Question about vertical centripital force

In summary, we are discussing the tension of a string in vertical circular motion with a mass of 1kg and velocity of 10 m/sec. Using centripetal force and weight equations, we find that the tension at 30 deg below the horizon is 15 N, 0 N at the top of the loop, and 20 N at the bottom of the loop. However, this situation may not be realistic due to the need for careful engineering to maintain constant tangential speed.
  • #1
barryj
854
51

Homework Statement


What is the tension of the string at these places.
See figure. Vertical circular motion. m = 1kg, assume g = 10 m/s^2, velocity = 10 m/sec
a) what is the tension in the string when the mass is 30 deg below the horizon.
b)what is the tension at the top of the loop
c) what is the tension at the bottom of the loop[/B]

Homework Equations


centripetal force F = mv^2/r = 1(10)^2/10 = 10N
weight = mg = (1)(10) = 10 N[/B]

The Attempt at a Solution


I broke up the weight into two parts, one along the direction of the string, the other tangential to the circle. The component of the weight along the direction of the string is mgcos(60 deg) = 1(10)(.866) = 8.66N. So I figure the tension in the string must be the sum of the centripital force plus the component of the weight in the direction of the string or T = 10 + 8.66 = 18.66.
The tension at the top is 0 N and at the bottom 20 N.[/B]
 

Attachments

  • img545.jpg
    img545.jpg
    14.2 KB · Views: 397
Physics news on Phys.org
  • #2
Is the cosine of 30##^\circ## 0.866 ?
 
  • #3
Yes, the cosine of 30 deg is 0.866 but I think I should have used sin of 30 deg = 0.5 instead. This would make the tension at 30 deg = 15 N,yes??
 
  • #4
You're right. o:) Twice :smile:
I looked at the angles and got confued (your 30##^\circ## angle and 60##^\circ## angle look distractingly alike)
My approach was that the contribution to T would have to go to 0 if the angle went to 0.

Well done.
 
  • #5
It is good to know I did it correct. Circular motion in the vertical plane is a bit confusing.
 
  • #6
barryj said:
Circular motion in the vertical plane is a bit confusing.
Made more confusing by unrealistic questions, perhaps. Maintaining constant tangential speed for an object tied to a string in a vertical circle would require some careful engineering.
 
  • Like
Likes BvU

Related to Question about vertical centripital force

What is vertical centripital force?

Vertical centripital force is the force that acts towards the center of a circular motion in the vertical direction. It is responsible for keeping an object moving in a circular path and preventing it from flying off in a straight line.

How is vertical centripital force calculated?

The formula for calculating vertical centripital force is F = (mv^2)/r, where F is the force, m is the mass of the object, v is the velocity, and r is the radius of the circular path.

What are some examples of vertical centripital force?

Some examples of vertical centripital force include the force that keeps a roller coaster car moving along a vertical loop, or the force that keeps a satellite in orbit around the Earth.

What happens if there is not enough vertical centripital force?

If there is not enough vertical centripital force, the object will not be able to maintain its circular motion and will fly off in a straight line. This is known as centrifugal force, which is the outward force that arises from an object's inertia.

How does vertical centripital force relate to gravity?

Vertical centripital force is closely related to gravity, as it is the force that keeps objects in orbit around a larger body, such as the Earth. Gravity provides the necessary centripital force to keep objects in circular motion.

Similar threads

  • Introductory Physics Homework Help
2
Replies
38
Views
6K
  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
225
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
Replies
6
Views
967
  • Introductory Physics Homework Help
Replies
4
Views
411
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
435
Back
Top