Circular Motion and maximum tension Problem

In summary, the conversation discusses making a communication device based on an invention by the Australian Aborigines. The design involves a noise-maker on a string and the question is whether the string is strong enough for the device. The tension in the string is calculated using the formula for centripetal force and the mass and acceleration of the device. The conversation also mentions finding the tangential velocity, but an easier way to calculate the tension is to simply use the mass and acceleration to find the force.
  • #1
desichick07
18
0

Homework Statement


yes this question is a bit ridiculous but stick with it please...

After watching the movie "Corcodile Dundee" you and some friends decide to make a communications device invented by the Austrailian Aborigines. It consists of a noise-maker swung in a vertical circle on the end of a string. Your design calls for a 400 gram noisemaker on a 60 cm string. You are worried about whether the string you have will be strong enough, so you deicde to calculate the tension in the string when the device is swung with an aceleration which has a constant magnitude of 20 m/s2. You and your friends can't agree whether the maximum tension will occur when the noise maker is at the highest point in the circle, at the lowest point in the circle, or is always the same. To settle the argument you decide to calculate the tension at the highest point and at the lowest point and compare them.

Given:
.4 kg
.06 m string
a = 20 m/s^2

Homework Equations


Top: F(n) = mv^2 - mg
r

bottom: F(n) = mv^2 + mg
r

The Attempt at a Solution



T = F(n), where t = tension in the string and F(n) = normal force
F(n)

the centripetal force F(c) = ma(c)

TOP:
mv^2 - mg = ma(c)
r

the masses cancel and you're left with
v^2 - g = a(c) (*+*)
r

and solving for v, the velocity, you get:
v = [r*(a(c) + g)]^.5
v = [.6*(20+9.8)]^.5
= 4.23 m/s^2

T = F(n)
then use the above equation (*+*)with 4.23 for the velocity.
upon solving I got -1.1 N. I triple checked my calculations and they seem right. Can anyone spot my error?
Also is there an easier way to do this?
 
Last edited:
Physics news on Phys.org
  • #2
The centripetal acceleration is 20m/s^2 so that the centripetal force [itex]F_c=ma_c[/itex]
so just use the mass and multiply by the acceleration to get the force. the weight is always constant;[itex]w=mg[/itex]. So then all you need to do is sub it into the two equations. But why did you find the tangential velocity for?EDIT: Even though the question asks for a comparison of values, all you can alternately do is compare the formulas.
 
  • #3




I would first like to commend your creative thinking and enthusiasm in trying to replicate a communication device from a movie. However, I would like to address the issue of circular motion and maximum tension in your design.

Firstly, let's clarify some concepts. In circular motion, the tension in the string is not constant throughout the motion. It varies depending on the position of the object in the circle. This is because the centripetal force, which is responsible for keeping the object in circular motion, changes with the object's position in the circle.

Now, to answer your question, the maximum tension in the string will occur when the object is at the lowest point in the circle. This is because at this point, the centripetal force is equal to the weight of the object, which is the maximum force acting on the string. This can also be seen in your equations, where the tension at the bottom is greater than the tension at the top.

In terms of your calculations, I believe you have made an error in the equation for the velocity. The correct equation should be v = √(r*a(c)). Using this equation, I got a velocity of 4.47 m/s^2, which gives a tension of 2.2 N. This is a positive value, indicating that the tension is indeed greater at the bottom.

Furthermore, as a scientist, I would suggest using the concept of centripetal force, F(c) = mv^2/r, to calculate the tension in the string. At the bottom, F(c) = mg + T, where T is the tension in the string. This will give you the same result as using the equation for normal force, but it is a more direct approach.

In conclusion, the maximum tension in the string will occur at the bottom of the circle, and the tension at the top will be lower. This is due to the changing centripetal force acting on the object. I hope this clarifies your doubts and helps you in your design. Keep up the creative thinking!
 

1. What is circular motion?

Circular motion is the movement of an object along a circular path. This type of motion involves a constant change in direction, but the magnitude of the object's velocity remains the same.

2. How is centripetal acceleration related to circular motion?

Centripetal acceleration is the acceleration that points towards the center of a circular path. In circular motion, this acceleration is necessary to keep the object moving along the circular path without flying off in a straight line.

3. What is the maximum tension problem in circular motion?

The maximum tension problem in circular motion is a physics problem that involves calculating the maximum tension that a string or rope can withstand before breaking when attached to an object moving in a circular path.

4. How is the maximum tension calculated in circular motion?

The maximum tension can be calculated using the centripetal force equation, which states that the maximum tension is equal to the product of the mass of the object, its velocity squared, and the radius of the circular path, divided by the radius of the object itself.

5. What factors affect the maximum tension in circular motion?

The maximum tension in circular motion is affected by the mass of the object, its velocity, and the radius of the circular path. A larger mass or higher velocity will result in a greater maximum tension, while a larger radius will decrease the maximum tension.

Similar threads

  • Introductory Physics Homework Help
Replies
14
Views
546
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
27
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
618
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
Back
Top