Kinematics and communications

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Kinematics often considers the movement of absolutely rigid bodies, that is, bodies whose distances between any two points remain constant. At the same time, there are methods that significantly forgive the solution of kinematic problems. We will now get acquainted with one of them.

Let the bodies touch each other when moving, and there is no sliding between them. then the velocities of both bodies at the point of contact completely coincide (Fig. 1). If there is slippage between the bodies, then only the projections of the velocities on the peredicular to the tangent at the point of contact coincide. In this case, it is enough that the tangent exists for at least one of the sliding surfaces (Fig. 2).

Let’s look at a few examples.

1. The rod $OA$ rotates clockwise

arrow with angular velocity $\omega$, driving the brick $ABCD$ from the side $a$ (Fig. 3). Find the dependence of the speed of the brick $v$ on the angle $\alpha$.

Solved and e. The rod and the brick touch at the point $A$. Therefore, the velocities of the brick and the rod at this point in the direction $MN$ ($MN \perp \bot OA$) coincide. Thus,

$$v \cos(90^\circ - \alpha) = \omega \cdot OA,$$

or

$$v = \frac{\omega}{\sin \alpha} \cdot \frac{a}{\mathrm{tg} \alpha} = \frac{\omega a \cos \alpha}{\sin^2 \alpha}.$$

2. The light source $S$ is located at a distance $l$ from the screen $MN$ (Fig. 4). At the initial moment of time, a flat object of height $h$ begins to move uniformly with a speed $v$ from the source to the screen. Find the dependence of the speed of movement of the edge of the shadow on the screen on time.

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Solution. In this problem, the role of the rod is played by the ray $SB$. At point $A$ the beam “touches” the limit, and at point $B$ - with the screen, forming the boundaries of the shadow. Let’s create two equations connecting the projections of velocities at points $A$ and $B$: $\omega \cdot SA = v \cos(\pi - \alpha)$ is the projection onto $LK$, $\omega \cdot SB = u \cos \alpha$ is the projection onto $MN$. Here $\omega$ is the angular velocity of the beam rotation. Dividing the second equality by the first and taking into account that

$$\frac{SB}{SA} = \frac{l}{SL} = \frac{l}{vt} \quad \text{и} \quad tg\alpha = \frac{h}{SC} = \frac{h}{vt},$$

we get

$$u = \frac{lh}{vt^2}.$$

Let now the rod $AB$ of a given length $l$ (Fig. 5) move arbitrarily. The velocities $\vec{v}_A$ and $\vec{v}_B$ of its ends can be different, but since the length of the rod does not change, the projections of these velocities $v''_A$ and $v''_B$ must be equal: $v''_A = v''_B$. The velocity projections $v'_A$ and $v'_B$ define the circular motion of the rod with an angular velocity $\frac{v'_B - v'_A}{l}$ (check this is the most accurate).

Let’s solve two problems.

3. The rod $AB$ rests its ends on the sides of the obtuse angle $\beta$ (Fig. 6). The upper end of the rod is pulled at speed $v$ along side $AO$. Find the dependence of the speed $u$ of point $B$ on the angle $\alpha$.

Solved and e. Since the length of the rod $AB$ is unchanged, the projections of the velocities of its ends onto the direction of the rod are the same:

$$u \cos \alpha = v \cos(\pi - \alpha - \beta),$$

or

$$u = -\frac{\cos(\alpha + \beta)}{\cos \alpha} v.$$

Let us consider the case when the length of the rod changes during movement (“the rod” can serve, for example, as a segment connecting two given points, the distance between which varies). Then

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the correspondence connecting the process of velocities of the ends of the rod takes the form

$$|v_A'' - v_B''| = u,$$

where $u$ is the rate of change in the length of the rod. (The module is needed here, since it is not known which speed is greater.)

4. A boat from a round berg is pulled by a rope at a constant speed $v$. Find the dependence of the speed of the boat $u$ on the angle $\alpha$.

Solution. In this case, we are interested in the part of the rope $AB$. The speed of its contraction is $v$. The velocity vectors of rope ends $A$ and $B$ are shown in figure 7. According to the statement above, we have

$$v_A \cos \alpha = v_B,$$

or since $v_A = u$, $v_B = v$,

$$u = \frac{v}{\cos \alpha}.$$

Try to solve similar problems yourself.

1. The rod $OA$ rotates clockwise with an angular velocity $\omega$, driving a cylinder of radius $r$ (Fig. 8). There is no slip between the cylinder and the plane. Find the dependence of the speed of the cylinder $v$ on the angle $\alpha$.

2. The crank $AO$ of length $r$ (Fig. 9) rotates with an angular velocity $\omega$, the length of the shaft $AB$ is $l$. Find the speed $v$ of point $B$ of the shatur, if $\angle ABO = \alpha$.

3. The ball, having previously untwisted around its axis, is placed on a horizontal surface. The friction coefficient of the ball on the surface is different from zero. Under the influence of friction, the ball changes its initial rotational motion and begins to somehow move along the surface. Describe how this movement will occur.