Animated Solution for Physics - Rotational Motion: Comprehension Passage

When you first look at this problem, it seems like a nightmare of 3D rotations. A disc welded to a stick, tilted at weird angles, spinning around a vertical axis... it feels like you need a supercomputer to track the motion of every particle!

But physics is often about finding the hidden simplicity behind the chaos. The key here is the word welded.

Because the disc is rigidly fixed to the stick, the entire setup—the stick and the disc—acts as a single, unbreakable rigid body. And rigid bodies have a beautiful, almost magical property when it comes to rotation.

Let's strip away the disc and the stick for a moment and look at the pure math. The fundamental theorem of rigid body kinematics states that the velocity of any point $P$ on a rigid body can be expressed as the sum of two parts:

1. The translational velocity of the center of mass, ${\vec{v}}_{cm}$. 2. The rotational velocity of point $P$ relative to the center of mass.

Mathematically, this is written as:

Here, ${\vec{\omega }}_{cm}$ is the angular velocity of the body about an axis passing through its center of mass, and ${\vec{r}}^{\prime }$ is the position vector of $P$ relative to the center of mass.

Now, let's look at our specific problem. We know the entire rigid body is rotating about the fixed Z-axis with a constant angular velocity $\vec{\omega }=\omega \hat{k}$.

Because it's a fixed-axis rotation, the velocity of any point $P$ can also be written simply as:

But wait! The position vector ${\vec{r}}_P$ is just the vector sum of the center of mass position and the relative position: ${\vec{r}}_P={\vec{r}}_{cm}+{\vec{r}}^{\prime }$. Let's substitute this in:

Look closely at the first term, $\vec{\omega }\times {\vec{r}}_{cm}$. That is exactly the definition of the velocity of the center of mass, ${\vec{v}}_{cm}$! So our equation becomes:

Compare this with our master kinematics equation. The conclusion is inescapable:

This is a profound result. It tells us that the angular velocity vector is a free vector. It doesn't matter if you measure the rotation about the fixed Z-axis or about an instantaneous axis passing through the center of mass; the angular velocity vector is exactly the same!

Armed with this revelation, the questions collapse instantly.

For Question 1, we are asked for the angular speed about the instantaneous axis passing through the center of mass. Since ${\vec{\omega }}_{cm}=\vec{\omega }$, the magnitude is simply $\omega$. It doesn't matter if the disc is vertical, horizontal, or tilted at $45^{\circ }$. The internal orientation of the disc is irrelevant because it is welded! Thus, the answer is $\omega$ for both cases.

For Question 2, we need the direction of that instantaneous axis. Again, since ${\vec{\omega }}_{cm}=\vec{\omega }=\omega \hat{k}$, the direction is purely along the Z-axis. Therefore, the instantaneous axis passing through the center of mass is perfectly vertical for both cases.

The complexity was an illusion. The rigid body rotates as one, and its angular velocity is a universal constant for the entire system!