# Insect climbing edge of disk

## Homework Statement

(Important disclaimer: I came up with this problem out of curiosity in a classical mechanics course, so it might not be in a textbook, in the right place, or even solvable for that matter. However, I feel like it should be solvable, and most likely with elementary methods. It also reads very much like a homework problem, hence why I am posting it here.)

A disk of mass $M$ and radius $R$ is free to pivot about its center. An insect of mass $m$ crawls at constant linear speed $v$ up the edge of the disk, starting from its lowest point. What minimum value of $v$ allows the insect to reach the top?

In the hope of making life simpler, I'm choosing the direction of the insect's motion to be counterclockwise. The initial velocity of the insect is denoted by $v_0$ (taken to be positive in the CCW direction). The initial and final angular velocities are denoted by $\omega_0$ and $\omega$, respectively (taken to be positive in the CW direction).

## Homework Equations

I would like the freedom to replace the disk with a solid / hollow sphere, so I use a somewhat general expression for the moment of inertia:
$$I=\beta MR^2.$$
Here $\beta=\frac{1}{2}$ for the given disk. Also, the rotational expressions
$$J=I\omega,\quad K_\mathrm{rot}=\tfrac{1}{2}I\omega^2,$$
and, for a point particle with $\mathbf{v}$ perpendicular to $\mathbf{r}$,
$$J=mvr,\quad K_\mathrm{lin}=\tfrac{1}{2}mv^2.$$

## The Attempt at a Solution

I've tried lots of approaches for this problem and made good progress, but I'm genuinely stumped for the final stretch. Any help (preferably vague) would be appreciated. Here's my current approach:

Initially the disk and insect are stationary so $J=0$. Then the insect moves CCW, forcing the disk to rotate CW. When the insect reaches the top of the disk, it has no leftover kinetic energy, so its linear speed (in the lab frame) is zero. But it moves with constant speed relative to the disk, so the angular velocity of the disk is $\omega=v/R$.

When the insect is in the downward position and begins to move, there are no external torques yet, so angular momentum should be conserved:
$$mv_0R-I\omega_0=0.$$
Note that the angular momenta are in opposite directions.
The linear velocity (in the CW direction) of a point on the rim of the disk is $R\omega_0$, while that of the insect is $v_0$. By relative velocity relations, $v_0=v-R\omega_0$, so
$$m(v-R\omega_0)R-\beta MR^2\omega_0=0 \Rightarrow\omega_0=\frac{m}{m+\beta M}\left(\frac{v}{R}\right).$$
This also nicely gives the initial linear velocity $v_0$:
$$v_0=v-R\omega_0=\left(\frac{\beta M}{m+\beta M}\right)v.$$
Here's where I'm running into trouble; the initial kinetic energy of the system (in the lab frame) includes translational and rotational contributions:
$$K_0=\frac{1}{2}mv_0^2+\frac{1}{2}I\omega_0^2=\frac{1}{2}m\left(\frac{\beta M}{m+\beta M}\right)^2v^2+\frac{1}{2}\beta MR^2\left(\frac{m}{m+\beta M}\right)^2\frac{v^2}{R^2},$$
which, after some algebra, simplifies into
$$K_0=\frac{1}{2}\left(\frac{\beta Mm}{m+\beta M}\right)v^2=\frac{1}{2}\beta Mv^2\left(\frac{m}{m+\beta M}\right).$$
When the insect reaches the top of the disk, the kinetic energy is purely due to the disk's rotation:
$$K=\frac{1}{2}I\omega^2=\frac{1}{2}\beta Mv^2>\frac{1}{2}\beta Mv^2\left(\frac{m}{m+\beta M}\right)=K_0.$$
But the insect has also gained gravitational PE $U=2mgR$. So energy seems not to be conserved:
$$\frac{1}{2}\left(\frac{\beta Mm}{m+\beta M}\right)v^2\neq\frac{1}{2}\beta Mv^2+2mgR.$$

I'm now trying to find $v$, but without energy conservation I think I'm stuck. I do have a couple ideas: Maybe gravity is doing work on the system (there is a nonzero torque as soon as the insect is displaced from the vertical) that I somehow haven't accounted for in the PE term. Messing around with that idea got me $\dot{J}=\tau=-mgR\sin\theta$ and $J=\beta MR^2\omega(t)+m(v-R\omega(t))R$. However, I can't Taylor-expand the sine, and $\theta$ itself depends on the integral of $v-R\omega(t)$ which I don't have, and can't get out of the $\sin\theta$ expression via FTC magic. If I could do anything on this end, I suspect it would involve solving a differential equation to find an explicit time dependence, which certainly seems like overkill considering the problem statement.

Another possibility is that some amount of work is being done against the insect because its kinetic energy is decreasing (namely, to zero). But I don't know how to make this abstract notion tangible, and I'm doubtful it's even valid considering that the kinetic energy of the disk is increasing (so this may be due to internal forces / torques).

For what it's worth, the angular momentum decreases from $0$ to $-I\omega=-\beta MRv$ while the insect climbs to the top. I can only see this being useful for constant torque, which is not the case here, but I don't know.

Finally, I'm also pondering the centripetal acceleration of the insect, even though it is perpendicular to $\mathbf{v}$ and does not seem like it should do work since $R$ does not change.

Most generally, I'm trying to figure out what external work $W_\mathrm{ext}$ to throw in that will make energy conservation valid:
$$\frac{1}{2}\left(\frac{\beta Mm}{m+\beta M}\right)v^2+W_\mathrm{ext}=\frac{1}{2}\beta Mv^2+2mgR.$$
Every time I've seen something like this, an integral or otherwise tedious work calculation can be avoided using kinetic energy considerations, but I just can't make that work if I'm also solving for velocity. And I can't find any other ways to isolate $v$. Does anybody have any ideas or suggestions?

Thanks,
0/x

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## Answers and Replies

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haruspex
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I don't think conservation of work is useful here. The longer the insect runs, the more work it does, but even then its rate of doing work depends on where it is on the disk and how fast the disk is rotating.
You could figure out the angular momentum as a function of time, perhaps.
I would go back to first principles (torque and acceleration) and obtain the differential equation.

rude man
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Got me baffled. You mean, no matter how thin the disk, how massive the disk M, how light the insect m, it can't crawl up the disk's height at any arbitrarily low linear speed? Why not? The disk is not rotating when the insect is at the bottom, right? No initial rotation rate being given. For that matter I don't see why the disk would be brought into rotation at all.

jbriggs444
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Got me baffled. You mean, no matter how thin the disk, how massive the disk M, how light the insect m, it can't crawl up the disk's height at any arbitrarily low linear speed? Why not? The disk is not rotating when the insect is at the bottom, right? No initial rotation rate being given. For that matter I don't see why the disk would be brought into rotation at all.
Think hamster wheel. The hamster cannot get to the top of the wheel by walking slowly. But if he's fast enough and the wheel is massive enough, he could get there by running.

haruspex
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The equation for time taken gets nasty, involves $\int (A\cos(\theta)+B)^{-\frac 12}d\theta$, but fortunately you don't need to solve that to answer the question as posed. I get $v^2>4mgR^3(I+mR^2)I^{-2}$, where I is the moment of inertia of the disk.
Now, Zero Divisor, let's see an attempt at a differential equation.

rude man
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Think hamster wheel. The hamster cannot get to the top of the wheel by walking slowly. But if he's fast enough and the wheel is massive enough, he could get there by running.
Oh, OK. I thought of the spin axis as vertical, not horizontal.
Based on conservation of angular momentum I came up with:
v can be any arbitrarily low number as long as 2m/M < 1. Then
v = πR/t(1 - 2m/M).
t = time for the insect to reach the top.
If 2m/M > 1 there is no value of v that will get the bug to the top.

haruspex
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Oh, OK. I thought of the spin axis as vertical, not horizontal.
Based on conservation of angular momentum I came up with:
v can be any arbitrarily low number as long as 2m/M < 1. Then
v = πR/t(1 - 2m/M).
t = time for the insect to reach the top.
If 2m/M > 1 there is no value of v that will get the bug to the top.
It seems very clear to me that there must be a minimum speed, whatever the mass ratio. If the bug takes too long, the disk will just spin faster and faster. Of course, the bug could solve that by reversing direction each time it passes the lowest point. With that strategy, it can always get there.

rude man
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It seems very clear to me that there must be a minimum speed, whatever the mass ratio. If the bug takes too long, the disk will just spin faster and faster. Of course, the bug could solve that by reversing direction each time it passes the lowest point. With that strategy, it can always get there.
That''s violating conservation of angular momentum: mvR = Iω, measured in an inertial coordinate system.
Thus, ω and v are related by v = (I/mR)ω The slower the bug's v, the slower the disk's ω. The product vt = constant = π/(1/R - mR/I).

EDIT: I see a problem with what I did. I computed inertial velocity for the bug when what was asked for was velocity relative to the rotating disc. Have to look at it some more.

OK, after that correction I still get vT = constant = (πR)/[1 - 2m/(M + 2m)].
My premises:
mviR = Iω, I = MR2/2
θ(t) = vit/R = vrt/R - ωt
vi = bug velocity relative to inertial coord. system
vr = bug velocity relative to disk
θ(t) = inertial angle between bottom radius and bug position. So 0 < θ(t) < π.
At time t=T, θ = π.

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jbriggs444
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That''s violating conservation of angular momentum: mvR = Iω, measured in an inertial coordinate system.
Thus, ω and v are related by v = (I/mR)ω The slower the bug's v, the slower the disk's ω. The product vt = constant = π/(1/R - mR/I).
There is an external torque from gravity. The bug is weighing one side of the wheel down.

Again, think hamster wheel. Ordinarily the hamster never reaches the top.

rude man
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There is an external torque from gravity. The bug is weighing one side of the wheel down.
Good point. That makes it too complicated for comfort!

Thanks to everybody who replied! I ended up toughing it out with a partial solution of a differential equation (not to be confused with the solution of a partial differential equation).
The equation for time taken gets nasty, involves $\int (A\cos(\theta)+B)^{-\frac 12}d\theta$, but fortunately you don't need to solve that to answer the question as posed. I get $v^2>4mgR^3(I+mR^2)I^{-2}$, where I is the moment of inertia of the disk.
Now, Zero Divisor, let's see an attempt at a differential equation.
I'm pretty sure my answer is equivalent to this. Here is my solution:
I set $\theta$ as the CCW angular position of the insect, and $\phi$ as the CW angular position of the disk. Relative motion gives $R\theta+R\phi=vt$, so
$$R\dot\theta+R\dot\phi=v.$$
The insect has angular momentum $J_1=mR^2\dot\theta$ and the disk has $J_2=-(\beta MR^2)\dot\phi=\beta MR(R\dot\theta-v)$, so
$$J=mR^2\dot\theta+\beta MR^2\dot\theta-\beta MRv\quad\Rightarrow\quad\dot{J}=(m+\beta M)R^2\ddot\theta,$$
since $\beta MRv$ is of course constant in time.
The torque due to gravity is due only to the insect, so $\tau=-mgR\sin\theta$ (the sign indicates the torque opposes the direction of motion). Thus
$$(m+\beta M)R^2\ddot\theta=-mgR\sin\theta\quad\Rightarrow\quad (m+\beta M)R^2\ddot\theta+mgR\sin\theta=0.$$
The trick to this differential equation is to multiply through by the integrating factor $\dot\theta$;observing that $\dot\theta\ddot\theta=\frac{\mathrm{d}}{\mathrm{d}t}[\frac{1}{2}\dot\theta^2]$ and $\dot\theta\sin\theta=\frac{\mathrm{d}}{\mathrm{d}t}[1-\cos\theta]$ allows us to integrate the entire equation neatly:
$$\frac{1}{2}(m+\beta M)R^2\dot\theta^2+mgR(1-\cos\theta)=E,$$
where $E$ is a (suggestive) constant of integration.

At the top of the disk, we have $\theta=\pi$ and $\dot\theta=0$ (since we started with minimal velocity), so
$$mgR(1-\cos\pi)=2mgR=E.$$
On the other hand, at $t=0$ the insect had just started climbing from the bottom of the disk, so $\theta=0$ and $\dot\theta=v_0/R$, where
$$v_0=\left(\frac{\beta M}{m+\beta M}\right)v$$
was found in the OP. Therefore
$$\frac{1}{2}\left(\frac{\beta^2M^2}{m+\beta M}\right)v^2=E,$$
and since $E=2mgR$ we can solve to obtain
$$v=\left[\frac{4m(m+\beta M)gR}{\beta^2M^2}\right]^{1/2}.$$
I think this can be shown to agree with your result upon multiplying by $R^4/R^4$ inside the radical.
-------------
Granted, we've arrived at the same result, but there's still something that feels a little strange: The equation
$$\frac{1}{2}(m+\beta M)R^2\dot\theta^2+mgR(1-\cos\theta)=E.$$
This looks very, very much like a conservation of energy equation, except (1) we've established that energy is not conserved, and (2) for a conservation of energy equation, the rotational term would then be $\frac{1}{2}\beta M(R^2\dot\theta^2-2Rv\dot\theta+v^2)$ instead of $\frac{1}{2}\beta MR^2\dot\theta^2$. The term proportional to $v^2$ is constant and can be absorbed into $E$, but the term proportional to $\dot\theta$ cannot. The only way this can make sense to me is if $\beta MRv\dot\theta$ is related to the work done and thus cancels from the equation.

As for the differential equation for $\theta(t)$, I freely admit it's too much work for me right now. However, I think a half-angle identity and some trickery with integrating secant in an invertible way should give the equation and it might come to me at some point. Stay tuned!

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haruspex
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Here is my solution:
That appears to be exactly how I solved it.
Granted, we've arrived at the same result, but there's still something that feels a little strange: The equation
$$\frac{1}{2}(m+\beta M)R^2\dot\theta^2+mgR(1-\cos\theta)=E.$$
This looks very, very much like a conservation of energy equation, except (1) we've established that energy is not conserved, and (2) for a conservation of energy equation, the rotational term would then be $\frac{1}{2}\beta M(R^2\dot\theta^2-2Rv\dot\theta+v^2)$ instead of $\frac{1}{2}\beta MR^2\dot\theta^2$. The term proportional to $v^2$ is constant and can be absorbed into $E$, but the term proportional to $\dot\theta$ cannot. The only way this can make sense to me is if $\beta MRv\dot\theta$ is related to the work done and thus cancels from the equation.
That interpretation looks right.

Note that the bug's strategy of reversing direction creates other interesting questions...
- exactly when should the direction reverse - is it always at the lowest point or is there something better?
- how many reversals will be needed?
- what's the total time in general?
- under what parameter combinations is it quicker to use the reversing strategy than to keep going one way?