Transcript moment of inertia of a disk about a tangent axis.ppt

A uniform thin disk has radius “R” and mass “M”. Find the
moment of inertia of the disk about an axis tangent to the disk.
axis of
rotation
Request: try to solve the problem without
using references.
R
M
Hint: there are smart ways to solve the
problem, and there are ‘conventional’
methods [i.e. through integrating r2 dm].
Suggestion: solve the problem BOTH
ways. Of course, you should get the same
result both ways!

axis of
rotation
First, the smart way:
1- We all know that the moment of inertia of a uniform disk (M,R) about an
axis (z1) through its center (O) perpendicular to the plane of the disk [see
red dot in the figure] is Iz1 = ½ M R2.  one can easily integrate to find this
out, but all physicists probably memorize it by heart!
y1
y
2- You should also know that for any planar body, the moment of inertia
about an axis perpendicular to the plane of the body is equal to the sum of
the moments of inertia about any two orthogonal axes in the plane of the
body where the three axes intersect  this is sometimes called the
perpendicular axis theorem (it applies only for planar bodies): Hence:
R
O
M
x1
Ix1 + Iy1 = Iz1
3- From symmetry in our situation, Ix1= Iy1.
Therefore,
2 Iy1 = ½ M R2
Therefore,
Iy1 = ¼ M R2
4- Noticing that O is at the center of mass use the parallel axis theorem:
:z1
Iy = Iy1 + M D2 = Iy1 + M R2
Therefore,
Iy = 5/4 M R2
Now, the direct
integration method with
the help of Mathematica
y
axis of
rotation
R
M
x
The red strip has an area da = height times width
The height is (2 y) and the width is dx
Now, the direct
integration method
without the help of
Mathematica 
y
axis of
rotation
R
M
x