Momentof inertia and center of mass

AI Thread Summary
The discussion centers on the concepts of moment of inertia and center of mass as presented in Serway's textbook. It highlights the confusion regarding the integration process used to derive these concepts, particularly questioning how the integration leads to seemingly trivial results. The moment of inertia for a rigid body is clarified as not being simply I = mR^2, as this only applies to point masses at a uniform distance from the axis. Additionally, the center of mass calculation is scrutinized for its treatment of mass distribution and the implications of variable distances in the integration process. Overall, the thread emphasizes the need for a deeper understanding of how integration applies to these physical concepts.
The Trice
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momentof inertia and center of mass!

in book of serway : he says moment of inertia of a body is m(r^2).---->(1)
is mass of the body and r is the distance of the body.
but for a rigid body we will divide into particles of very small masses so i = E(Mi * (Ri)^2)
E() is submission function to number if i Th particles.
and he says if we decrease this mi or delta m i to very small amount
so I = lim(mi->0) mi*(ri)^2 which will equal integration( r^2 dm).
SO WHAT DID HE DO IF I SOLVE THE INTEGRATION IT WILL BE I=MR^2 the same as the first equation(1) !?
and also i have the same problem in center of mass
he says if we want to find x coordinate of center of mass so:
x=E(mi*xi)/M and again if we deal with a rigid body of infinite of particles it will be :
x=lim(mi->0) (mi*xi)/M = integration( x dm)/M
so what! if we solve this integration it will be xcm= (x* M)/M
so xcm=x (and u don't have x because x is xcm that u want to find so what did he do with this stupid integration !)
 
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and also why did he make it integration( x dm)/M it seems tome as as mass in changing, but mass is constant for this particles!and even don't we get the constants out of integrtion soit would be xcm=x* integration(dm)/M so it would be 1=1!
and also in I=intefration(r^2 dm) why did he only change mi to dm and not also r^2 i mean he made the r is equal for all mass particles!
 
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The Trice said:
in book of serway : he says moment of inertia of a body is m(r^2).---->(1)
is mass of the body and r is the distance of the body.
I = mr2 is true for a point mass, where r is the distance from the axis of rotation.
but for a rigid body we will divide into particles of very small masses so i = E(Mi * (Ri)^2)
E() is submission function to number if i Th particles.
and he says if we decrease this mi or delta m i to very small amount
so I = lim(mi->0) mi*(ri)^2 which will equal integration( r^2 dm).
SO WHAT DID HE DO IF I SOLVE THE INTEGRATION IT WILL BE I=MR^2 the same as the first equation(1) !?
No. In general it's not true that I = mR2 for a rigid body. It would be true if all the mass is at the same distance from the axis.
and also i have the same problem in center of mass
he says if we want to find x coordinate of center of mass so:
x=E(mi*xi)/M and again if we deal with a rigid body of infinite of particles it will be :
x=lim(mi->0) (mi*xi)/M = integration( x dm)/M
so what! if we solve this integration it will be xcm= (x* M)/M
so xcm=x (and u don't have x because x is xcm that u want to find so what did he do with this stupid integration !)
Careful. Σ(xi*mi) ≠ Σ(xi)*Σ(mi), except in special cases.
 


so r is a function in terms of variable mass right!?
like when i say impulse is integration(f dt) so that means as time chnages the force changes ,soin inertia it means as mass changes!? the r changes!
 

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