Rotation and fictitious forces

Question

A bug eats through an apple and forms a vertical, infinitesimally thin canal parallel to the vertical diameter at a distance $\frac{R}{2}$ from the center. The apple rotates at angular velocity around that center equal to $\sqrt{\frac{g}{R}}$. As the bug attains the other side of the apple it slips and falls through that canal, whose coefficient of friction is $\mu$, and I am asked to find its velocity as it passes through the bottom.

Now, the length of the canal is obviously $R\sqrt{3}$. The location of the bug may be written as: $(\frac{R}{2},y)$ wrt to the center of the apple. Coriolis would have an effect in the negative $x$ direction whereas the centrifugal force, unless I am mistaken, should have an effect in both $x$ and $y$ directions. Hence, $$N = m \omega^2 \frac{R}{2} - 2m \omega \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)$$ and $$-mg + \mu N + m \omega^2 y=m\left(\frac{\mathrm{d}^2y}{\mathrm{d}t^2}\right)$$

Is this how this problem ought to be approached?

Answer 1

The problem is much more simple than you think. The Coriolis force vanishes in this case (this is simple to see, since the motion of the bug is circular with constant speed), therefore only the centrifugal has an effect on friction. Then, the speed on exit can be easily found if you use the conservation of energy, it is $\sqrt{ 2\sqrt{3} gR ( 1-\mu/2)}$. If you need the full velocity vector, you'll need to add the velocity due to rotation.