Uniform circular motion - finding velocity

[reminiscent]

Homework Statement

Homework Equations

radial acceleration = v^2/R

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.

 

[TSny]

Tcable = mg/cos(theta). In the radial direction, I have Tr = ma(radial) > Tsin(theta) = mv^2/L1

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

 

[reminiscent]

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

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

 

[TSny]

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

Yes it is.

 

[reminiscent]

Yes it is.

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 canceled out, how come in the problem they asked for the expression to contain the mass?

 

[TSny]

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?

Yes.

But the mass canceled out, how come in the problem they asked for the expression to contain the mass?

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

 

[reminiscent]

Yes.

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

Okay thank you so much!

 

[reminiscent]

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

 

[TSny]

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

 

[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 ...