Two blocks connected by springs moving in a circle

In summary, on a frictionless table, two blocks connected by two springs with masses of 0.45 kg and 0.32 kg respectively are in uniform circular motion with spring lengths of 0.80 m and 0.50 m. The linear speed of block B is 2.8 m/s and the distance spring 2 stretches is 0.070 m. By using the equations F=ma, a=v^2/r, and F=kx, the spring constant of spring 2 can be determined to be (mBvB^2)/rB = kxB. This is the same as if spring 1 and block A were replaced by an inextensible massless string of the same length
  • #1
ConorDMK
25
0

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-T2For block B (From FBD)
Resolving (y-direction): FBy=mBaBy=NB-mBg=0

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

⇒(mBvB2)/rB=T2And 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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  • #2
ConorDMK said:
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
haruspex said:
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
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
ConorDMK said:
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
haruspex said:
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.
 

Related to Two blocks connected by springs moving in a circle

1. What is the concept of "two blocks connected by springs moving in a circle"?

The concept refers to a physical system in which two blocks are connected by springs and are moving in a circular motion. The blocks are connected by a fixed center point and the springs act as the restoring force to keep the blocks moving in a circular path.

2. What is the significance of studying this system?

Studying this system allows us to understand the principles of circular motion, oscillations, and the behavior of springs. It also has real-life applications in various fields such as mechanical engineering and physics.

3. How does the mass of the blocks affect the motion of the system?

The mass of the blocks affects the period of oscillations of the system. Heavier blocks will have a longer period, while lighter blocks will have a shorter period. However, the mass does not affect the shape of the circular motion.

4. Can the spring constants of the two springs be different?

Yes, the spring constants of the two springs can be different. This will affect the restoring force acting on each block and may result in a non-uniform circular motion.

5. How is the velocity of the blocks related to the amplitude of the circular motion?

The velocity of the blocks is directly proportional to the amplitude of the circular motion. This means that as the amplitude increases, the velocity of the blocks also increases. However, the frequency of the oscillations remains constant.

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