Circular Motion and Friction of brass block

[nahya]

The coefficient of friction between a certain brass block and a large revolving turntable is µ = 0.23. How far from the axis of rotation can the block be placed before it slides off the turntable if it is rotating at 33 1/3 rev/min?

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the only force on the block is the friction in the same direction as the centripetal acceleration, right?
so...

f = m * a(centripetal)
0.23 * N = m * v^2 / r = mRω^2, where ω = 2*pi*F(frequency), F = 1/(33 1/3) = 0.03 and N = mg.

so, simplifying,
0.23*m*g = 100/3*m*R*ω^2,
0.23*g = 100/3*R*w^2
0.0069*g = R(w^2)
R = 0.06762 / w^2, where w^2 = 0.03553
R = 1.9032 meters, which, apparently, is wrong. but why?

my work seems right, and the answer sounds right...
what have i done wrong?

 

[pseudovector]

General tip - always do the algebra with letters only and substitue numerical values only at the end.
Your calculation should look like this:
miu * m* g = m * R * w^2
miu * g = R * w^2
R=(miu * g)/w^2 = (0.23 * 9.8)/(2 * pi * (33.33/60))^2=0.185m=18.5cm

 

[nahya]

how did you get 33.33/60?
where did 60 come from?
is that converting it to rev/hour? if so, why? it's not part of the question btw. I'm just wondering. =)

 

[Doc Al]

f = m * a(centripetal)
0.23 * N = m * v^2 / r = mRω^2, where ω = 2*pi*F(frequency), F = 1/(33 1/3) = 0.03 and N = mg.

It looks like you are using F = 1/T, where T = period. No need for that, since you are given the frequency (you just have to convert it to standard units).

33 1/3 is the frequency in rev/min; you need the frequency in rev/sec. (That's where the (33.33)/60 comes from.)