In the configuration of the planar four-bar mechanism at a certain instant as shown in the figure, the angular velocity of the 2 cm long link is 𝜔₂ = 5 rad/s. Given the dimensions as shown, the magnitude of the angular velocity 𝜔₄ of the 4 cm long link is given by _____ rad/s (round off to 2 decimal places).

The correct answer is 1.25.

Step-by-step Explanation:

1. Identification of Links and Joint Centers:
Let us label the links of the planar four-bar mechanism shown in the figure:
- Link 1: Fixed ground link connecting the two pivots.
- Link 2: Input link of length 2 cm, rotating at angular velocity ${\omega }_2=5rad/s$.
- Link 3: Coupler link of length 5 cm.
- Link 4: Output link of length 4 cm, rotating at angular velocity ${\omega }_4$.
The pin joints (instantaneous centers of rotation) are:
- $I_{12}$: Pivot connecting fixed Link 1 and rotating Link 2.
- $I_{23}$: Joint connecting Link 2 and coupler Link 3.
- $I_{34}$: Joint connecting coupler Link 3 and output Link 4.
- $I_{14}$: Pivot connecting output Link 4 and fixed Link 1.

2. Instantaneous Center Analysis using Kennedy's Theorem:
According to the Aronhold-Kennedy Theorem of three centers, the three instantaneous centers $I_{12}$, $I_{14}$, and $I_{24}$ must lie on a single straight line. Thus, $I_{24}$ lies on the line passing through $I_{12}$ and $I_{14}$ (the horizontal ground line).
Similarly, the three instantaneous centers $I_{23}$, $I_{34}$, and $I_{24}$ must also lie on a single straight line. Thus, $I_{24}$ lies on the line passing through $I_{23}$ and $I_{34}$.

At the instant shown in the figure, Link 2 (line segment $I_{12}{I}_{23}$) is horizontal and collinear with the ground link line ($I_{12}{I}_{14}$). This means that the point $I_{23}$ lies directly on the horizontal line connecting $I_{12}$ and $I_{14}$.
Since $I_{24}$ must be the intersection of the line $I_{12}{I}_{14}$ and the line $I_{23}{I}_{34}$, and these two lines intersect exactly at $I_{23}$, the instantaneous center $I_{24}$ coincides with $I_{23}$:
$I_{24}=I_{23}$

3. Determining Distances from $I_{24}$:
- The distance from $I_{24}$ to $I_{12}$ is:
$\left(I_{24}{I}_{12}\right)=\left(I_{23}{I}_{12}\right)=2cm$
- The distance between the ground pivots $I_{12}$ and $I_{14}$ is given as 10 cm.
- The distance from $I_{24}$ to $I_{14}$ is therefore:
$\left(I_{24}{I}_{14}\right)=\left(I_{23}{I}_{14}\right)=10cm-2cm=8cm$

4. Calculating Angular Velocity:
Using the angular velocity ratio relation between Link 2 and Link 4 via their common instantaneous center $I_{24}$:
${\omega }_4\times \left(I_{24}{I}_{14}\right)={\omega }_2\times \left(I_{24}{I}_{12}\right)$

Substitute the given values (${\omega }_2=5rad/s$, $\left(I_{24}{I}_{12}\right)=2cm$, and $\left(I_{24}{I}_{14}\right)=8cm$):
${\omega }_4\times 8=5\times 2$
${\omega }_4\times 8=10$
${\omega }_4=\frac{10}{8}=1.25rad/s$

Thus, the magnitude of the angular velocity ${\omega }_4$ of the 4 cm link is 1.25 rad/s.