$F=ma$ contest problem on angular momentum [closed]

Question

Problem #25 from the 2009 $F=ma$ contest:

Two discs are mounted on thin, lightweight rods oriented through their centers and normal to the discs. These axles are constrained to be vertical at all times, and the discs can pivot frictionlessly on the rods. The discs have identical thickness and are made of the same material, but have differing radii $r_1$ and $r_2$. The discs are given angular velocities of magnitudes $\omega_1$ and $\omega_2$, respectively, and brought into contact at their edges. After the discs interact via friction it is found that both discs come exactly to a halt. Which of the following must hold? [...] Ignore effects associated with the vertical rods.

Diagram here: http://www.aapt.org/physicsteam/2010/upload/2009_F-maSolutions.pdf

Probably conservation of angular momentum is useful here, but I am not clear how it can be applied. There are two distinct axes of rotation, one for each disc, so we cannot talk about conservation of angular momentum about one axis. If the two disks rotated about the same axes then the solution is straightforward, but this is not the case.

Answer 1

The friction force is equal and opposite on the two disks, so there is a net impulse $J$ (force over time) acting on each disk edge netting a change in angular moment of $r_1 J$ on one disk and $r_2 J$ on the other.

If the initial angular momentum was $H_1 = I_1 \omega_1$ and $H_2 = I_2 \omega_2$ then after they come to a halt you have

$$ \begin{aligned} I_1 \omega_1-r_1 J & =0 \\I_2 \omega_2-r_2 J & =0 \end{aligned} $$

Given the same material you have that $I_1 = \frac{m_1}{2} r_1^2 = \propto r_1^4$ and $I_2 =\frac{m_2}{2} r_2^2 = \propto r_2^4$

Use the above two equations to find that $J = K \omega_1 r_1^3 = K \omega_2 r_2^3$ where $K$ is the proportionality constant in the mass moment of inertias.

Answer 2

$\Delta(L_A)=\tau*\Delta(t)$

Yet for collisions (conserved)

$\Delta(L_A)=\tau$

where $\tau=r \times F$

The angular momentum type being analyzed is rotational thus the equation $L_A$$_,$$_i$$=I\omega$ holds true for both object.

The finial state of the system can be found via $L_A$$_,$$_f$$=I\omega$$+rF=0 \leftarrow$ Because they both halt.

Let's better define $I_d$$_i$$_s$$_k$$=\frac12 MR^2$ $ \rho V=M$, where $\rho=$density & $V=$volume of a disk

$\therefore$$I_d$$_i$$_s$$_k$=$\frac12\rho \pi hR^4$

Initial States:Spinning no interaction with each other

$L_1$$_,$$_i$$=\frac12\rho \pi hR_1^4\omega_1$

$L_2$$_,$$_i$$=\frac12\rho \pi hR_2^4\omega_2$

Final State:Colision takes place, Imposing equal and opposite forces upon each other

$L_1$$_,$$_f$$=\frac12\rho \pi hR_1^4\omega_1+R_1 F_1$$,_2$$=0$

$L_1$$_,$$_f$$=R_1(\frac12\rho \pi hR_1^3\omega_1+ F_1$$,_2$$)=0$

$\frac12\rho \pi hR_1^3\omega_1+ F_1$$,_2$$=0$

$F_1$$,_2=$Force imposed on 1 by 2

$L_2$$_,$$_f$$=\frac12\rho \pi hR_2^3\omega_2+ F_2$$,_1$$=0$

$F_1$$,_2=-\frac12\rho \pi hR_1^3\omega_1$

$F_1$$,_2=F_2$$,_1$ substitute

$L_2$$_,$$_f$$=\frac12\rho \pi hR_2^3\omega_2-\frac12\rho \pi hR_1^3\omega_1=0$

$\frac12\rho \pi hR_2^3\omega_2=\frac12\rho \pi hR_1^3\omega_1$

The problem states 'the discs have identical thickness and are made of the same material [same density]', so those terms cancel.

$R_2^3\omega_2=R_1^3\omega_1$

Answer 3

Note: This answer is an extended comment that intentionally does not solve the problem. There are two correct solutions already. This answer instead addresses the issue of conservation of angular momentum, which was raised in the question as a mechanism for solving this problem and continued in the discussion under the question.

Conservation of angular momentum is not useful here. The angular momentum of a set of object with respect to an origin is the vector sum of the angular momenta of the objects about their centers of mass and the angular momenta of the objects as a whole with respect to the chosen origin: $$\mathbf L_\text{tot} = \sum_i \left(\mathbf L_{i,\text{cm}} + \mathbf r_{i,\text{cm}}\times m_i\mathbf v_{i,\text{cm}}\right)$$

I see three ways to attack this problem:

1. As a collision that saps the rotational motion of the two objects. The objects go scooting away from each other after the collision. Here, rotational angular momentum is converted to angular momentum due to translational motion. Angular momentum is conserved, but it's not useful in solving the problem. All that angular momentum does is tell you how fast the objects are moving away from one another at the end of the interaction. It doesn't help you solve the problem.

2. As a forced interaction. The wording of the problem ("these axles are constrained to be vertical at all times" and "the discs are ... brought into contact at their edges") suggests that some external mechanism is at play. External forces, presumably applied through those vertical axles, are needed to keep the tops in contact. These external forces are equal but opposite in nature (but not 3rd law pairs) and are separated by a distance $r_1+r_2$. This is a pure couple that acts on the top₁+top₂ system; angular momentum of this non-isolated system is is not conserved. The angular momentum of the top₁+top₂ system is transferred to the Earth.

3. It doesn't matter what setup enables this to happen. All that matters is that the sole interaction that governs the slowdown is friction due to contact, and that the spinning motion of the two tops come to a halt.

In all cases, conservation of angular momentum doesn't help in solving the problem. What does help is knowing that an external torque changes angular momentum. Numoru appears to have looked at the problem as a collision (case #1) while ja72 took the perspective of "who cares?" (case #3).