What is the block's angular velocity after 10 rev?

Question

A 500 g steel block rotates on a steel table while attached to a 2.0-m-long massless rod. Compressed air fed through the rod is ejected from a nozzle on the back of the block, exerting a thrust force of 3.5 N. The nozzle is 70 degrees from the radial line. The block starts from rest.

What is the tension in the rod after 10 rev?

Answer 1

Fθ = 3.5sin70°

Aθ = M/Fθ = (0.5 kg) / (3.5 N)sin70° ≈ 0.152025396 m/s^2

α ≈ (0.152025396 m/s^2) / (2.0 m) ≈ 0.0760126980 rad/s^2

At 10 revs

ω ≈ √[(2)(10)(2π)(0.0760126980 rad/s^2)

ω ≈ 3.090637 rad/s

Fr ≈ (2.0 m)(0.5 kg)(3.090637 rad/s)^2 - 3.5cos70°

Fr ≈ 9.552037 N - 1.197070 N

Fr ≈ 8.354967 N ≈ 8.35 N