Finding number of rotations till arm breaks (circular motion)

[Micah]

Homework Statement

A 400 g steel block rotates on a steel table while attached to a 1.20 m -long hollow tube. Compressed air fed through the tube and ejected from a nozzle on the back of the block exerts a thrust force of 4.91 Nperpendicular to the tube. The maximum tension the tube can withstand without breaking is 60.0 N . Assume the coefficient of kinetic friction between steel block and steel table is 0.60. (Figure 1)

Homework Equations

∑F_t=thrust-ƒ_k
∑F_r=mv²/r
a_t=αr
v=ωr
ω_f²=ω_i²+2αΔΘ
2π radians=1 revolution

The Attempt at a Solution

∑F_t=ma_t=4.91N-(.4kg*.6*9.81m/s²
a_t=2.5556kg m/s² /.4kg
a_t=6.839m/s²

α=6.839m/s²/1.2m
α=5.324s^-2

I assume radians are assumed as units in α..not sure on this point.

∑F_r=mv²/r
Since 60N is the max ∑F_r can be
60N=.4kgv²/1.2m
180 m²/s²=v²
v=13.416 m/s

v/r=ω
13.416m/s/1.2m
ω=11.18s^-1

Im assuming radians are implicit here. Again not sure. ω_f²=ω_i²+2αΔΘ

We know ω_i² is zero so:

ω_f²=2αΔΘ
ω_f²/2α=ΔΘ
(11.18s^-1)²/(2*5.324s^-2)=ΔΘ
ΔΘ=11.73

Again I'm assuming ΔΘ is in radians. So

11.73 * (1 revolution)/(2π radians)

1.86 revolutions.

My answer is wrong. That's all I got.

 

[haruspex]

1.86 revolutions.

I make it closer to 1.87, but other than that I agree. I see nothing wrong in your logic.
But I strongly recommend working entirely symbolically until the final step. This has many advantages. In the present problem, you would have discovered that the radius was irrelevant, and you would have avoided the need to take a square root.