# Physics of two orbiting masses connected with cable when the cable is cut

## The Problem:

Figure 1 - connected masses

In a 2D world without gravity or friction:

Lets say we have two masses, m1 and m2. They are both connected with some massless cable and the whole system is rotating about their center of mass cm with some angular velocity ω0. (See figure 1)

Q: When the cable is cut what is the resulting linear and angular velocity of each mass? (See figure 2)

Figure 2 - Cut cable for connected masses

## My attempt:

Linear velocity is simple enough. I can use the equation for tangential velocity: v_⊥ = ω*r, where ω is the angular velocity of the system and r is the distance between the mass and center of mass of the system before separation.

v1_⊥ = ω0*r1  # Linear velocity of m1
v2_⊥ = ω0*r2  # Linear velocity of m2


Angular velocity is the one I'm struggling with. The angular momentum of the system must be conserved.

• Lbefore = Lafter
• Lbefore = Lm1 + Lm2
• Using the formula for angular momentum L=I*ω
• Im1&m2*ω0 = Im1*ω1 + Im2*ω2
• Using the formula for moment of inertia I=m*r^2
• (m1*r12 + m2*r22)*ω0 = m1*r12*ω1 + m2*r22*ω2
• And this is where I get stuck.

The closest I could find to my problem is this post on Physics StackExchange: Will two bodies initially connected to and revolving around each other, start spinning when disconnected?

• You didn't factor in the angular momentum coming from the linear motion of two masses after the rope is cut. Feb 4, 2022 at 9:16

When the string is cut, both masses wil keep on rotating with angular velocity $$\omega_0$$. Before the cut they both rotate with this angular velocity around their central axis and there is no torque changing that. The angular momentum due to the rope is turned to linear momentum.

$$L = r \times p$$
$$v_1$$ = (0, $$\omega_0 r_1$$) and $$v_2$$ = (0, $$-\omega_0 r_2$$)
The positions of the two partciles are also quite easy. They are both travelling up or down the page (y-diretion) so the $$x$$ component of position is constant (at $$-r_1$$ and $$+r_2$$ for the two masses respectively). The angular momentum afterwards comes out as... exactly the same formula you wrote for the angular momentum beforehand (with the $$\omega_0$$ multiplying $$r^2$$).