Nonuniform Circular Motion of a steel table

AI Thread Summary
A 600 g steel block rotates on a steel table, attached to a 1.20 m hollow tube, with a thrust force of 4.30 N acting perpendicular to the tube. The maximum tension the tube can withstand is 60.0 N, prompting calculations to determine how many revolutions the block makes before the tube breaks. Initial calculations yielded a tangential acceleration of 1.287 m/s² and an incorrect period of 8.513 seconds, leading to confusion about the total number of revolutions. After clarification, the correct number of revolutions was determined to be approximately 6.1. The discussion highlights the importance of accurately calculating time and acceleration in nonuniform circular motion scenarios.
pcmarine
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A 600 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 4.30 N perpendicular to the tube. The maximum tension the tube can withstand without breaking is 60.0 N.

If the block starts from rest, how many revolutions does it make before the tube breaks?

knight_Figure_07_55.jpg



I've done some calculations and have come up with:
Tangential Acceleration: 1.287 m/s^2
Omega: 9.13
Period: 8.513 seconds

plugged these numbers into Theta=((a)/(2*r))*(period)^2
and got 38.953 rads, or 61.19 revolutions... which was wrong :mad:
 
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pcmarine said:
A 600 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 4.30 N perpendicular to the tube. The maximum tension the tube can withstand without breaking is 60.0 N.

If the block starts from rest, how many revolutions does it make before the tube breaks?

knight_Figure_07_55.jpg



I've done some calculations and have come up with:
Tangential Acceleration: 1.287 m/s^2
Omega: 9.13
Period: 8.513 seconds

plugged these numbers into Theta=((a)/(2*r))*(period)^2
and got 38.953 rads, or 61.19 revolutions... which was wrong :mad:


I am assuming that you don't mean "period", you mean the total time it takes to reach the final omega?? (it does not make sense to talk about a period since the rotation is not at a constant omega). And I am guessing that you mean 6.119 revolutions.

Can you just show how you got your tangential acceleration?
 
Wow thanks, yeah you were right, it was actually a decimal over... 6.1 revolutions.
 
pcmarine said:
Wow thanks, yeah you were right, it was actually a decimal over... 6.1 revolutions.

So, do you have the right answer now?
 
yep, thanks
 
how the heck are you getting that time total of 8.513?
 
bump? Any help there?
 
Fierofiend said:
how the heck are you getting that time total of 8.513?
What do you think it should be?
 
No idea, tried for hours, couldn't come up with that answer. I figured it would be something with omega and accel, but nothing computes.
 
  • #10
Fierofiend said:
No idea, tried for hours, couldn't come up with that answer. I figured it would be something with omega and accel, but nothing computes.
Perhaps because that answer i wrong. It does have something to do with acceleration, and omega if you choose to use that. What did you try?
 
Last edited:
  • #11
my final velocity was 10.95 m/s and I tried to divide by my accel
 
  • #12
Fierofiend said:
my final velocity was 10.95 m/s and I tried to divide by my accel
Let's say that is correct. What did dividing by acceleration give you? Where do you go from there?
 

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