Vibration of Rigid Body with One Spring: Understanding Torque Equations

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SUMMARY

The discussion focuses on the torque equations for a rigid body with one spring, specifically addressing the misunderstanding of how weights affect torque calculations. The correct torque equation is established as M = I*a - k*0.4sin(theta)*0.4cos(theta), indicating that the weights of the bar and cube do not contribute to the torque in this scenario. Instead, the equilibrium condition kδ = mgcosθ is highlighted, showing that the weight cancels out when considering displacements. This clarification is crucial for solving related differential equations accurately.

PREREQUISITES
  • Understanding of torque equations, specifically M = I * α
  • Familiarity with spring mechanics and Hooke's Law
  • Basic knowledge of rigid body dynamics
  • Ability to solve differential equations in physics
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  • Study the application of Hooke's Law in dynamic systems
  • Learn about the derivation and application of torque equations in rigid body dynamics
  • Explore methods for solving differential equations in mechanical systems
  • Investigate the effects of equilibrium conditions on system behavior
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Students and educators in physics, particularly those focusing on mechanics, as well as engineers working with dynamic systems involving springs and rigid bodies.

Alex Santos
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Homework Statement



Rigid body with one spring.

Homework Equations


Torque equations ( M=I * "alpha")

The Attempt at a Solution


So I think I can solve this problem but there is only one thing I do not understand.

When I am writing my torque equation I would think that the weight of the bar and the cube would affect the torque
M = I * a + Mb * g *0.4 cos (theta) + Mw * g * 0.4 cos (theta) - k*0.4sin(theta) *0.4 cos(theta)
but apparently the equation should be
M = I*a -k*0.4sin(theta)*0.4cos(theta)

why does the mg of both the objects not affect the torque, whenever I try to include them the differential equation becomes unsolveable
 

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From the equilibrium position, your forces would be along the lines of kδ=mgcosθ (component of the weight)

So that when the mass is displaced again by some distance x, the spring force becomes k(x+δ). And the weight cancels out.
 
Alright thanks for the answer. Think I understand this now :)
 

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