# Calculating the Angular velocity and momentum of a rotating cuboid

1. Aug 21, 2013

### tomphys

1. The problem statement, all variables and given/known data

A concrete block is a uniformly dense cuboid of dimensions 40 x
20 x 10 cm, with mass M. It is constrained to rotate about an axis passing
through two opposite corners and its centre of mass, with constant angular
speed ω.

Calculate the direction of the angular momentum of the block.
Explain why an external torque must be exerted to sustain motion around
this axis. Hint: you will need to find the angular momentum component
along each principal axis.

2. Relevant equations

Moment of inertia perp. to block face = I = 1/12 m(a^2+b^2)
where a and b are the lengths of the sides of the face.

L = Iω

3. The attempt at a solution

I understand that in order to find the angular momentum, I should first find the angular velocity for each principal axis, and then using the above formulas find the angular momentum for each axis, then use these components to find the total angular momentum.

I have calculated the angular momentum for each principal axis as:

L = 10000/3 mω - 10 by 20 face
L = 40000/3 mω - 10 by 40 face
L = 160000/3 mω - 20 by 40 face

I am unsure how to combine these components to find the total angular momentum, though I imagine it is through the use of cos(theta) and cos(phi).

Note: This is my first post, and I am unsure if this is the correct place to post it. I am in my second year of an undergraduate course, so this seemed appropriate.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 21, 2013

### TSny

Hello tomphys. Welcome to PF!

Would you mind showing how you got the following results? I get something different. Maybe I'm overlooking something.

3. Aug 22, 2013

### tomphys

Using the formula shown above: I = 1/12 m(a^2+b^2)
where a and b are 10, 20, 40 for the various faces. Then L= Iω.
I have double checked them and am fairly certain they are numerically correct, but I may be making a method error.

4. Aug 22, 2013

### TSny

Let x, y, z axes correspond to the principle axes. To find the x-component of angular momentum, you should use Lx = Ixωx. So, you need to find the x-component of the angular velocity vector.

5. Aug 22, 2013

### Filip Larsen

Notice that in general you have L = Iω as a vector equation (L and ω are vectors, I is a matrix), meaning that the direction of L may not coincide with the direction of ω.

Also, when the involved values are given with coordinates relative to the body (so that the elements of I can be considered constant), the elements of both L and ω represent the instantaneous values relative to the body, not with respect to inertial space. Since ω for this problem also has constant element in body coordinates ("fixed rotation axis"), it means that L must also be constant in body coordinates, or equivalently, that L over time will move in a small circle around the ω-axis (that is, to the extend that L are not aligned with ω). Now you only need yet another equation (namely the rotational version of Newtons second law) in order to model and calculate the required torque.