Solving Physical Pendulum Homework Equation

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SUMMARY

The discussion focuses on solving the physical pendulum equation, specifically deriving the period of oscillation for a pendulum using the relationship between torque and angular acceleration. The user integrates the equation of motion, starting with the second derivative of angular displacement, \(\partial^2\vartheta/\partial t^2 = \alpha = \frac{mgdT\vartheta}{I}\). The final result shows that the period \(T\) is given by \(T = 2\pi\sqrt{\frac{I}{mgl}}\), where \(I\) is the moment of inertia, \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(l\) is the distance from the center of mass to the axis of rotation.

PREREQUISITES
  • Understanding of angular motion and torque
  • Familiarity with the concepts of moment of inertia
  • Knowledge of simple harmonic motion (SHM)
  • Basic calculus for integration
NEXT STEPS
  • Study the derivation of the moment of inertia for various shapes
  • Learn about the principles of simple harmonic motion in rotational systems
  • Explore the effects of damping on pendulum motion
  • Investigate the relationship between torque and angular acceleration in different contexts
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of pendulum motion and its applications in mechanics.

shyta
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Homework Statement


http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html#c2

i'm trying to prove
pendp4.gif
to be
pendp6.gif


Homework Equations



Letting d = Lcm
now we already know [tex]\partial^2\vartheta[/tex]/[tex]\partial t^2[/tex] = [tex]\alpha[/tex] = mgdT[tex]\vartheta[/tex] / I


I tried integrating the whole equation wrt dt

so [tex]\partial\vartheta[/tex]/[tex]\partial t[/tex] = [tex]\int mgd\vartheta/ I dt[/tex] (with limits 0->T) = mgd[tex]\vartheta[/tex]/ I

I only need help with this step. How do I deal with the [tex]\vartheta[/tex]?
 
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I have a better method-

Let l be the distance of COM from the axis of rotation. For the equilibrium, the COM should be vertically below the axis.
Now let us rotate the body through an angle θ. The torque of forces acting on the body about the axis is τ= -mglsinθ (The only force is due to its weight).
τ = Iα and for small angles sinθ is app. θ
so α = -mglθ/I
Comparing it with the equation of angular SHM,
α = -ω²θ,
T = 2π/ω = 2π(I/mgl)^0.5
 

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