A compound pendulum consists of a thin rod of length 2 m and a disc of radius 0.

Question

A compound pendulum consists of a thin rod of length 2 m and a disc of radius 0.21 m. The centre of the disc is attached to the end of the rod and the pendulum pivots about the opposite end of the rod. Both the mass of the rod and the mass of the disc are the same, each being 2.4 kg. What is the period of the pendulum?

[g=9.81 ms/2, MI rod about its end 1/3ml^2, MI disc through centre, perpendicular to the plane 1/2mr^2]

I can not figure out how to do this. I just know that I need to use T=2pi?I/(mgl).but I can not sure how to calculate I and l for this question.ps:can anyone tell me what MI rod and MI disc through centre mean?

Explanation / Answer

First you have to find out the center of mass of the system.

center of mass of rod will be at its mid point ..

so

xcm = m1*(0) + m2*x / (m1+m2)

where x is the distance from the center of the rod to teh center of mass

m1 = m2 = 2.4

Now you have to find moment of inertia of hte system about this point .

MI of the rod about center of the rod = Ml^2/12

MI about a point x distance from center = ML^2/12 + Mx^2

MI of disc about a point which is (L/2- x ) from it = mr^2/2 + m(L/2-x)^2

total MI = sum of MIs

I= ML^2/12 + Mx^2 +mr^2/2 + m(L/2-x^2)

m = M = 2.4

T = 2pi*sqrt(I/mgl)

here l = L/2 + x

all values are given...so you can find out

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