Rotational Motion + Tension Problem

[the7joker7]

Homework Statement

A 4-kg mass is attached to a vertical rod by two strings as in the figure shown. The strings are under tension when the rod rotates about it's axis. If the speed of the mass is a constant 6 m/s in the horizontal plane, find the tension in the upper and lower strings. DRAW FORCES.

http://img152.imageshack.us/img152/6571/diagrambf6.png

Homework Equations

The Attempt at a Solution

tan[tex]^{-1}[/tex](3/4) = 36.87 degrees.
2sin(36.87) = 1.2
[tex]\omega[/tex] = 5.31
V[tex]_{t}[/tex] = 1.2*5.31 = 6.376
a[tex]_{c}[/tex] = 36/r = 30

Force on upper = 6.372 * 4 sin(36.87) = 15.29N
Force on lower = 6.372 * 4 sin(36.87) + mg = 54.49N

Not sure what to do after that, and I get the feeling I'm not even close. =/

Help?
Force on the upper =

 

[Hootenanny]

tan[tex]^{-1}[/tex](3/4) = 36.87 degrees.

This is not correct, where does the 4 come from?

 

[the7joker7]

The length of the rod = 3

The length of the two strings = 4

 

[Hootenanny]

The length of the rod = 3

Good.

The length of the two strings = 4

Not so good. Can you show me the triangle in that diagram with sides of length 3m and 4m?

 

[the7joker7]

So you're saying I have to do it piece by piece, with 3/2 instead of 3/4?

 

[Hootenanny]

So you're saying I have to do it piece by piece, with 3/2 instead of 3/4?

Yes, but note that,

[tex]\theta = 2\cdot\tan^{-1}\left(\frac{3}{2}\right)[/tex]

Where [itex]\theta[/itex] is the [internal] angle between the two strings.

 

[the7joker7]

Okay, so I get

the angle = 112.61 degrees

Where do I go from there? Did I have the right idea otherwise?

 

[rohanprabhu]

Where do I go from there? Did I have the right idea otherwise?

Well.. you do have the right idea though.. you need to find the force acting on the particle.. and then the force acting on the strings.. then equate them in vector notation.. Even though the object is not in translational motion, the centripetal force is provided by the tension in the two strings.

So.. what you need to do now, is find the force acting on the bob due to the rotation of the rod. Just find the centripetal force on the bob. What would the direction of this force be?

 

[the7joker7]

So...angle between the two strings is 112.61.

Meaning the angles between the pole and the strings are both 33.7 degrees.

So we now have...

2sin(33.7) = 1.11

A[tex]_{c}[/tex] = (36)/1.11

A[tex]_{c}[/tex] = 32.432

[tex]\omega[/tex] = 6/1.11 = 5.4054

Force on upper string = 32.432 + mg(sin(56.3)) = 32.612N
Force on lower string = 32.432 + mg(sin(33.7)) = 21.745N

Does this seem alright?

 

[the7joker7]

Wait...wait...what about...

θ = arcsin (1.5 / 2)
sinθ = 3/4
θ = 48.59

(T1 + T2) = mv^2 / 2(1 - sin^2 θ) = 1152/7
T1 - T2 = 52.32

T1 = 108.45 N (top)
T2 = 56.13 N

Which one is correct?

 

[Hootenanny]

I'm not entirely sure what your doing in either case, so if you don't mind we'll start from the top. Can you calculate for me the centripetal force acting on the particle?