Solving Harmonic Pendulum Homework

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The discussion focuses on solving a homework problem involving a pendulum made of a solid sphere and a rod. To find the period of small oscillations, the relationship between torque, moment of inertia, and angular acceleration is emphasized. The torque is calculated based on the gravitational forces acting on both the sphere and the rod, while the moment of inertia is derived using the parallel axis theorem. The differential equation for the system is established, and simplifications are suggested to align it with the standard form of simple harmonic motion. Ultimately, the goal is to identify the terms that determine the oscillation period.
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Homework Statement




A pendulum consists of a solid, uniform sphere of mass M and radius
R attached to one end of a thin, uniform rod of mass m and length L.
The pendulum swings freely about the other end of the rod. Find the
period of small oscillations of this pendulum.


Homework Equations





The Attempt at a Solution



T = -mhgsinQ = Ia = Id2Q/dt2

d2Q/dt2 +mghsinQ/I = 0

does sinQ = Q for small approx

should i also be able to find the actual I
 

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joemama69 said:

Homework Statement




A pendulum consists of a solid, uniform sphere of mass M and radius
R attached to one end of a thin, uniform rod of mass m and length L.
The pendulum swings freely about the other end of the rod. Find the
period of small oscillations of this pendulum.

It almost seems from the way the problem is worded that you are expected to use

\tau = I \alpha

where tau is the torque, I is the moment of inertia, and alpha is the angular acceleration. Once you have set up the differential equation that goes with this, you can pick off what you need to answer the question.

Assuming that is the case, the torque is the force (due to gravity) times the distance from the center of mass of the ball to the pivot point, plus the force acting on the center of mass of the rod times the distance between the center of mass of the rod to the pivot point:

Mg sin(\theta) (R+L) + mg sin(\theta) (\frac{L}{2})


The moment of inertia for the system is the sum of the moment of inertia of the ball about the distant pivot point (use the parallel axis theorem) plus the moment of inertia of the rod about the end point (pivot point). I'll let you figure that out. (One of those two moments of inertia you can look up.)

Then your differential equation is:

(I_{total}) \frac{d^2 \theta} {d^2 t} = \tau


One of your tasks will be to simplify the terms for I total and the torque into something neater. Since it's a small angle oscillation, replace the sines with just the angle. Simplify the way the equation looks and try to make it match the equation for simple harmonic motion. Check and see what terms in the D.E. for SHM determine the period, and figure out your answer.
 
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