IPhO 1968, Теория и експеримент — Задача 1. Problem 1

• 3.25 m/s²

If the cord is stressed the cylinder and the block are moving with the same acceleration $a$. Let $F$ be the tension in the cord, $S$ the frictional force between the cylinder and the inclined plane (Fig. 2). The angular acceleration of the cylinder is $a/r$. The net force causing the acceleration of the block:

and the net force causing the acceleration of the cylinder:

The equation of motion for the rotation of the cylinder:

($I$ is the moment of inertia of the cylinder, $S\cdot r$ is the torque of the frictional force.)
Solving the system of equations we get:

The moment of inertia of a solid cylinder is $I = \dfrac{m_1 r^2}{2}$. Using the given numerical values:

Discussion (See Fig. 3.)

The condition for the system to start moving is $a > 0$. Inserting $a = 0$ into (1) we obtain the limit for angle $\alpha_1$:

For the cylinder separately $\alpha_1 = 0$, and for the block separately $\alpha_1 = \tan^{-1}\mu = 11.31^{\circ}$.

If the cord is not stretched the bodies move separately. We obtain the limit by inserting $F = 0$ into (3):

The condition for the cylinder to slip is that the value of $S$ (calculated from (2) taking the same coefficient of friction) exceeds the value of $\mu m_1 g\cos\alpha$. This gives the same value for $\alpha_3$ as we had for $\alpha_2$. The acceleration of the centers of the cylinder and the block is the same: $g(\sin\alpha - \mu\cos\alpha)$, the frictional force at the bottom of the cylinder is $\mu m_1 g\cos\alpha$, the peripheral acceleration of the cylinder is $\mu\cdot\dfrac{m_1 r^2}{I}\cdot g\cos\alpha$.