# Radial acceleration of a particle in a groove on a turntable

## 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

$\dot{\vec{e}_r} = \dot{\theta}\vec{e}_\theta$
$\dot{\vec{e}_\theta} = -\dot{\theta}\vec{e}_r$

From a previous part of the question:
$\dot{\theta}^2 = \omega^2\frac{2}{3}(1-cos\theta)$

## 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 $\vec{e}_r$ along O'm and $\vec{e}_\theta$
Position of m $p_m = r\vec{e}_r$
Velocity of m $v_m = v_{O'} + v_{O'm} = \frac{r}{3}\omega\vec{e}_r + r\dot{\vec{e}_r}= \frac{r}{3}\omega\vec{e}_r + r(\dot{\theta} + \omega)\vec{e}_{\theta}$

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

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

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

vela
Staff Emeritus
Homework Helper
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 $\vec{e}_r$ along O'm and $\vec{e}_\theta$
Position of m $p_m = r\vec{e}_r$

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

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

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

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From the attached solution, the radial acceleration should be $a_r = -r(\dot{\theta} + \omega)^2 + \frac{r\omega^2}{3}cos\theta$
One problem I see is that your expression for the position doesn't depend on ##\omega t##. You should also have ##\vec{v} = \dot{\vec{r}}##, and given this, it doesn't appear your two expressions are consistent, e.g., where did ##\vec{v}_{O'}## come from? Your expression for ##\vec{v}_{O'}## isn't correct either. It doesn't point in the ##\hat{e}_r## direction.

What I would do set the origin of your coordinate system at O and write ##\vec{r} = \vec{r}_{O'} + \vec{r}'## where ##\vec{r}_{O'}## is the position of O' and ##\vec{r}'## is the position of the mass relative to O'. (You might have done this already in deriving the expression for T). Then the acceleration along O'm is given by ##\hat{r}' \cdot \ddot{\vec{r}}##.

Setting O as the origin of the coordinate system and $\vec{e}_r$ in the direction of O to O', would the following components be correct?
$r_O' = \frac{r}{3} \vec{e}_r$
$r_{O'm} = -r cos(\theta + \omega t) \vec{e}_r$
$r_{OM} = r_O' + r_{O'm}$

vela
Staff Emeritus
Homework Helper
No. The way you defined ##\hat{e}_r##, it points along O'm, so why would ##\hat{r}'_O## point in the same direction? ##\hat{r}'_O## is directed along the line connecting O and O'. Also, ##\vec{r}_{O'm}## should have magnitude ##r## since the mass is constrained to a circle about O'. Your vector's magnitude varies with time because of the cosine factor.

I'd suggest working in cartesian coordinates.