Problem Statement: A disc rotates on a horizontal floor divided into two sections with different coefficients of friction $\mu_1 = 0.78$ ($x>0$) and $\mu_2 = 0.67$ ($x<0$). We need to find the initial acceleration of the center of mass.

Consider a small element of mass $dm$ at $(r, \theta)$. The velocity $\vec{v}$ is tangential. The kinetic friction force $d\vec{f}$ acts opposite to velocity (tangential, clockwise if $\omega$ is CCW).

The direction of friction is $-\hat{\theta} = \sin\theta \hat{i} – \cos\theta \hat{j}$.

Magnitude: $df = \mu g dm$.
y-component: $df_y = -\mu g dm \cos\theta$.

X-components: The x-component of friction depends on $\sin\theta$. For every point $(x, y)$, there is a symmetric point $(x, -y)$ where the x-component cancels out. Thus, $F_x = 0$.

Y-components: The y-component depends on $-\cos\theta$.
For the right half ($\mu_1$), $\cos\theta > 0$, so force is downwards ($-\hat{j}$).
For the left half ($\mu_2$), $\cos\theta < 0$, so force is upwards ($+\hat{j}$).

Let mass density $\sigma = \frac{M}{\pi R^2}$. Mass element $dm = \sigma r dr d\theta$.

Force from Region 1 (Right, $-\pi/2 \to \pi/2$):

$$ F_{y1} = \int_{-\pi/2}^{\pi/2} -\mu_1 g (\sigma r dr d\theta) \cos\theta $$ $$ F_{y1} = -\mu_1 g \sigma \left( \int_0^R r dr \right) \left( \int_{-\pi/2}^{\pi/2} \cos\theta d\theta \right) $$ $$ F_{y1} = -\mu_1 g \left(\frac{M}{\pi R^2}\right) \left(\frac{R^2}{2}\right) [2] = -\frac{\mu_1 Mg}{\pi} $$

Force from Region 2 (Left, $\pi/2 \to 3\pi/2$):

$$ F_{y2} = -\frac{\mu_2 Mg}{\pi} (\text{result of integral is -2}) = +\frac{\mu_2 Mg}{\pi} $$

Net Force $F = F_{y1} + F_{y2} = \frac{Mg}{\pi}(\mu_2 – \mu_1)$.

Newton’s Law: $F = Ma$. Thus $a = \frac{g}{\pi}(\mu_2 – \mu_1)$.

Substituting values ($g=10, \pi=22/7, \mu_1=0.78, \mu_2=0.67$):

$$ a = \frac{10}{22/7} |0.67 – 0.78| = \frac{70}{22} (0.11) $$ $$ a = \frac{70}{22} \times \frac{11}{100} = \frac{70}{200} = 0.35 \, \text{m/s}^2 $$