# Tension of two strings with centripetal acceleration?

1. Oct 17, 2009

### Elenor

1. The problem statement, all variables and given/known data:

A 4.00 kg object is attached to a vertical rod by two strings (as below). The object rotates in a horizontal circle at constant speed 6.00 m/s.

mass = 4.00 kg
V = 6.00 m/s

2. Relevant equations

centripetal acceleration = Ac
Ac = V^2/r

$$\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}$$

3. The attempt at a solution

mass = 4.00 kg
V = 6.00 m/s

Using trig, the radius (r) = 1.3229m

The lower angle of the top triangle (composed of the half the length of the rod, the length of the string, and the radius) should be 48.5904 degrees. The upper angle would then be 41.4096 degrees.

Using (Ac = V^2/r): (6.00m/s)^2/1.3229m = Ac = 27.2129m/s^2

T1 is the upper string, and
T2 in the lower string.
g = 9.8 m/(s^2)

Here is my main concern--do I have the following two equations right?

$$(\vec{F}_{net})_x = \Sigma F_x = 0$$ = m(Ac) - T1cos(41.4096) - T2cos(41.4096)

$$(\vec{F}_{net})_x = \Sigma F_x = 0$$ = T1sin(41.4096) - T2sin(41.4096) - mg

Therefore:

X direction: m(Ac) = T1cos(41.4096) + T2cos(41.4096)

and

Y direction: mg = T1sin(41.4096) - T2sin(41.4096)

At this point I have all the values except T1 and T2, so I can solve for them . . .

T1 = [m(Ac) - T2cos(41.4096)]/cos(41.4096)

mg = {[m(Ac) - T2cos(41.4096)]/cos(41.4096)} sin(41.4096) - T2sin(41.4096)

mg = m(Ac)[sin(41.4096)]/cos(41.4096) - T2sin(41.4096) - T2sin(41.4096)

-mg + m(Ac)[sin(41.4096)]/cos(41.4096) = T2[sin(41.4096) + sin(41.4096)]

{m(Ac) sin(41.4096)/cos(41.4096) - mg} / 2sin(41.4096)] = T2

T2 = 42.9353N

And . . . using that value of T2, T1 = 102.2001N

Now, I've resorted to working the problem with these numbers, as I have an answer to the problems with these numbers . . . and my problem is that this is not right, and I can't figure out where I went wrong.

With these numbers, I know that the correct answers are: T1 = 109N and T2 = 56.4N

When I figure out what I'm doing wrong, and how to do it right . . . I'll switch to the other set of numbers, for my problem.

Is my method correct? Frequently my problems are due to some little think like forgetting to square something . . . If that's all it is, I'm sorry to bother you guys with it. However, I'll be glad for any help you can give me with this problem! Thank you very much!

Last edited: Oct 17, 2009
2. Oct 17, 2009

### Elenor

Please let me know if there's something I should clarify. :-)

3. Oct 18, 2009

### Elenor

I'm not sure how to mark this solved, but at any rate, I've finally figured it out. :-) I was simply using the wrong angle. Oh well . . . at least I had it mostly right . . . ;)

4. Oct 18, 2009

### cruisx

Seems like a question from MDHS >_> =)

5. Oct 18, 2009

### Elenor

MDHS? Would that be one of these:

MDHS Mississippi Department of Human Services
MDHS Methods for the Determination of Hazardous Substances (United Kingdom)
MDHS Mater Dei High School (Santa Ana, CA)
MDHS McDonnell Douglas Helicopter Systems
MDHS Markham District High School (Canada)
MDHS Monsignor Donovan High School (Ocean County, NJ)
MDHS Milton District High School (Canada)
MDHS Mount Dora High School (Lake County, FL)
(http://acronyms.thefreedictionary.com/MDHS)

Or something else entirely?

6. Oct 18, 2009

### cruisx

that would be

MDHS Milton District High School (Canada)

had the same sort of question on my assignment.