The figure shows a wheel rolling without slipping on a horizontal plane with angular velocity 𝜔₁. A rigid bar PQ is pinned to the wheel at P while the end Q slides on the floor. What is the angular velocity 𝜔₂of the bar PQ?

The correct answer is:
$${\omega }_2=0.25{\omega }_1$$

Step-by-Step Explanation:

We can solve this problem using two different elegant methods: the Instantaneous Center of Rotation (I-Center) Method using Kennedy's Theorem and the Vector Velocity Method.

Method 1: Instantaneous Center (I-Center) Method (Kennedy's Theorem)
Let us define the links in the system:
• Link 1: The fixed ground (surface)
• Link 2: The rolling wheel (disc)
• Link 3: The rigid bar PQ

1. Identify the I-centers of the individual links:
• Since the wheel (Link 2) rolls without slipping on the ground (Link 1), the instantaneous center of rotation between Link 1 and Link 2, denoted as $I_{12}$, lies at the point of contact R.
• The pin joint P connects the wheel (Link 2) and the bar (Link 3). Hence, the relative I-center between Link 2 and Link 3 is located at the pin P, denoted as $I_{23}$.
• The end Q of the bar PQ slides along the horizontal floor (Link 1). Therefore, the velocity of Q is purely horizontal, meaning its perpendicular line of motion is vertical. Thus, the I-center of the bar relative to the ground, denoted as $I_{13}$, must lie on the vertical line passing through Q.

2. Apply Kennedy's Three-Center Theorem:
Kennedy's theorem states that if three bodies have relative motion, their three relative I-centers must lie on a straight line. Therefore, the I-centers $I_{12}$ (at R), $I_{23}$ (at P), and $I_{13}$ must be collinear.
This collinear line passes through the point of contact R $\left(I_{12}\right)$ and the pin joint P $\left(I_{23}\right)$, and extends to meet the vertical line through Q at $I_{13}$.

3. Use Geometry and Similar Triangles:
From the given image:
• The radius of the wheel is $3\text{ m}$, so the vertical height of P above R is $3\text{ m}$.
• The horizontal distance from the center O to P is $2\text{ m}$.
• The horizontal distance from the vertical line through P to the point Q is $8\text{ m}$.
Let B be the projection of P on the ground line. The triangle formed by $I_{12}$, B, and P is similar to the triangle formed by $I_{23}$ (P), A (horizontal projection of Q at the level of P), and $I_{13}$.
Using similar triangles:
$$\frac{I_{12}{I}_{23}}{I_{23}{I}_{13}}=\frac{I_{12}B}{PA}$$
Substituting the horizontal lengths:
$$\frac{I_{12}P}{P{I}_{13}}=\frac{2}{8}=0.25$$

4. Calculate Angular Velocity Relationship:
Since P is the common point between the wheel (rotating with ${\omega }_1$ about $I_{12}$) and the bar (rotating with ${\omega }_2$ about $I_{13}$), the velocity of point P can be written as:
$$v_P={\omega }_1\times \left(I_{12}P\right)={\omega }_2\times \left(P{I}_{13}\right)$$
Rearranging the equation:
$${\omega }_2={\omega }_1\times \frac{I_{12}P}{P{I}_{13}}=0.25{\omega }_1$$

Method 2: Vector Velocity Method
Let us place a coordinate system at the center of the wheel O, with unit vectors $\hat{i}$ pointing horizontally to the right and $\hat{j}$ pointing vertically upwards.
• The contact point with the floor is $R=(0,-3)$.
• The pin joint P is located at $P=(2,0)$.
• The velocity of point P on the wheel, relative to the instantaneous center of rotation R, is:
$${\overrightarrow{v}}_P={\overrightarrow{\omega }}_1\times {\overrightarrow{r}}_{P/R}$$
Since the wheel rotates counterclockwise with ${\overrightarrow{\omega }}_1={\omega }_1\hat{k}$ and the position vector is ${\overrightarrow{r}}_{P/R}=2\hat{i}+3\hat{j}$:
$${\overrightarrow{v}}_P=\left({\omega }_1\hat{k}\right)\times (2\hat{i}+3\hat{j})=-3{\omega }_1\hat{i}+2{\omega }_1\hat{j}$$
The magnitude of the velocity of P is:
$$v_P={\omega }_1\sqrt{3^2+2^2}=\sqrt{13}{\omega }_1$$

Now consider the rod PQ rotating with angular velocity ${\omega }_2$ about its own instantaneous center $I_{13}$:
• Since Q slides on the horizontal floor, the I-center $I_{13}$ has the coordinates $(10,y)$ relative to O, or relative to P is $(8,y)$.
• The velocity vector of P must be perpendicular to the position vector ${\overrightarrow{r}}_{I_{13}/P}=8\hat{i}+y\hat{j}$:
$${\overrightarrow{v}}_P·{\overrightarrow{r}}_{I_{13}/P}=0$$
$$(-3{\omega }_1)\left(8\right)+\left(2{\omega }_1\right)y=0\Rightarrow y=12\text{ m}$$
Thus, the distance from P to the I-center of the bar is:
$$d=\sqrt{8^2+{12}^2}=\sqrt{64+144}=\sqrt{208}=4\sqrt{13}\text{ m}$$
Therefore, the angular velocity of the bar ${\omega }_2$ is:
$${\omega }_2=\frac{v_P}{d}=\frac{\sqrt{13}{\omega }_1}{4\sqrt{13}}=0.25{\omega }_1$$