Bug moving on a turntable rotating with constant omega

1. Oct 4, 2016

Dustgil

1. The problem statement, all variables and given/known data
A cockroach crawls with constant speed in a circular path of radius b on a phonograph turntable rotating with constant angular speed omega. The circle path is concentric with the center of the turntable. If the mass of the insect is m and the coefficient of static friction with the surface is the table is mu sub s, how fast, relative to the turntable, can the cockroach crawl before it starts to slip if it goes (a) in the direction of rotation and (b) opposite to the direction of rotation?

2. Relevant equations

3. The attempt at a solution

Best I saw was to pick the reference frame of the turntable. In this fram, the cockroach walks in a circle of radius b.

$$r^{'} = bcos\theta i+bsin\theta j$$
$$v^{'} = -bsin\theta i+bcos\theta j$$
$$a^{'} = -bcos\theta i - bsin\theta j$$

The transverse force is zero (the reference frame is rotating with constant angular velocity), so the equation of motion reads as

$$F - 2m\omega \times v^{'}-m\omega \times (\omega \times r^{'}) = ma^{'}$$

Omega is entirely in the k direction, and the only force acting on the cockroach in the reference frame is the frictional force, so by evaluating the cross products the equation becomes

$$F_{f}= -2m\omega b(cos\theta i+ sin\theta j)-m\omega ^{2}b(cos\theta i -sin\theta j) -mbcos\theta i - mbsin\theta j$$

We can then separate this into components and solve for the max forces in either direction. Since the normal force and the graviational force oppose each other, the frictional force must be less than $$\mu_{s}mg$$

so, in the x direction for example,

$$-2m\omega b cos\theta - m\omega^{2} b cos\theta-mbcos\theta < \mu_{s}mgcos\theta$$
$$-2\omega b - \omega^{2} b-b < \mu_{s}g$$

So, if we solve this for b we can put that into our equation for velocity to find the max velocity the cockroach can travel in either direction. But it doesn't seem totally correct to me. It seems messy...plus, when i used to same process for the cockroach traveling opposing the direction of rotation, it was possible to make b undefined for some values of omega. So, is my approach correct?

2. Oct 4, 2016

Staff: Mentor

Since the bug is moving on a path concentric with the axis of rotation of the turntable, why not describe his speed as an angular velocity with respect to the rotating frame of the turntable? Wouldn't that save a lot trouble adding and subtracting velocities?

3. Oct 6, 2016

rude man

Compute the bug's angular speeds (CW and CCW) in inertial space. This is the algebraic sum of its speed relative to the table plus the table speed. That gives you the centrifugal force acting on the bug to keep it in its track. Compare to the static friction force.

4. Oct 7, 2016

Dustgil

Sure.

$$r^{'}=re_{r}=be_{r}$$
$$v^{'}=\dot{r}e_{r}+r\dot{\theta}e_\theta = b\omega_{t}e_{\theta}$$
$$a^{'}=(\ddot{r}-r\dot{\theta}^{2})e_{r}+(r\ddot{\theta}+2\dot{r}\dot{\theta})e_\theta=b\omega_{t}^{2}e_{r}$$

where omega is the angular velocity of the bug in the reference frame of the turntable. I took r=b and theta = omega*t.

so, given that the angular velocity of the turntable itself is in the k direction, the cross products evaluate to

$$\omega \times v^{'}=-b\omega \omega_{t}e_{r}$$
$$\omega \times \omega \times r^{'}=-b\omega \omega_{t} e_{r}$$

then the equation of motion in the radial direction is (solved for the force of static friction)

$$F_s=mb\omega_{t}^{2}-3mb\omega \omega_{t}=\mu_{s}mg$$

but wouldn't that imply there were two maximum angular velocities that would cause the bug to just overcome static friction? (since omega sub t is squared). seems to me that there should only be one.

5. Oct 7, 2016

Indeed:

6. Oct 7, 2016

rude man

Why? Solving for ω your way gives only one value.

Anyway, I think you're going about this the hard way. Just compute centripetal forces on the bug going with & against the turntable. But you need to find two values of ω, not just one.

Anyway, I got different results.