27-10-09 Prove Perpendicular and Parallel axis theorem.

Prove Perpendicular and Parallel axis theorem.

Hint: for perp axis thm: Think of why Ix+Iy=Iz is valid only on one plane :P

I was teaching recently and suddenly wondered why i never asked this as a QOD here :P

7 Answers

1
Philip Calvert ·

perpendicular axis theorem is for laminar bodies.. and is more or less a direct result of the pythagoras theorem.

parallel axis proof is also kind of simple....feeling very lazy now.
maybe someone who doesnt know this should try first.

19
Debotosh.. ·

PERPENDICULAR AXIS THEOREM :
x2 + y2 = z2
=> mx2 + my2 = mz2
=> Σ mx2 + Σmy2 = Σmz2
=> IX + IY = IZ

THIS THEOREM IS NOT APPLICABLE FOR A BODY WHICH IS SYMMETRIC ABOUT ALL THREE AXES !
(similarly for parallel axis theorem !)

62
Lokesh Verma ·

is Ix = Σ m x2 ??

Think of a rod placed along the Y axis...

x = 0 there.. then the moment of inertia of the rod about the X axis should have been zero!

1
Philip Calvert ·

hmmm I agree that the notation is a bit confusing perhaps

however;
we define the second moment of inertia of a body about an axis O
such that

I_0=\sum{m_ir^2}

where r is the _|_ distance of the ith particle from the axis O.

so particularly for laminar bodies it is clear that
I_x=\sum{m_iy^2} \\and \ I_y=\sum{m_ix^2}
where x and y are the 'x' and 'y' coordinates respectively of the ith particle in the respective equations
now the _|_ distance from the z axis of any point (x,y) on the x, y plane is √x2+y2

so I_z=\sum{m_i(x^2+y^2)}=I_y+I_x

62
Lokesh Verma ·

That's COrrect philip [1]

19
Debotosh.. ·

sorry for the trouble ! [2] [2] [2]

1
Philip Calvert ·

Anyone trying parallel axis theorem.... ?

Hint : Start from the definition I gave in #5

the rest is algebra and common sense

Your Answer

Close [X]