Parallel-Axis Theorem Derivation
How to calculate the rotational inertia of a body around a parallel axis of rotation that is
through a point other than the center of mass.
y
dm
R
r
P
cm
h
a
y b
b
xa
x
Parallel-Axis Theorem Derivation
How to calculate the rotational inertia of a body around a parallel axis of rotation that is
through a point other than the center of mass.
I   r 2 dm
I    x  a    y  b   dm
2
2
I   x 2  2ax  a 2  y 2  2by  b 2  dm






I   x 2  y 2  2ax  2by  a 2  b2 dm


I   x 2  y 2 dm  2a  x dm  2b  y dm   a 2  b 2 dm
Parallel-Axis Theorem Derivation
How to calculate the rotational inertia of a body around a parallel axis of rotation that is
through a point other than the center of mass.




I   x 2  y 2 dm  2a  x dm  2b  y dm   a 2  b 2 dm
But
2a  x dm  2axcm  0
2b  y dm  2bycm
cm  0




I   x 2  y 2 dm   a 2  b 2 dm
I   R 2 dm   h 2 dm
Parallel-Axis Theorem Derivation
How to calculate the rotational inertia of a body around a parallel axis of rotation that is
through a point other than the center of mass.
I   R dm   h dm
2
2
I  I cm   h2dm
But h2 is constant.
I  I cm  h2  dm
I  I cm  Mh
Where
2
Parallel-Axis Theorem
M  total mass of the body
h  distance between the center of mass
and the parallel axis of rotation
Parallel-Axis Theorem Example
Two equal masses of 2.0 kg each are separated by a thin, massless rod 4.0 m long.
What would its rotational inertia be about an axis perpendicular to the rod and through
one end?
Method 2
Parallel-Axis Theorem
2.0 kg
2.0 kg
L
Method 1
I  I cm  Mh2
2
I
2
I   mi ri2
I
I  mr12  mr22
I
i 1
I  mL2
I  32 kgm2
I
1 2
1 

 mL  2m  L 
2
2 
1 2
1 2

 mL  2m  L 
2
4 
1 2 1 2
 mL  mL
2
2
 mL2
I  32 kgm2