Moment of inertia of a regular triangle

AI Thread Summary
The discussion centers on calculating the moment of inertia of a triangle about the OY axis. The original poster considered splitting the triangle into rods and using the Huygens-Steiner theorem but encountered confusion regarding the correct method of splitting. After some back and forth, it was clarified that using strips parallel to the y-axis was the correct approach, while the alternative method led to errors. The poster acknowledged their mistake and expressed gratitude for the guidance received. The conversation highlights the importance of method selection in physics calculations.
Who_w
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Homework Statement
Calculate the moment of inertia of a triangle relatively OY
Relevant Equations
Given the height of the triangle
Please, I need help! I need to calculate the moment of inertia of a triangle relatively OY. I have an idea to split my triangle into rods and use Huygens-Steiner theorem, but after discussed this exercise with my friend, I have a question: which of these splits are right (picture 1 and 2)? Or maybe my idea is wrong?
 

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Each can work, but it would be more natural to use the one where the strip is parallel to the axis.
 
I found a mistake in the first split, thanks a lot! I would be very grateful if you tell me where the error is (attached the file). With another split, I got a different answer. I solved it 3 times in this way, but did not find an error.
 

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Last edited:
Who_w said:
I found a mistake in the first split, thanks a lot! I would be very grateful if you tell me where the error is (attached the file). With another split, I got a different answer. I solved it 3 times in this way, but did not find an error.
(l-x)dx? Both of those distances are parallel to the x axis. It is not the area of a rectangle in the figure.
Also, you seem to be taking a strip parallel to the x axis, which is not what I recommended. Are you just checking it can be done both ways?
 
haruspex said:
(l-x)dx? Both of those distances are parallel to the x axis. It is not the area of a rectangle in the figure.
Also, you seem to be taking a strip parallel to the x axis, which is not what I recommended. Are you just checking it can be done both ways?
Oh! I understood my mistake! Thank you a lot :)
Yes, I'm tried to check both ways. The first way worked (strip parallel to the y axis), but it was really interesting for me why I couldn't to solve it the second way.
Thanks again!
 
Who_w said:
Oh! I understood my mistake! Thank you a lot :)
Yes, I'm tried to check both ways. The first way worked (strip parallel to the y axis), but it was really interesting for me why I couldn't to solve it the second way.
Thanks again!
I commend your inquisitiveness.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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