Moment of inertia of a right triangle

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SUMMARY

The discussion revolves around calculating the moment of inertia of a right triangle with respect to the y-axis, given a mass of 1 kg. The user initially calculated the moment of inertia using the formula \( I = mr^2 \) with \( r \) as the y-coordinate of the center of mass, which they determined to be \( \frac{5}{3} \). However, this approach was questioned, indicating a misunderstanding of how to apply the moment of inertia formula correctly for the specified axis.

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munther
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in my homework the question wants from me to find the moment of inertia of
right triangle with the y-axis

as shown in the attatchment

note: the mass of the plate is 1 kg

i found the y coordinate of the center of mass of the plate it is 5/3
and i supposed it is the r

so the inertia will be mr^2
=1*(5/3)^2
=25/9
=2.77777 kg m^2

is that right??
 

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If you're calculating the moment of inertia with respect to the y-axis, then it doesn't look right.
 
Thank you >
but Why??

can you give me a hint??
 
munther said:
i found the y coordinate of the center of mass of the plate it is 5/3
and i supposed it is the r

I think it would be better for you to show us how you found that. Also, why do you suppose it is "the r"?
 
e(ho0n3 said:
I think it would be better for you to show us how you found that. Also, why do you suppose it is "the r"?


i found it from the first part of the question
 
munther said:
i found it from the first part of the question

Now that you found it, please post it.
 

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