Ball Suspended from Ceiling (Uniform Circular Motion)

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a ball suspended from a ceiling, moving in a horizontal circular path. The problem requires determining the net force acting on the ball and analyzing how changes in mass affect the angle of the string while maintaining constant speed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of Free Body Diagrams to analyze forces acting on the ball. There is an exploration of equations related to net force and tension, as well as attempts to derive a mathematical proof regarding the relationship between mass, speed, and angle.

Discussion Status

Some participants have provided guidance on manipulating equations to find relationships between variables. There is an acknowledgment of a mistake regarding the use of trigonometric functions, leading to further clarification and exploration of the correct relationships.

Contextual Notes

The original poster expresses confusion about the requirement for a mathematical proof and the implications of changing mass while keeping speed constant. There is an emphasis on the need for clarity in deriving relationships without directly solving the problem.

Gotejjeken
Messages
27
Reaction score
0

Homework Statement



A ball of mass M = 0.5kg is suspended from a string whose other end is attached to the ceiling. The ball travels in a horizontal circle of radius R = 1.5m at a constant speed of v = 2 m/s.

http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/oldexams/exam1/fa07/fig25.gif

A. What is the magnitude of the net force Fnet on the mass?

B. If the mass of the ball were increased while the speed of the ball was kept the same, how would the angle theta change?

Homework Equations



Fnet = m * v2/r

The Attempt at a Solution



A. I used a Free Body Diagram here with Tension pointing diagonally in the first quadrant, and weight pointing down on the y-axis. I came up with these equations:

(Fnet)x: Tx = Mb * (v2/r)
(Fnet)y: Ty - W = 0

Using the first of these equations I was able to solve for Tension and find that the magnitude of the net force is 1.33N.

B. Here is where I am a little confused at how to approach the problem. The part specifically asks for a mathematical proof in order to be correct, however I am unable to think of how to come up with such a proof.

I was able to come up with a basic idea by using the (Fnet) equations from above and the Pythagorean Theorem. I solved for (Fnet)x and (Fnet)y, then set up a right triangle and found theta to be 74.81 degrees. Then I doubled the mass and again set up a triangle with the new (Fnet)x and (Fnet)y values and found theta to be 74.60 degrees.

While this leads me to believe that theta will stay the same when the mass is changed and the velocity is left the same, it is not the mathematical proof that is asked for. Could someone please point me in the right direction to such a proof?
 
Last edited by a moderator:
Physics news on Phys.org
I think it will be very helpful to solve the 2nd equation for T, then sub into the first. The mass cancels so you'll get a relationship between v and theta.
 
Ah, wow. Thank you, it's the little things that are so often overlooked :smile:. I solved for T as you suggested and subbed it into get:

v = sqrt(g * r * cot(theta))

I suppose the proof would then be the steps leading up to this conclusion and the fact that the result is independent of mass, so no matter what mass the ball is the angle will still remain the same if the velocity is kept constant.
 
Right!
Check that again - I'm getting tan where you have cot.
 
Doh! I was using the wrong angle, and thus got cot(theta) instead of tan(theta). Thanks for the help.
 
Most welcome! Thanks for the interesting problem.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
6
Views
4K
  • · Replies 14 ·
Replies
14
Views
5K