Nonuniform Circular Motion of a steel table

In summary, the conversation discusses a scenario where a 600 g steel block is rotating on a steel table attached to a hollow tube. Compressed air is fed through the tube and exerts a thrust force, and the maximum tension the tube can withstand is also mentioned. The conversation then goes on to discuss the number of revolutions the block will make before the tube breaks, with calculations and attempts being shared. Ultimately, there is confusion about the correct answer and the conversation ends without a clear resolution.
  • #1
pcmarine
6
0
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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  • #2
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?
 
  • #3
Wow thanks, yeah you were right, it was actually a decimal over... 6.1 revolutions.
 
  • #4
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?
 
  • #5
yep, thanks
 
  • #6
how the heck are you getting that time total of 8.513?
 
  • #7
bump? Any help there?
 
  • #8
Fierofiend said:
how the heck are you getting that time total of 8.513?
What do you think it should be?
 
  • #9
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?
 

1. What is nonuniform circular motion?

Nonuniform circular motion refers to the movement of an object in a circular path at varying speeds. This means that the object is changing its velocity, either in magnitude or direction, as it moves along the circular path.

2. How is the motion of a steel table considered nonuniform circular motion?

A steel table can be considered to be in nonuniform circular motion if it is rotating at varying speeds. This could happen if the table is being pushed or pulled by an external force, causing it to accelerate or decelerate as it rotates.

3. What is the difference between nonuniform and uniform circular motion?

The main difference between nonuniform and uniform circular motion is that in uniform circular motion, the object moves at a constant speed along the circular path, while in nonuniform circular motion, the speed varies. Additionally, in uniform circular motion, the acceleration is always perpendicular to the direction of motion, whereas in nonuniform circular motion, the acceleration can be in any direction.

4. How is the acceleration of a steel table determined in nonuniform circular motion?

In nonuniform circular motion, the acceleration of a steel table can be determined using the formula a = v^2/r, where v is the speed of the table and r is the radius of the circular path. This acceleration is always directed towards the center of the circle.

5. What factors can affect the nonuniform circular motion of a steel table?

The nonuniform circular motion of a steel table can be affected by various factors such as the magnitude and direction of the external force acting on the table, the mass and shape of the table, and the friction between the table and the surface it is rotating on. These factors can cause the table to accelerate or decelerate, leading to changes in its velocity and acceleration along the circular path.

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