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## Homework Statement

A horizontal turntable rotates at a constant rate ω about a fixed vertical axis through its center

*O*. A particle of mass m can slide in a fictionless circular groove of radius

*r*centered at

*O'*which is

*r*/3 from

*O*. What is the radial acceleration of m in direction

*O'm?*

Derived from a problem in MIT OCW 16.61 (http://web.mit.edu/course/16/16.61/www/pdfs/SampleQuiz-2.pdf) and Princples of Dynamics by Greenwood.

## Homework Equations

[itex]\dot{\vec{e}_r} = \dot{\theta}\vec{e}_\theta[/itex]

[itex]\dot{\vec{e}_\theta} = -\dot{\theta}\vec{e}_r[/itex]

From a previous part of the question:

[itex]\dot{\theta}^2 = \omega^2\frac{2}{3}(1-cos\theta)[/itex]

## The Attempt at a Solution

I am trying to understand expressing the position, velocity, and acceleration of the particle in cylindrical coordinates and am currently stuck on how to express the acceleration in terms of cylindrical coordinates.

Define cylindrical coordinate system at

*O'*with [itex] \vec{e}_r [/itex] along O'm and [itex] \vec{e}_\theta [/itex]

Position of m [itex]p_m = r\vec{e}_r[/itex]

Velocity of m [itex]v_m = v_{O'} + v_{O'm} = \frac{r}{3}\omega\vec{e}_r + r\dot{\vec{e}_r}= \frac{r}{3}\omega\vec{e}_r + r(\dot{\theta} + \omega)\vec{e}_{\theta}[/itex]

Acceleration of m [itex]a_m = \frac{r}{3}\omega\dot{\theta}\vec{e}_\theta + r\ddot{\theta}\vec{e}_{\theta} + r(\dot{\theta} + \omega)\dot{\theta}\vec{e}_{r}[/itex]

Radial accleration [itex]a_r = r\dot{\theta}^2 + r\omega\dot{\theta} = r(\omega^2\frac{2}{3}(1-cos(\theta)) + \omega \dot{\theta}[/itex]

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From the attached solution, the radial acceleration should be [itex]a_r = -r(\dot{\theta} + \omega)^2 + \frac{r\omega^2}{3}cos\theta[/itex]