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

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. 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.

## closed as off-topic by John Rennie, Neuneck, Pranav Hosangadi, ACuriousMind♦, Kyle KanosJan 7 '15 at 13:27

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• Angular momentum conservation has nothing to do with axes, however, it's easy to see that angular momentum is not conserved. Both disks are rotating in the same direction, angular momentum is additive, so it's not the same before the disks contact (it's positive) and after (when it is zero). – CuriousOne Jan 7 '15 at 5:04
• Actually angular momentum is with respect to a chosen axes. Suppose you chose the axes passing through the centers of each disk. If you bring the disks together along this axes, angular momentum actually is conserved. – Joshua Benabou Jan 7 '15 at 5:07
• You can pick the axis of the larger disk as your reference. The process of bringing the smaller disk into contact with the large one will result in a torque on the small disk, so that without an external torque the small disk will end up rotating around the large one. Or maybe I don't understand the setup correctly? – CuriousOne Jan 7 '15 at 5:12
• I agree with that. But you said "its easy to see that it is not conserved" without specifying an axis. As I said, you can choose an axes such that it is conserved, but that would not be helpful for solving the problem. – Joshua Benabou Jan 7 '15 at 5:20
• You don't need to specify an axis to compare before and after. You can, if you want to. I picked the center of mass system in which both bodies rotate in the same direction before they get in contact and neither rotates after. – CuriousOne Jan 7 '15 at 5:41

## 3 Answers

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.

• The only thing I don't understand why is the change in angular momentum given by $r_1J$ and $r_2J$? Isn't the change in momentum equal to the impulse $J$? – Joshua Benabou Jan 7 '15 at 5:31
• Angular momentum is like torque to linear momentum. It is $\vec{r} \times \vec{J}$ but you do it by component it works out to $r_1 J_1$ and $(-r_2) (-J_2)$. – ja72 Jan 7 '15 at 5:47
• Okay now I understand! So I would phrase this as follows: torque is the derivative of angular momentum so integrating the torque on one disc over the period of interaction between the two discs, we get the change in linear momnetum on the disc. The torque is the product of the frictional force and the radius of the disc so it is equal to $rJ$. – Joshua Benabou Jan 7 '15 at 18:42
• Integral of torque is change in angular momentum not linear. But yes, $$\Delta (I \omega) = \int \vec{r} \times \vec{F} \,{\rm d} t = \vec{r} \times \vec{J}$$ – ja72 Jan 7 '15 at 18:50

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

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$$=\frac12\rho \pi hR_2^3\omega_2-\frac12\rho \pi hR_1^3\omega_1=0 \frac12\rho \pi hR_2^3\omega_2=\frac12\rho \pi hR_1^3\omega_1 The problem states 'the discs have identical thickness and are made of the same material [same density]', so those terms cancel. R_2^3\omega_2=R_1^3\omega_1 Note: This answer is an extended comment that intentionally does not solve the problem. There are two correct solutions already. This answer instead addresses the issue of conservation of angular momentum, which was raised in the question as a mechanism for solving this problem and continued in the discussion under the question. Conservation of angular momentum is not useful here. The angular momentum of a set of object with respect to an origin is the vector sum of the angular momenta of the objects about their centers of mass and the angular momenta of the objects as a whole with respect to the chosen origin:$$\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).