What is the moment of inertia of the pendulum

AI Thread Summary
The discussion revolves around calculating the moment of inertia for a pendulum composed of a solid sphere, a bar, and a block, each with a mass of 4 kg. The axis of rotation is positioned halfway between the top of the sphere and its center. The moment of inertia is determined using the formula I = I(sphere) + I(bar) + I(block) and applying the parallel axis theorem twice. The user initially questions whether to plug in the equations directly but later resolves the problem, arriving at a total moment of inertia of 124.6 r^2. The discussion highlights the importance of correctly applying the parallel axis theorem in such calculations.
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Homework Statement


A pendulum constructed of a solid sphere, bar, and block, each with mass m=4 kg
The sphere has radius R, The bar attaches the block and sphere and has length 2R, and the block has a center of mass R away from the end of the bar. The axis of rotation is halfway between the top of the sphere and its center (R/2)
What is the moment of inertia of the pendulum (note: you will need to use the parallel axis theorem twice!)

Homework Equations


I = I(sphere) + I (bar) + I (block)
I(sphere) = (2mr^2)/5
I(bar) = (mL^2)/12
I(block) = mr^2 (treat as point mass bc you know center of gravity)
Parallel axis theorem: Iaxis = Icm + mr^2

The Attempt at a Solution


Should I simply solve for I, plugging in the equations above?

I = [(2mr^2)/5 + m(r/2)^2] + (m(2r)^2)+m(2.5r)^2 + m(4.5r)^2 = 124.6 r^2
 
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Nevermind, figured the problem out.
 
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