Solving Odd Dynamics Problem with 2 lb/ft Spring & 10 lb Bob

[Lancelot59]

The spring has a 2 lb/ft, and the bob has a tangental velocity of 6 ft/s, and a weight of 10 lb.

I need to find the angle theta.

I decided to try getting the centripetal acceleration first:
$$a_{c}=\frac{v^{2}}{r}=\frac{36}{r}$$

which led me nowhere. I tried a mess of things.

The best I could come up with was a triangle with three unknowns, the hypotenuse of which was the spring.

The solution manual is quite confusing.
Let l be the length of the spring
It sets up the force of the spring as this:
$$F_{spring}=ks=20(l-2) lb$$
I don't get where they got l-2 from.

Then it goes on to this:
$$a_{c}=\frac{v^{2}}{r}=\frac{36}{0.5+lsin(\theta)}$$

The rest of it is simple enough to understand, it's just messing with the net forces. The setup of this is what I don't get. How did they arrive at that formula for the spring force, and then stick it into r like that?

 

[tiny-tim]

Hi Lancelot59!

I don't get where they got l-2 from.

2 feet must be the unstretched length of the spring (so l - 2 is the extension)

 

[Lancelot59]

Hi Lancelot59!


2 feet must be the unstretched length of the spring (so l - 2 is the extension)

I see. So that's the spring's displacement.

How did they derive that expression for r then? I understand where the lsin(theta) term comes from, but why are they adding 0.5? Would it not be 6?

 

[tiny-tim]

why are they adding 0.5? Would it not be 6?

erm …

feet! ​

 

[Lancelot59]

erm …

feet! ​

Oh! I see. I don't usually work with the imperial system. Thanks!