# Homework Help: Radial acceleration of a turntable

1. May 17, 2007

### Office_Shredder

Staff Emeritus
This is from an example in a book:

On a horizontal turntable that is rotating at constant angular speed, a bug is crawling outward on a radial line such that its distance from the center increases quadratically $$r=bt^2, \theta=\omegat$$ where $$b, \omega$$ are constants. The example then solves for the acceleration of the bug (using r, $$\theta$$ unit vectors)

$$dr/dt = 2bt; d^2r/dt^2 = 2b; d\theta/dt = \omega; d^2\omega/dt^2=0$$

So
$$a=e_r(2b-bt^2\omega^2) + e_{\theta}(0 + 2(2bt)\omega)$$
$$=b(2-t^2\omega^2)e_r + (4bt\omega)e_{\theta}$$

So as t goes to infinity, the acceleration in the radial component becomes negative. But the velocity in the radial direction is just 2bt, which increases with time.

How does this work (the book just says to note the radial acceleration is negative :grumpy: )???

2. May 17, 2007

### Dick

Even a stationary bug on the turntable experiences a negative radial acceleration of magnitude r*w^2. Your moving bug has a constant radial acceleration component from his/her movement outwards but as r gets larger the negative component from the rotation will overcome it. This doesn't mean that the bug will turn backwards - it just means it had better grip the turntable harder to keep from being thrown off.

3. May 17, 2007

### Office_Shredder

Staff Emeritus
Right... because he has centripetal acceleration to move in the circle. Makes sense now. Thanks