Can anyone calculate the moment of inertia of a cube about an axis passing through its centre of mass?
I don't know the answer but got it as ML2/12. So, please calculate it and tell me the right answer.
Hi Sooraj! The right answer is ML2/6. I don't know how you solved it, but this is what I did:
Take an axis perpendicular to any plane of the cube and passing through the COM. Choose the COM as the origin. Take an element square of thickness dx from a distance 'x' from the origin. MI (dI) of the element square is ML2 /6.
Mass per unit volume is M/L3. Now integrate dI from -L/2 to L/2. You'll get ML2/6.
Hope it's clear!
I think u made a mistake here.... MI of square element will be of the form ML2/12 !!!
To Ankul - Hey, I think you forgot that the square element is rotating about an axis at a distance 'x' from its own axis, and not rotating about its own axis. Therefore, we have to use parallel axes theorem, right?
Vibhavari Dasagi's reply is correct.
However if you assume the axis to be a diagonal of any plane of the cube ( a square) which passes through the COM then your answer is right. [USING PERPENDICULAR AXIS THEOREM]
The answer is supposed to be ML2/6 for ANY axis passing through COM. I don't know the proof. If you get it, please tell me.
By the way, Asif, how can you use perpendicular axis theorem here? Please give the solution.
I got it! Take an element square of mass dM at a dist. x from origin (COM). Let the axis of rotation be a diagonal of a square passing through origin. Let the MI of element square be dI.
dI = dM.L2/12 + dM.x2 (using parallel axes theorem)
Then integrate from -L/2 to L/2. You will get ML2/6.
Therefore, for ANY axis through COM, MI is ML2/6.
Er, I am sorry. U Cant apply perpendicular axis theorem in this case for the whole cube.
However consider an elemental plane of thickness dx. Then applying perpendicular axis theorem we get I = MLdx/6 for a diagonal axis.
Then apply parallel axis theorem integrating from +L/2 to -L/2 & u should get the MI to be (ML2)/3.
i AM NOT VERY SURE WHEther i can do this but it seems logical. If u think that my rational is wrong plz point out the mistake.
I think what Vibhavari replied is correct, it is coming ML2/6 about axis perp. to and passing thro' COM but what Asif has replied also seems to be correct i.e. about the diagonal its coming ML2/12 using perp. axis theorem.I think his logic is allrite. But then finally its ML2/6 as we wanted it about axis perp. to plane of cube and passing thro' COM. If anybody's sure of the ans. about a diagonal then please reply. Anywayz, thanx guys 4UR replies.
To Asif - I didn't understand how u got MLdx/6 using perpendicular axis theorem, nor did I understand how u applied parallel axis theorem, integrated and got ML2/3. Do u mind giving the steps? Thanks.
By the way, can anyone tell me where I went wrong in my solution for a diagonal axis? Thanks a lot!!!!!!!!
Suppose we take our origin at COM and consider an element on the surface of the cube{I have assumed the cube to hollow since it is not mentioned whether the cube is hollow or solid} whose coordinates are [xi,yi,zi]. THEN, I x = summation [ m i ( y2 i + z2 i )]
Similarly we can write Iy and Iz . Since mass is distri buted uniformly along every axes, thus I = I x + I y + I z / 3 . You will get answer as 2ML2/3. Why are answers are coming different?
I think I was wrong in my earlier reply because xi 2 + y2i + z2i shud be equal to L/2 and then ans. will be ML2/6.