Two blocks connected by springs moving in a circle

  • Thread starter ConorDMK
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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.


Homework Equations


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.
 

Answers and Replies

  • #2
haruspex
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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.
 
  • #3
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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?
 
  • #4
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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.
 
  • #5
haruspex
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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.
I don't see how string 2 could tell the difference if string 1 and block A were replaced by an inextensible massless string of the same (current) length.
 
  • #6
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I don't see how string 2 could tell the difference if string 1 and block A were replaced by an inextensible massless string of the same (current) length.
Thank you for clearing that up for me in my mind. I've been doing this question for the past three days and I kept thinking to myself that the answer couldn't be as simple as I thought it was.
 

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