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Steel Block Rotating On A Steel Table

  • Thread starter TonkaQD4
  • Start date
  • #1
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A 0.5kg steel block rotates on a steel table while attached to a 1.2meter long hollow tube. Compressed air fed through the tub and ejected from a nozzle on the back of the block exerts thrust force of 4.0N perpendicular to the tube. The maximum tension the tub can withstand without breaking is 50N. If the block starts from rest, how many revolutions does it make before the tube breaks?

u_k = 0.6

Components:

F_y:

n - mg = 0

n= mg

F_x:

F_t - u_kn = ma_x

F_t - u_kmg = ma_x

4.0N - (.6)(.5kg)(9.8m/s^2) = ma_x

4.0N - 2.94N = ma_x

1.06N / .5kg = a_x

a_x = 2.12m/s^2

I found the acceleration of the block, now I need to find the speed at which the tube breaks....

F = ma = (mv^2 / r)

v^2 = (Fr / m)

v = SqRt [(50N)(1.2m) / .5kg ]

v= 10.95m/s

Now do I use a kinematic equation to find distance?

v_f^2 = v_i^2 + 2a(deltaD)

(10.95m/s)^2 = 0 +2(2.12m/s^2) D

D= 10.95 / 4.24

D= 2.58 m

Now this is where I get stuck. If I have done everything correctly up to this point, which I am not sure if I have (correct me if it's wrong), how do I put everything together? What equation do I use?

Any help would be great!!
 

Answers and Replies

  • #2
Shooting Star
Homework Helper
1,977
4
You have not said how the tube is lying with respect to the rotating table. A rough diagram or a more vivid description of the set up will help in getting responses.
 

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