particles.

2 Helpful 2 Interesting 2 Brilliant 0 Confused

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