Moment of Inertia of a Rectangular Picture Frame

AI Thread Summary
The discussion focuses on calculating the moment of inertia of a rectangular picture frame made of four thin wooden pieces. The key point is understanding how to derive the distance R as Lsin(angle) for the rotational inertia calculation. A small element of the frame is analyzed, where its distance from the axis is expressed as xsin(theta), leading to the moment of inertia formula involving integration. The approach simplifies by recognizing that the frame can be treated as having an effective length of Lsin(theta) perpendicular to the axis. This method clarifies the relationship between the frame's geometry and its rotational inertia.
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?
 
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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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