Skip to content

Moments of Inertia

January 31, 2008

Moment of Inertia is defined as the product of mass and the square of the distance from the axis of rotation.

While less emphasized in STPM Physics, moment of inertia is studied in depth in A-Level Further Mechanics. (Derivation for the moment of inertia using integration is not required at this level.)

Some basic knowledge that you will need to know:

a) Equations for the moment of inertia for loop, solid cylinder, uniform sphere and rod.

(You can find the equations in any STPM Physics reference book)

b) Parallel Axis Theorem

c) Perpendicular Axis Theorem

Parallel and Perpendicular Axis Theorem

We know that the moment of inertia of a disc rotating around axis through the center of disc is given by 1/2 MR². But what if the disc is rotating around an axis perpendicular to the rim of the disc?

The parallel axis theorem states that if I’ is the moment of inertia of a body of total mass M about any axis, and I is the moment of inertia about an axis passing through the center of the mass and parallel to the first axis but a distance h away , then

I’ = I + Mh²

In the above case, I of the disc about axis Z is

I’ = 1/2 MR² + MR² (since h = R in this case)

= 3/2 MR²

The second theorem, Perpendicular Axis Theorem states that the sum of the moments of inertia of a plane body about any two perpendicular axes in the plane of the body is equals to the moment of inertia about an axis through their point of intersection perpendicular to the plane of the object.

Sounds confusing? Refer to the diagram above. The moment of inertia of the disc about the green axis is equals to the sum of the moments of inertia about the blue and red axis.

By symmetry, we can deduce that the moment of inertia about the blue axis and the red axis are equal. (Since the disc is uniform)

Hence, the moment of inertia of the disc about the red/blue axis, I is given by

I + I = 1/2 MR²

I = 1/4 MR²

Note : Proving of Perpendicular Axis Theorem and parallel Axis Theorem is not required at this stage.

 

 

 

 

 

 

 

 

Advertisements
No comments yet

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: