# Acceleration with respect to rotating frame and a rotating object

Consider a space, free from gravity, etc. and there is a cylinder (radius $R$) rotating about its axis in this space with an angular velocity $\vec{\omega}$. I'm standing (rest wrt cylinder) inside on the floor (circular face), at a distance $r$ from the center, and an object is there inside the cylinder (distance $z$ from center) which does not move with respect to a stationary observer outside the cylinder.

Consider the acceleration of the object with respect to a frame fixed to the rotating cylinder (and rotating with it). The pseudoforce is $m\omega^2z + 2m\vec{\omega}\times\vec{v}$, $\vec{v}$ being its velocity with respect to the rotating frame. Note that this is independent of $r$.

Consider me. With respect to ground frame, I'll have an acceleration of $-\omega^2\vec{r}$, $\vec{r}$=my position vector. So with respect to me acceleration of object = $\omega^2\vec{r}$, dependent on $r$.

Now I'm fixed wrt rotating frame. So shouldn't these two accelerations be equal?

(I'm might be missing something totally obvious. Please correct me.)

• The expression $m\,\omega^2 z + 2 m\,\vec \omega \times \vec v$ is a force, not an acceleration. Dividing by mass yields the acceleration; the result is $\omega^2 z + 2 \vec \omega \times \vec v$. Now substitute $z=r$, $\vec v = 0$ into $\omega^2 z + 2 \vec \omega \times \vec v$. What's the result? – David Hammen Dec 16 '16 at 8:32
• Why should $z = r$? $z$ = distance of object. $r$ = my distance, so they are not necessarily equal. – Mriganka Basu Roy Chowdhury Dec 16 '16 at 16:10
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