Question

A 600 g steel block rotates on a steel table (μ_k = 0.6) while attached to a 1.4-m-long hollow tube as shown in the figure below. Compressed air fed through the tube and ejected from a nozzle on the back of the block exerts a thrust force of 4.0 N perpendicular to the tube. The maximum tension the tube can withstand without breaking is 50 N. If the block starts from rest, how many revolutions does it make before the tube breaks?

Answer 1

The linear (tangential) acceleration is

a_t = F / m = 4.0N / 0.600kg = 6.67 m/s²

and after time "t" the velocity is

v = a_t * t = 6.67m/s² * t

and so the centripetal acceleration is

a_c = v²/r = (6.67m/s² * t)² / 1.4m = 31.75m/s⁴ * t²

The tension is

T = m*a_c, so

50.0 N = 0.600kg * 31.75m/s⁴ * t²

t = 1.62 s

ω = v / r = 6.67m/s² * 1.62s / 1.4m = 7.72 rad/s

so the average angular velocity ωavg = 3.86 rad/s

Θ = ωavg*t = 3.86rad/s * 1.62s = 6.25 rads = 0.99 revolutions (nearly equal to 1)