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.
The Attempt at a Solution
y-direction is perpendicular to surface and x-direction is parallel to surface
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
For block B (From FBD)
Resolving (y-direction): FBy=mBaBy=NB-mBg=0
Resolving (x-direction): FBx=mBaBx=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.