4. Condition: Two weightless cylindrical rollers lie on a horizontal surface parallel to each other and have radii $R=1$ m and $r=75$ cm. A heavy flat slab of mass $m=75$ kg is placed on them such that it is inclined to the horizontal by $\alpha=\arccos (0.98)$. 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: 1 m/s; $\arcsin 0.1;$

Solution: Let's transition to a coordinate system where the center of the (left) wheel is at rest. In this system, all points on the rim move with the same speed $v$. That is, the surface moves horizontally to the left with this speed. And the point of contact with the plate 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 plate moves at an angle $\alpha$ to the right. That is, the plate moves along itself (but only in this coordinate system!)

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

Next, from the law of conservation of energy: if the plate 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: $a = \sqrt{\frac{1 - \cos \alpha}{2}}$

## Criteria: