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## 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 [itex]M[/itex] and radius [itex]R[/itex] is free to pivot about its center. An insect of mass [itex]m[/itex] crawls at constant linear speed [itex]v[/itex] up the edge of the disk, starting from its lowest point. What minimum value of [itex]v[/itex] 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 [itex]v_0[/itex] (taken to be positive in the CCW direction). The initial and final angular velocities are denoted by [itex]\omega_0[/itex] and [itex]\omega[/itex], 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:

[tex]I=\beta MR^2.[/tex]

Here [itex]\beta=\frac{1}{2}[/itex] for the given disk. Also, the rotational expressions

[tex]J=I\omega,\quad K_\mathrm{rot}=\tfrac{1}{2}I\omega^2,[/tex]

and, for a point particle with [itex]\mathbf{v}[/itex] perpendicular to [itex]\mathbf{r}[/itex],

[tex]J=mvr,\quad K_\mathrm{lin}=\tfrac{1}{2}mv^2.[/tex]

## 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 [itex]J=0[/itex]. 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 [itex]\omega=v/R[/itex].

When the insect is in the downward position and begins to move, there are no external torques yet, so angular momentum should be conserved:

[tex]

mv_0R-I\omega_0=0.

[/tex]

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 [itex]R\omega_0[/itex], while that of the insect is [itex]v_0[/itex]. By relative velocity relations, [itex]v_0=v-R\omega_0[/itex], so

[tex]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).[/tex]

This also nicely gives the initial linear velocity [itex]v_0[/itex]:

[tex]v_0=v-R\omega_0=\left(\frac{\beta M}{m+\beta M}\right)v.[/tex]

Here's where I'm running into trouble; the initial kinetic energy of the system (in the lab frame) includes translational and rotational contributions:

[tex]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},

[/tex]

which, after some algebra, simplifies into

[tex]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).[/tex]

When the insect reaches the top of the disk, the kinetic energy is purely due to the disk's rotation:

[tex]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.[/tex]

But the insect has also gained gravitational PE [itex]U=2mgR[/itex]. So energy seems not to be conserved:

[tex]\frac{1}{2}\left(\frac{\beta Mm}{m+\beta M}\right)v^2\neq\frac{1}{2}\beta Mv^2+2mgR.[/tex]

I'm now trying to find [itex]v[/itex], 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 [itex]\dot{J}=\tau=-mgR\sin\theta[/itex] and [itex]J=\beta MR^2\omega(t)+m(v-R\omega(t))R[/itex]. However, I can't Taylor-expand the sine, and [itex]\theta[/itex] itself depends on the integral of [itex]v-R\omega(t)[/itex] which I don't have, and can't get out of the [itex]\sin\theta[/itex] 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 [itex]0[/itex] to [itex]-I\omega=-\beta MRv[/itex] 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 [itex]\mathbf{v}[/itex] and does not seem like it should do work since [itex]R[/itex] does not change.

Most generally, I'm trying to figure out what external work [itex]W_\mathrm{ext}[/itex] to throw in that will make energy conservation valid:

[tex]\frac{1}{2}\left(\frac{\beta Mm}{m+\beta M}\right)v^2+W_\mathrm{ext}=\frac{1}{2}\beta Mv^2+2mgR.[/tex]

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 [itex]v[/itex]. Does anybody have any ideas or suggestions?

Thanks,

0/x

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