Moment of Inertia of a Rectangular Picture Frame

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SUMMARY

The discussion focuses on calculating the moment of inertia (MoI) of a rectangular picture frame constructed from four thin wooden pieces, each with a uniform mass per unit length (λ). The key equation used is I = mr², specifically applying the formula for a rod at the edge, I = (1/3)mr². The derivation of the distance R as Lsin(angle) is clarified through the integration of a small element dx, leading to the conclusion that the MoI can be simplified by treating the spars as having an effective length of Lsin(θ).

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kepherax
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Homework Statement
A picture frame consists of 4 thin pieces of wood glued together. Each wooden piece has the same mass per unit length lambda. The dimensions of each piece is given below. Determine the rotational inertia of the frame about the dashed axis.
Relevant Equations
I = mr^2, for rod at edge I = 1/3 mr^2
I total = I1+I2+I3....+In
https://www.physicsforums.com/attachments/250905

I know the answer, but am not certain how they got Lsin(angle) for R?
 
Last edited:
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kepherax said:
Homework Statement: A picture frame consists of 4 thin pieces of wood glued together. Each wooden piece has the same mass per unit length lambda. The dimensions of each piece is given below. Determine the rotational inertia of the frame about the dashed axis.
Homework Equations: I = mr^2, for rod at edge I = 1/3 mr^2
I total = I1+I2+I3...+In

https://www.physicsforums.com/attachments/250905

I know the answer, but am not certain how they got Lsin(angle) for R?
The link does not work for me. Please post an attempt, per forum rules.
 
I don't know why it won't let me edit this, but here is the problem and my attempt.

frame.JPG


20191010_082036.jpg
 
kepherax said:
how they got Lsin(angle) for R?
Consider a small element dx of one of the spars length L, distance x from the end at the axis.
Its distance from the axis is ##x\sin(\theta)##, so its MoI about the axis is ##\lambda dx (x\sin(\theta))^2##. Then integrate, or observe that this is the same as if the spar were the same mass but length ##L\sin(\theta)## and normal to the axis.
 

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