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!