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?

[tex]Rate = 100 rev/sec[/tex]

[tex]a = 0.3g[/tex]

[tex]r=?[/tex]

Homework Equations

[tex]\omega=\frac{\Delta\theta}{\Delta t}[/tex]

[tex]\omega=\frac{2 \pi}{T}[/tex]

[tex]v= \omega r[/tex]

[tex]T= \frac{2 \pi}{\omega}[/tex]

[tex]a=\frac{v^{2}}{r}[/tex]

[tex]a=\omega^{2}r[/tex]

The Attempt at a Solution

I've worked out that the time period, [tex]T = 0.6s[/tex]

[tex]a=0.3g=2.94 ms^{-2}[/tex]

then use: [tex]\omega=\frac{2 \pi}{T}[/tex]

[tex]\omega=\frac{2 \pi}{0.6} = 10.47 rad s^{-1}[/tex]

then I rearrange this formula: [tex]a=\omega^{2}r[/tex]

[tex]r= \frac{a}{\omega^{2}}[/tex]

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

*fixed*

This is incorrect. What am I doing wrong?

 

[Quinzio]

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

This is incorrect. What am I doing wrong?

[tex]2.7*10^{-2}m = 27mm[/tex]

 

[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!