# Rotational Motion + Tension Problem

1. Mar 10, 2008

### the7joker7

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

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 [Broken]

2. Relevant equations

3. The attempt at a solution

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

Last edited by a moderator: May 3, 2017
2. Mar 10, 2008

### Hootenanny

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

3. Mar 10, 2008

### the7joker7

The length of the rod = 3

The length of the two strings = 4

4. Mar 10, 2008

### Hootenanny

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

5. Mar 10, 2008

### the7joker7

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

6. Mar 10, 2008

### Hootenanny

Staff Emeritus
Yes, but note that,

$$\theta = 2\cdot\tan^{-1}\left(\frac{3}{2}\right)$$

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

7. Mar 10, 2008

### the7joker7

Okay, so I get

the angle = 112.61 degrees

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

8. Mar 10, 2008

### rohanprabhu

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?

9. Mar 10, 2008

### 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$$_{c}$$ = (36)/1.11

A$$_{c}$$ = 32.432

$$\omega$$ = 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?

10. Mar 10, 2008

### the7joker7

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?

11. Mar 10, 2008

### Hootenanny

Staff Emeritus
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?