# Horizontal Circular Motion-Tension, Speed- Quick Help Needed

1. Feb 28, 2005

### cde42003

Can anyone give me some clues as how to do this problem. I have been on it for ever and am stumped. Any help greatly appreciated. Thanks

Two wires are tied to the 2.0 kg sphere shown in the figure . The sphere revolves in a horizontal circle at constant speed.

For what speed is the tension the same in both wires?(in m/s)
What is the tension?

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2. Feb 28, 2005

### Andrew Mason

I posted an outline for the solution to this same problem a few weeks ago. See: https://www.physicsforums.com/showthread.php?t=64510

Here are the details then:

Let the top wire = L1 with tension T; the other is L2, T:

$$Tcos(30) + Tcos(60) = mg$$

(1)$$T = 2*9.8/1.366 = 14.35 N$$

(2)$$Tsin(30) + Tsin(60) = m\omega^2r$$

(3)$$L_1cos(30) - L_2cos(60) = 1$$ (ie. 1 m.)

$$L_1sin(30) = L_2sin(60) = r$$

So:
$$L_2 = L_1sin(30)/sin(60)$$

Substitute for L_2 in (3):

$$L_1(cos(30) - sin(30)cos(60)/sin(60)) = 1$$

$$L_1 = 1/(.866 - .5 * .5 /.866) = 1.54 m$$

$$r = .77m$$

I'll leave the rest to you to work out

AM

3. Feb 28, 2005

### stunner5000pt

OK for the triangle give to you at first, find all teh angles and hence find all the sides

Now for the ball, what forces are acting on it? (one starts with C and the other with G)

G points straight down (as always), and C is 90 degrees to g , and points left

Draw these two vectors.
Now you have two triangles... (although it sounds more complicated it isnt)
Now write the force equations (considering the components of force points up and down, and left and right separately). You should have no trouble hereafter

4. Feb 28, 2005

### cde42003

Thanks to both of you. I got the answer correct

5. Oct 15, 2011

### ment2byours

Would it be for the same concept for the diagram below?

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