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Calculating Tension Force Problem

by Declension
Tags: force, tension
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Declension
#1
Nov26-11, 03:41 PM
P: 3
1. The problem statement, all variables and given/known data

A girl swings a 2.7kg ball attached to a 72.0-cm string in a horizontal circle above her head, which makes one revolution in .98s. What is the tension force, [tex]F_{T}[/tex], exerted on the string by the ball?

2. Relevant equations

Alright, I have never worked with Tension Force in my class, but we have worked with basic physic formulas, which I applied to this problem. Since mass and time are given, I assumed to use Displacement & Time (assuming 72cm [or 0.72m] is displacement instead of radius, since I'm unsure if it's radius or not...), which is:

[tex]x = x_0 + v_0 t + (1/2) a t^2[/tex]

For the second part, I applied Newton's Second Law to find Tension Force.

[tex]{F} = {m}{a}[/tex]

3. The attempt at a solution

[tex]x = x_0 + v_0 t + (1/2) a t^2[/tex]

or (assuming [tex]v_0[/tex] is 0):

0.72 = (1/2) (a) (0.98)^2

And for acceleration I got [tex]1.49 (m/s)^2[/tex].

Then, applying Newton's Second Law:

[tex]{F} = {m}{a}[/tex]
or
F = (2.7)(1.49) = 4.04 N.

However, my only available answers are 3.8N, 3*(10^3)N, 8*(10^1)N, and 92N.

I would of rounded 4.04N to 3.8N and moved on, but I have a strong feeling I did this entire problem wrong.
If I did, could someone explain what I did wrong and guide me in the right direction?
Thanks!
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rubenhero
#2
Nov26-11, 05:28 PM
P: 42
It would be help if you drew a picture or diagram of this problem and labeling all the forces acting on the ball. I'm curious as to where you got the equation x=x0+v0t+(1/2)at2 from.
Declension
#3
Nov26-11, 05:57 PM
P: 3
Quote Quote by rubenhero View Post
It would be help if you drew a picture or diagram of this problem and labeling all the forces acting on the ball.
This is what I drew on my paper. My diagram is what originally confused me (It's a bad one).



I assumed Force of Gravity (mg) is acting on the ball, but I'm unsure if any other forces are acting upon it. I do know there is tension on the string by the ball, because that is what I'm solving for.

I was unsure if there is any Normal Force, Friction Force, or Applied Force [tex](F_N, F_f, F_p)[/tex], so I resorted to the formula for Uniformly Accelerated Motion.

Quote Quote by rubenhero View Post
I'm curious as to where you got the equation x=x0+v0t+(1/2)at2 from.
I resorted to that formula for Uniformly Accelerated Motion, since I have Distance and Time, and I'm trying to find Acceleration so I can find Tension Force.

I could use Centripetal Acceleration to find acceleration, but I don't have a Velocity, and I'm instead given other variables (mass, time, distance [or radius, still not sure]).

I'm just completely lost and confused at this point.

rubenhero
#4
Nov26-11, 06:13 PM
P: 42
Calculating Tension Force Problem

I agree the diagram you drew is confusing. How I read the problem was the horizontal circle's radius is the length of the string. You can get the velocity since you have the period.
Doc Al
#5
Nov26-11, 06:14 PM
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This is a circular motion / centripetal acceleration problem. Hints: Realize that the string must make an angle with the horizontal. Apply Newton's 2nd law.
rubenhero
#6
Nov26-11, 06:27 PM
P: 42
hey Doc Al, are you referring to a tether ball diagram because I got confused after your response.
Doc Al
#7
Nov26-11, 06:29 PM
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Quote Quote by rubenhero View Post
hey Doc Al, are you referring to a tether ball diagram because I got confused after your response.
Yes, it would look something like a tether ball diagram. Same basic idea.
Declension
#8
Nov26-11, 08:00 PM
P: 3
Quote Quote by rubenhero View Post
I agree the diagram you drew is confusing. How I read the problem was the horizontal circle's radius is the length of the string. You can get the velocity since you have the period.
I'm assuming I use [tex]V = {x}/{t}[/tex]
with x = 0.72m?
rubenhero
#9
Nov26-11, 11:11 PM
P: 42
V = x/t is for linear motion, to find speed here you have to use angular velocity because it is circular/rotational motion.
Doc Al
#10
Nov27-11, 04:36 AM
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Quote Quote by Declension View Post
I'm assuming I use [tex]V = {x}/{t}[/tex]
with x = 0.72m?
Two problems here:
(1) The distance traveled in one revolution will equal the circumference of the circle.
(2) 0.72m is the length of the string, not the radius of the circular path.

This is quite an involved problem. As I pointed out earlier, while the circle is horizontal, the string is not. You'll need a bit of trig to relate the length of the string to the radius, in addition to what I mentioned earlier.

Start by drawing an accurate diagram.


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