4. Condition: Two weightless cylindrical rollers lie on a horizontal surface parallel to each other and have radii $R=1$ m and $r=40$ cm. A heavy flat slab of mass $m=150$ kg is placed on them such that it is inclined to the horizontal by $\alpha=\arccos (0.68)$. Find the magnitude and direction of the slab's acceleration before it touches the ground, assuming no slipping. With what acceleration do the rollers move? The acceleration due to gravity is $10 \mathrm{~m} / \mathrm{s}^{2}$.
Answer: $4 \mathrm{m} / \mathrm{c} ; \arcsin 0.4$

Solution: Let's move to a coordinate system where the center (of the left wheel) is at rest. In this system, all points on the rim move at the same speed $v$. That is, the surface moves horizontally to the left at this speed. And the point of contact with the slab also has this speed, but it is directed at an angle $\alpha$ to the horizontal to the right.

The same speed $v$ is also present at the point of contact of the second wheel with the surface. In this coordinate system, the center of the wheel is also at rest, and the point of contact with the slab moves at an angle $\alpha$ to the right. That is, the slab moves along itself (but only in this coordinate system!)

In the absolute coordinate system, associated with the surface, we must add the horizontal speed $v$ to all velocities. Therefore, the slab moves translationally to the right at an angle $\alpha / 2$ with a speed of $2 v \cos (\alpha / 2)$. The motion is translational and linear (the angle does not depend on the speed).

Next, from the law of conservation of energy: if the slab has descended by $h$ - from the law of conservation of energy, it will acquire a speed of $\sqrt{2 g h}$. Since its displacement over this time is $h / \sin (\alpha / 2)$, from the corresponding kinematic formula ($2 a s = v^2$), its acceleration is $a = g \sin (\alpha / 2)$.

Since the arccosine of $\alpha$ is given in the problem, the answer can be written as: $a = \sqrt{\frac{1 - \cos \alpha}{2}}$

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