Pendulum involving Spring and Rotating plank

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SUMMARY

The discussion focuses on calculating the period of small oscillations for a uniform rod of mass 0.24 kg, which is free to rotate about a vertical axis while connected to a spring with a force constant of 240 N/m. The relevant equations include torque, angular acceleration, and the formula for the period of oscillation, given by Period = 2π√(I/k). The solution involves relating the torque caused by the spring to the moment of inertia of the rod, leading to the equation a = 3(k/m) sinθ, with a noted correction regarding the sign in the final equation.

PREREQUISITES
  • Understanding of rotational dynamics and torque
  • Familiarity with the concept of moment of inertia
  • Knowledge of simple harmonic motion and oscillation period calculations
  • Basic trigonometric approximations for small angles
NEXT STEPS
  • Study the derivation of the moment of inertia for various shapes, particularly rods
  • Learn about the principles of torque and angular acceleration in rotational systems
  • Explore the approximation methods for small angle oscillations in physics
  • Investigate the effects of spring constants on oscillation periods in different mechanical systems
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of rotational motion and oscillatory systems involving springs.

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Homework Statement


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In the overhead view of the figure, a long uniform rod of mass m = 0.24 kg is free to rotate in a horizontal plane about a vertical axis through its center. A spring with force constant 240 N/m is connected horizontally between one end of the rod and a fixed wall. When the rod is in equilibrium, it is parallel to the wall. What is the period of the small oscillations that result when the rod is rotated slightly and released?

Homework Equations



Torque = rF
Torque = I*a(angular accel)
Period = 2pi sqrt(I/k)
w(angular vel.) = sqrt(k/m)

The Attempt at a Solution



So I started by trying to related the torque caused by the spring (Where L is the length of the rod):

T=r F = (L/2) (-kx)
x (is the extension of the spring) = (L/2) sinθ
T= (L/2) (-k)((L/2) sinθ) = (L^2/4) (-k) sinθ
T= Ia = ((1/12) mL^2) (a)

(L^2/4) (-kx) sinθ = ((1/12) mL^2) (a)
a = 3(k/m) sinθ
 
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Hello.

Your work looks good. I think you dropped a negative sign in the last equation. Since the oscillations are small, you can make the usual approximation for sinθ.
 

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