Tension and circular motion

In summary, a 500g steel block attached to a 1.2m long hollow tube is rotating on a steel table. Compressed air is fed through the tube and exerts a thrust force of 4.0N, perpendicular to the tube. The tube can withstand a maximum tension of 50N. In order to determine the number of revolutions the block makes before the tube breaks, the thrust force is divided by the weight and set equal to the acceleration times the distance times kinetic friction. However, when the numbers are changed, the resulting answer is incorrect. The correct method is to set the tension equal to the acceleration times the distance, and then multiply the result by 0.6, resulting in 3.75
  • #1
bpb85
2
0
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?
 
Physics news on Phys.org
  • #2
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.
 
Last edited:
  • #3


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
 

1. What is tension in circular motion?

Tension in circular motion refers to the force that is exerted by a string, rope, or any other flexible object that is attached to an object in circular motion. It is always directed towards the center of the circular motion and is responsible for keeping the object moving in a circular path.

2. How does tension affect circular motion?

Tension plays a crucial role in circular motion as it is the force that keeps the object moving in a circular path. Without tension, the object would move in a straight line due to the inertia of its motion. Therefore, the magnitude of tension must be sufficient to balance the centripetal force and maintain the circular motion.

3. Can tension be greater than the centripetal force in circular motion?

No, in circular motion, the tension must always be equal to or greater than the centripetal force. If the tension is less than the required centripetal force, the object will not be able to maintain its circular path and will move outwards. This is known as a centrifugal force.

4. How is tension related to the speed of an object in circular motion?

Tension is directly proportional to the speed of an object in circular motion. This means that as the speed of the object increases, the tension in the string or rope attached to it also increases. This is because a higher speed requires a greater centripetal force, which can only be provided by a higher tension in the string.

5. What factors influence the tension in circular motion?

The tension in circular motion is influenced by several factors, including the mass of the object, the speed of the object, and the radius of the circular path. A heavier object or a faster-moving object will require a higher tension to maintain its circular motion. Similarly, a smaller radius of the circular path will also require a higher tension.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
2
Replies
55
Views
653
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
4K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
676
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top