Equations of Motion w/ spring scale

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SUMMARY

The discussion centers on the application of Newton's second law and torque in analyzing the motion of a spring scale. The user attempts to relate the vertical displacement of the spring to the angular displacement of the mass using the equations ΣF = m a and Στ = Iθ". Key equations derived include (I0/L2)x" - k(L1/L2)x = mgL2, which requires correction due to sign errors and misinterpretation of the problem as static. The analysis concludes that the scale exhibits oscillatory motion without damping, contradicting the initial assumption of a static scenario.

PREREQUISITES
  • Understanding of Newton's second law (ΣF = m a)
  • Familiarity with torque and rotational dynamics (Στ = Iθ")
  • Knowledge of kinematics and angular motion
  • Basic principles of spring mechanics (Hooke's Law)
NEXT STEPS
  • Review the concept of mass moment of inertia in rotational dynamics
  • Study the effects of damping in oscillatory systems
  • Explore the relationship between linear and angular displacement
  • Practice solving problems involving spring dynamics and oscillations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators seeking to clarify concepts related to motion and forces in spring systems.

Nikstykal
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Homework Statement


http://imgur.com/a/X7mWA
QC78LeR.png

Homework Equations


1. ΣF = m a
2. Στ = Iθ"

The Attempt at a Solution


First , assuming small motion, all movement of the scale can disregard x components (so the spring stretches only in vertical direction without an impact from the x directional movement). I summed torque around the pin, point O, which would be equal to the mass moment of inertia * theta double dot. so...

mgL2 - kx1 = I0θ" (mass force positive since displacement is positive in the negative y direction)

x1 is the spring movement upwards, however we want to relate that to the x displacement downwards with the mass, which we'll just call x. Now, using like triangles, x1=(L1/L2)x. Then I know x = rθ, differentiate with time twice to get x" = r θ" so then θ" = x"/r, where r = L2. Now we have ...

(I0/L2)x" - k(L1/L2)x= mgL2 ===> (I0)x" + k(L1)x = mgL22

This answer just doesn't feel right. I feel like I may not have correctly applied Newtons 2nd law or maybe didn't correctly set up the kinematics. Any input would be helpful, thanks.
 
Last edited:
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You need not take into account the mass moment of Inertia of mass m, since it is a statics problem
 
Nikstykal said:

The Attempt at a Solution


First , assuming small motion, all movement of the scale can disregard x components (so the spring stretches only in vertical direction without an impact from the x directional movement). I summed torque around the pin, point O, which would be equal to the mass moment of inertia * theta double dot. so...

mgL2 - kx1 = I0θ" (mass force positive since displacement is positive in the negative y direction)

x1 is the spring movement upwards, however we want to relate that to the x displacement downwards with the mass, which we'll just call x. Now, using like triangles, x1=(L1/L2)x. Then I know x = rθ, differentiate with time twice to get x" = r θ" so then θ" = x"/r, where r = L2.
OK up to here.
Now we have ...
(I0/L2)x" - k(L1/L2)x= mgL2 ===> (I0)x" + k(L1)x = mgL22
Check your math, this has at least 2 mistakes in it. One is a sign error.
Also, what is I0? You need to express this in terms yuu were given.

Also BTW not only is this not a static question but, contary to the problem statement iself, the scale swings forever back & forth since no damping mechanism was introduced.
 
Last edited:

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