- #1
The Trice
- 7
- 0
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 !)
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 !)