# Two blocks connected by springs moving in a circle

## Homework Statement

Block A and block B are on a frictionless table as shown in Figure 2. Spring 1 connects block A to a frictionless peg at 0 and spring 2 connects block A and block B. Their masses are respectively mA=0:45 kg and mB=0:32 kg. When the blocks are in uniform circular motion about 0, the springs have lengths of lA=0.80 m and lB=0.50 m. The springs are ideal and massless, and the linear speed of block B is 2.8 m/s. If the distance that spring 2 stretches is 0.070 m, determine the spring constant of spring 2.

F=ma
a=v2/r
F=kx

## The Attempt at a Solution

y-direction is perpendicular to surface and x-direction is parallel to surface
rA=lA, rB=lA+lB
T1=Tension is spring 1, T2=Tension in spring 2
NA=Normal acting on block A, NB=Normal acting on block B
xB=0.07m (The extension of spring 2)

For block A (From FBD)
Resolving (y-direction): FAy=mAaAy=NA-mAg=0

Resolving (x-direction): FAx=mAaAx=T1-T2

⇒(mAvA2)/rA=T1-T2

For block B (From FBD)
Resolving (y-direction): FBy=mBaBy=NB-mBg=0

Resolving (x-direction): FBx=mBaBx=T2

⇒(mBvB2)/rB=T2

And this is where I have become stuck, because I know I can't simply say T2=kxB as this would give the spring constant for a single spring that has extended to a total length of 1.3m. And continuations from this point do not seem to go anywhere e.g. substituting centripetal force of block B into the equation for block A, does not help as far as I can see.

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haruspex
Homework Helper
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I know I can't simply say T2=kxB as this would give the spring constant for a single spring that has extended to a total length of 1.3m.
why would it do that? T2 is the tension, xB is the extension.

why would it do that? T2 is the tension, xB is the extension.
I thought of it originally, but then I thought it couldn't be because it doesn't require me to use anything from block A, and the block B would be under the same tension at that distance with a single spring, at least I think so. But surely it can't be as simple as to just equate (mBvB2)/rB with kxB?

I think I managed to convince myself that (mBvB2)/rB=kxB was wrong because it seemed too easy and didn't involve values from spring 1 or block A.

haruspex