Lancelot59
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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:
[tex]a_{c}=\frac{v^{2}}{r}=\frac{36}{r}[/tex]
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:
[tex]F_{spring}=ks=20(l-2) lb[/tex]
I don't get where they got l-2 from.
Then it goes on to this:
[tex]a_{c}=\frac{v^{2}}{r}=\frac{36}{0.5+lsin(\theta)}[/tex]
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?
I need to find the angle theta.
I decided to try getting the centripetal acceleration first:
[tex]a_{c}=\frac{v^{2}}{r}=\frac{36}{r}[/tex]
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:
[tex]F_{spring}=ks=20(l-2) lb[/tex]
I don't get where they got l-2 from.
Then it goes on to this:
[tex]a_{c}=\frac{v^{2}}{r}=\frac{36}{0.5+lsin(\theta)}[/tex]
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?
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