Help with tension problem?

Question

An 8kg ball is suspended .6m at the end of a horizontal bar which is attached to a vertical pole. The horizontal bar extends .8 meters from the top of the vertical pole. From the top of the pole, there is a second string that extends down to the ball at a 30 degree angle. The pole is rotating at a constant speed such that the linear velocity of the mass is 2.3 m/s. What is the tension of the vertical wire? Please show steps and formulas used to get solution.

Answer 1

First, there's a bit of trig to find the distance from the top of the pole to the ball and the vertical distance from the rod to the ball. Since there is no diagram given, I drew my interpretation
http://i142.photobucket.com/albums/r88/odu83/ball.jpg

I knew that the distance was not 1m since the tan(30)=0.577 not 0.75
sin(30)=h/d
cos(30)=(0.8+L)/d
L^2+h^2=0.6^2

after some algebra
(cos(30)*2*h-0.8)^2+h^2=0.6^2
solve for h
h=0.570 m
check
d=1.14 m
and
L=0.1873 m
That means the radius of angular motion is
0.8+0.1873
=0.9873 m

Note: There are two values for h because it is possible for the 0.6 m long wire to hang such that it closer to the pole instead of away from the pole and still have the wire from the top go to the ball at 30 degrees.
In that case h=0.1228 m

I will solve for h=0.570 m The technique is te same for the other case

Based on the radius, the centripetal acceleration is
2.3^2/0.9873
so the horizontal force is
8*2.3^2/0.9873
or 42.86 N. This is the horizontal force that must be balanced by both wires

Sum forces in the horizontal
42.68=T1*0.5+T2*0.1873/0.6
where T1 is the tension in the wire from the top of the pole and T2 is the tension in the wire from the end of the rod.

Sum the forces in the vertical
8*9.81=T1*cos(30)+T2*0.570/0.6

solve for T2

90.62=T1+T2*1.097
85.36=T1+T2*0.624

T2=11.13 N
check

T1=78.41 N

Checks

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