How Many Revolutions Before the Tube Breaks?

AI Thread Summary
The discussion revolves around calculating the number of revolutions a 500g steel block makes before a hollow tube breaks under tension. The block, attached to a 1.2m tube, experiences a thrust force of 4.0N, with the tube's maximum tension set at 50N. The initial approach mistakenly used the thrust to find acceleration, which does not contribute to the block's rotational motion. Instead, the correct method involves applying the centripetal force equation, setting the tension equal to the maximum limit, and calculating the number of revolutions based on the block's velocity and distance traveled. The correct approach leads to a solution that accurately reflects the physics of the scenario.
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A 500g steel block rotates on a steel table while attached to a 1.2m long hollow tube. Compressed air fed through the tube and ejected from the nozzle on the back of the block exerts a thrust force of 4.0N perpendicular to the tube. The maximum tension the tube can withstand without breaking is 50N. If the block starts from rest, how many revolutions does it make before the tube breaks?

This is a problem from the back of the book that I already know the answer to (3.75) but I think my way of getting it is wrong since I can't get the right answer to a similar problem with different numbers.

What I did first is divide the thrust by the weight 4N/.5kg= 8m/s^2, then I set the tension of 50N equal to the acceleration times the distance times kinetic friction

50N = 8m/s^2*distance*.6

this gets me the right answer of 3.75 revolutions but when the numbers are changed to 4.1N thrust and 60N max tension the answer I come up with (4.39) is wrong.

Anyone have any ideas?
 
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actually looking back, it's 50N= 8m/s^2*distance, so you get 6.25, then multiply that by .6

6.25*.6=3.75 revolutions.
 
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There are a few potential issues with the approach you have taken to solving this problem. First, it is important to note that the 4.0N thrust force is perpendicular to the tube, meaning it does not contribute to the rotational motion of the block. Therefore, dividing it by the weight of the block to find the acceleration is not necessary.

Additionally, the equation you have used to find the tension in the tube assumes that the block is moving with a constant velocity, which may not be the case in this scenario. Instead, we can use the equation for centripetal force, which is F = mv^2/r, where F is the force, m is the mass, v is the velocity, and r is the radius of the circular motion.

In this case, the force causing the circular motion is the tension in the tube, and we can set it equal to the maximum tension the tube can withstand without breaking, which is 50N. So, we have:

50N = mv^2/r

To find the velocity, we can use the fact that the block starts from rest and travels a distance of 1.2m in each revolution. This means that the velocity at the end of each revolution is equal to the circumference of the circle (2πr) divided by the time it takes to complete one revolution. We can find the time by dividing the distance traveled (1.2m) by the velocity (v), so we have:

v = 2πr/t

Plugging this into our first equation, we get:

50N = m(2πr/t)^2/r

Simplifying, we get:

50N = 4mπ^2r/t^2

Now, we can solve for the number of revolutions (n) it takes for the tube to break by rearranging the equation to solve for t:

t = √(4mπ^2r/50N)

And since we know that the block travels a distance of 1.2m in each revolution, we can set t equal to n times the time it takes for one revolution (1.2/v). So, we have:

n(1.2/v) = √(4mπ^2r/50N)

Solving for n, we get:

n = √(4mπ^2r/50N) * v/1.2

Plugging in the given
 
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