Uniform circular motion - finding velocity

1. Feb 13, 2016

reminiscent

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

2. Relevant equations

3. The attempt at a solution
I'm unsure if we have to mention the tension of the cable, but I'm guessing L2 will suffice? Correct me if I am wrong.
Anyways, I drew the diagram from a side view (kind of like a circular pendulum) but L1 and the angle is throwing me off. I also drew a free body diagram of the seat at the 3 o'clock position: Tcable (tension or L2?) point towards the center, Tz pointing north, Tr pointing left, and Fg downwards. In the z direction, I have Tz + Fg = 0, so Tcable = mg/cos(theta). In the radial direction, I have Tr = ma(radial) > Tsin(theta) = mv^2/L1(?) so far.
Help will be greatly appreciated.

2. Feb 13, 2016

TSny

OK for the tension in the cable. However, you do not have the correct radius of the circle that the rider is moving along.

3. Feb 13, 2016

reminiscent

Is it R = L1+L2x = L1 + L2sin(theta)?

4. Feb 13, 2016

TSny

Yes it is.

5. Feb 13, 2016

reminiscent

Okay so after that part, I plugged in T (from the z direction) and R and I found v = sqrt( (L1+L2sin(theta))*g*tan(theta)). Does that sound about right? But the mass cancelled out, how come in the problem they asked for the expression to contain the mass?

6. Feb 13, 2016

TSny

Yes.
I don't know. Maybe they didn't want to give away that the mass doesn't matter.

7. Feb 14, 2016

reminiscent

Okay thank you so much!

8. Feb 14, 2016

reminiscent

Yeah I didn't think we would need to consider the tangent acceleration in this case.

9. Feb 14, 2016

TSny

Right. There's no tangential acceleration in this case since the circular motion is "uniform".

10. Feb 14, 2016

KeithPickering

Consider the forces on the rider: there are three.
1. g, acting downward; acceleration g = 9.8 m/s²
2. centifugal force (call it C), acting outward; accleration C = v²/R
3. The force on the cable (call it T), which MUST close the triangle at the hypotenuse, hence T = sqrt (g² + (v²/R)²)

Then θ = arctan (C/g)

Therefore θ = arctan ([v²/R]/g)
tanθ = (v²/R)/g
g tanθ = v²/R
v² = R g tanθ
v = sqrt ( R g tanθ)

Now all you have to do is express r in the terms of the problem (L1, L2, etc) and plug it in ... I think you can figure that out ...