Horizontal Circular Motion-Tension, Speed- Quick Help Needed

[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?

 

[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

Then write the equation (2) for the horizontal force on the sphere (in terms of \omega and r)

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

Then find the lengths of the wires in order to find r.

(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

Substitute r into (2) (both tensions = T) and use T from (1) to find \omega

I'll leave the rest to you to work out

AM

 

[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

 

[cde42003]

Thanks to both of you. I got the answer correct

 

[ment2byours]

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