1. Locating the Instantaneous Axis of Rotation (IAR)

Let the square lamina be $ABCD$ with side length $a$. We are given that the velocity vectors $\vec{v}_A$ and $\vec{v}_B$ are perpendicular to each other.

The Instantaneous Axis of Rotation (IAR), denoted by point $I$, lies at the intersection of the lines perpendicular to the velocity vectors of the particles. Since $\vec{v}_A \perp \vec{v}_B$, the line $IA$ (perpendicular to $\vec{v}_A$) must be perpendicular to the line $IB$ (perpendicular to $\vec{v}_B$).

Therefore, $\angle AIB = 90^\circ$. This implies that the locus of point $I$ is a semicircle with $AB$ as the diameter.

2. Mathematical Formulation

Let the side of the square be $a$. Let’s set up a coordinate system with the origin at $D(0,0)$, such that $C=(a,0)$, $B=(a,a)$, and $A=(0,a)$.

Let the coordinates of the IAR $I$ be $(x, y)$.

• Since $I$ lies on the semicircle on diameter $AB$, its distance from the midpoint of $AB$, $P(a/2, a)$, is $a/2$. $$IP = \frac{a}{2}$$

3. Analyzing Velocity of Corner C

The velocity $\vec{v}$ of corner $C$ is perpendicular to the position vector $\vec{IC}$. Let $\alpha$ be the angle $\vec{v}$ makes with the vector $\vec{CD}$ (which lies along the negative x-axis).

Given: $\tan \alpha = 0.5 = 1/2$.

From geometry, if the velocity makes an angle $\alpha$ with the horizontal, the radius vector $IC$ (which is perpendicular to velocity) makes an angle $\alpha$ with the vertical. Alternatively, using slopes: $$ \text{Slope of } IC = \frac{y – 0}{x – a} = \frac{y}{x-a} $$ $$ \tan \alpha = \left| \frac{x-a}{y} \right| = \frac{1}{2} \implies |x-a| = \frac{y}{2} $$

Using the distance formula for $IC$: $$ r_C^2 = (x-a)^2 + y^2 = \left(\frac{y}{2}\right)^2 + y^2 = \frac{5}{4}y^2 $$ $$ r_C = \frac{\sqrt{5}}{2}y $$

The speed of C is given by $v = \omega r_C$: $$ \omega = \frac{v}{r_C} = \frac{2v}{\sqrt{5}y} $$

4. Calculating Velocity of Midpoint P

The midpoint $P$ of side $AB$ is the center of the semicircle containing $I$. Therefore, the distance $IP$ is exactly the radius of that semicircle: $$ r_P = IP = \frac{a}{2} $$ The velocity of P is: $$ v_P = \omega r_P = \left( \frac{2v}{\sqrt{5}y} \right) \left( \frac{a}{2} \right) = \frac{va}{\sqrt{5}y} \quad \dots(1) $$

5. Solving for y

Since $I(x,y)$ lies on the circle centered at $P(a/2, a)$ with radius $a/2$: $$ \left(x – \frac{a}{2}\right)^2 + (y – a)^2 = \left(\frac{a}{2}\right)^2 $$ Substituting $|x-a| = y/2$: We have two cases for $x$: $x = a – y/2$ or $x = a + y/2$. Let’s consider $x = a – y/2$ (IAR inside the strip): $$ \left(a – \frac{y}{2} – \frac{a}{2}\right)^2 + (y-a)^2 = \frac{a^2}{4} $$ $$ \left(\frac{a-y}{2}\right)^2 + (y-a)^2 = \frac{a^2}{4} $$ $$ \frac{1}{4}(y-a)^2 + (y-a)^2 = \frac{a^2}{4} $$ $$ \frac{5}{4}(y-a)^2 = \frac{a^2}{4} \implies (y-a)^2 = \frac{a^2}{5} $$ $$ y – a = \pm \frac{a}{\sqrt{5}} \implies y = a \left( 1 \pm \frac{1}{\sqrt{5}} \right) $$

6. Final Calculation

Substitute $y$ back into equation (1): $$ v_P = \frac{va}{\sqrt{5} \cdot a(1 \pm \frac{1}{\sqrt{5}})} = \frac{v}{\sqrt{5} \pm 1} $$ Rationalizing both possible values:

Case 1 (+): $$ \frac{v}{\sqrt{5} + 1} \times \frac{\sqrt{5}-1}{\sqrt{5}-1} = v \frac{\sqrt{5}-1}{4} $$

Case 2 (-): $$ \frac{v}{\sqrt{5} – 1} \times \frac{\sqrt{5}+1}{\sqrt{5}+1} = v \frac{\sqrt{5}+1}{4} $$