Mass Moments of Inertia
The answer is meant to be 6069. I can't see where I'm going wrong?
EDIT: I've just realised that I should divide by 3 instead of 12 when finding the individual moments of inertia. But that gives me an answer of 6277 (which is ridiculous close to the answer but yet so far...)
EDIT: I've just realised that I should divide by 3 instead of 12 when finding the individual moments of inertia. But that gives me an answer of 6277 (which is ridiculous close to the answer but yet so far...)
(edited 10 years ago)
Here's the updated working out (ignore the bottom attachment)
(edited 10 years ago)
bumpy
Original post by Serendreamers
bumpy
Some numbers for you to check against.
About their centres of mass the rectangles have moments of inertia:
I= 459 and I=640.
The squares of the distances of these individual centres of mass from the overall centre of mass:
0.000576m^2 and 0.005881m^2
Final answer in brief:
2576.32 x 2 + 916.48 = 6068.8 kg m^2
Edit: By the way, the question is ridiculous. What would these be made of if such a small lamina has a mass of hundreds of tonnes?
(edited 10 years ago)
Original post by BabyMaths
...
Thanks.
Apparently, Osmium is the densest element with 22,570 kg/m^3 and this lamina has 100 kg/mm^3 (assume it has a small thickness)...
Original post by Serendreamers
Thanks.
Apparently, Osmium is the densest element with 22,570 kg/m^3 and this lamina has 100 kg/mm^3 (assume it has a small thickness)...
Apparently, Osmium is the densest element with 22,570 kg/m^3 and this lamina has 100 kg/mm^3 (assume it has a small thickness)...
Osmium doesn't come close to this. 100 kg/mm^3 = 10^11 kg/m^3.
Still, the question can be answered. Did you get there in the end?
Original post by BabyMaths
Osmium doesn't come close to this. 100 kg/mm^3 = 10^11 kg/m^3.
Still, the question can be answered. Did you get there in the end?
Still, the question can be answered. Did you get there in the end?
Yes, I did. Thanks very much for the help.
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