Centripetal acceleration fan problem

[Adsy]

Homework Statement

A ceiling fan is turning at a rate of 100 revolutions per minute. A spiders is clinging to a blade of the fan. If the spider experiences a centripetal acceleration greater than 0.3g, it will lose its grip on the blad and be flung off. How far from the centre of the fan can the spider safely go?

$$Rate = 100 rev/sec$$

$$a = 0.3g$$

$$r=?$$

Homework Equations

$$\omega=\frac{\Delta\theta}{\Delta t}$$

$$\omega=\frac{2 \pi}{T}$$

$$v= \omega r$$

$$T= \frac{2 \pi}{\omega}$$

$$a=\frac{v^{2}}{r}$$

$$a=\omega^{2}r$$

The Attempt at a Solution

I've worked out that the time period, $$T = 0.6s$$

$$a=0.3g=2.94 ms^{-2}$$

then use: $$\omega=\frac{2 \pi}{T}$$

$$\omega=\frac{2 \pi}{0.6} = 10.47 rad s^{-1}$$

then I rearrange this formula: $$a=\omega^{2}r$$

$$r= \frac{a}{\omega^{2}}$$

then put in the known values to find r
$$r= \frac{2.94}{10.47^{2}} = 2.7*10^{-2}m$$

*fixed*

This is incorrect. What am I doing wrong?

 

[Quinzio]

then put in the known values to find r
$$r= \frac{2.94}{10.47^{2}} = 2.7*10^{2}m$$

This is incorrect. What am I doing wrong?

$$2.7*10^{-2}m = 27mm$$

 

[Adsy]

Oh, that's just a mistake with TeX.
That answer is still incorrect.

The correct answer is 1.05m but how do you work it out?

 

[Quinzio]

I think there's a mistake in your "correct" solution.
The solution is 27mm.
The problem is rather simple.
Other readers may double check the result.

 

[Adsy]

I've asked my friend about this question. He also said the answer is 2.7cm.
Hmm... maybe my textbook has an incorrect solution...

 

[Adsy]

Yep. My answer is correct. I've Googled the problem and found other people struggling with the same question. The book is incorrect. Silly Edexcel...
Thanks anyways, Quinzio!