Assoc. Prof. G. D. Terskov.

Motion of a Body on an Inclined Plane with Vertical Circular Oscillations

Introduction

An inclined plane with vertical circular oscillations is used for transporting products with simultaneous sorting in machines for harvesting cereal grains and for the beneficiation of minerals.

The character of the motion of a body along such a plane has not yet been clarified to such an extent that it would be possible consciously to design the working members of machines operating according to the indicated principle.

The task of the present work is to determine the character of the motion of a body in the general case for various kinematic data; moreover, to simplify the problem, the coefficient of friction is assumed constant.

1. Possibility of the Onset of Sliding of the Body

If the plane \(AB\) is fastened to two cranks, \(O_1A\) and \(O_2B\), of equal length, then all points of this plane will describe identical trajectories—circles of radius

\[ r = O_1A = O_2B. \]

We shall take the origin of coordinates to be located at the center of the circle described by some point \(M\), and direct the \(x\)-axis (Fig. 1) parallel to the plane downward; in this case the motion of point \(M\) will be determined by the equations

\[ x = r \cos \alpha; \quad y = r \sin \alpha . \tag{1} \]

where: \(\alpha\) is the angle of rotation of the crank. With uniform rotational motion of the crank and a given angular velocity \(\omega\),

\[ \alpha = \omega t . \tag{2} \]

In this case the velocity of the point will be determined by the equations

\[ U_x = - \omega r \sin \alpha; \quad U_y = \omega r \cos \alpha \ . . . . . . . . . \tag{3} \]

Fig. 1.

and the acceleration by the equations

\[ J_x = - \omega^2 r \cos \alpha; \quad J_y = - \omega^2 r \sin \alpha \ . . . . . . . . . \tag{4} \]

If some body lies on the plane at the point \(M\), then its motion will also be determined by expressions (1–4), and therefore it will be acted upon by the centrifugal force of inertia

\[ I_x = m\omega^2 r\cos\alpha; \quad I_y = m\omega^2 r\sin\alpha \tag{5} \]

In quadrants I and IV this force of inertia will tend to displace the body downward; in order for the displacement to be possible, it is necessary that

\[ m\omega^2 r\cos\alpha + mg\sin\beta > f(mg\cos\beta - m\omega^2 r\sin\alpha). \]

After the corresponding transformations, this expression gives the condition ensuring the possibility of the onset of downward sliding, in the form

\[ \omega^2 r\cos(\alpha - \varphi) > g\sin(\varphi - \beta) \tag{6} \]

The left-hand side of this inequality has its maximum value when \(\cos(\alpha - \varphi)=1\), or \(\alpha=\varphi\); therefore it is evident that the beginning of downward sliding of the body is possible at all only in the case when

\[ \omega^2 r > g\sin(\varphi-\beta). \tag{6a} \]

If this condition is fulfilled, then downward sliding can begin only in the case when condition (6) holds; in other words, the body can begin to slide downward only when the crank occupies a position between

\[ \alpha_3 = 2\pi + \varphi - \arccos\left[\frac{g}{\omega^2 r}\sin(\varphi-\beta)\right] \tag{7} \]

and

\[ \alpha_0 = \varphi + \arccos\left[\frac{g}{\omega^2 r}\sin(\varphi-\beta)\right]. \tag{8} \]

In quadrants II and III the force of inertia tends to displace the body upward; this displacement can occur only in the case when (Fig. 2)

Fig. 2.

\[ -m\omega^2 r\cos\alpha - mg\sin\beta > f(mg\cos\beta - m\omega^2 r\sin\alpha) \]

or

\[ -\omega^2 r\cos(\alpha+\varphi) > g\sin(\varphi+\beta). \tag{9} \]

Reasoning analogously to the preceding case, we find that the beginning of upward sliding can take place only in the case when

\[ \omega^2 r > g\sin(\varphi+\beta). \tag{9a} \]

and, if this condition is present, only when the crank occupies a position between

\[ \alpha_1=\pi-\varphi-\arccos\left[\frac{g}{\omega^2 r}\sin(\varphi+\beta)\right] \ldots \ldots \ldots (10) \]

and

\[ \alpha_2=\pi-\varphi+\arccos\left[\frac{g}{\omega^2 r}\sin(\varphi+\beta)\right] \ldots \ldots \ldots (11) \]

In addition, from Figs. 1 and 2 it is clear that the force of inertia tends to tear the body away from the plane; obviously, separation can occur only in the case when

\[ m\omega^2 r\sin\alpha > mg\cos\beta \ldots \ldots \ldots \ldots (12) \]

and in general can take place only if

\[ \omega^2 r > g\cos\beta \ldots \ldots \ldots \ldots (12a) \]

If this condition is satisfied, then separation can occur only for positions of the crank between

\[ \alpha'=\arcsin\left[\frac{g}{\omega^2 r}\cos\beta\right] \ldots \ldots \ldots (13) \]

and

\[ \alpha''=\pi-\arcsin\left[\frac{g}{\omega^2 r}\cos\beta\right] \ldots \ldots \ldots (14) \]

Thus, for example, if \(\beta=12^\circ\), \(f=0.4\), \(\varphi=21^\circ 50'\), then the body cannot begin to slide at all when \(\omega^2 r<1.68\ \mathrm{m/sec^2}\); it can begin to slide downward when \(1.68<\omega^2 r<5.47\ \mathrm{m/sec^2}\); it can begin to slide both upward and downward when \(\omega^2 r>5.47\ \mathrm{m/sec^2}\); and, finally, it can separate from the plane when \(\omega^2 r>9.6\ \mathrm{m/sec^2}\).

2. Motion of the body for \(g\sin(\varphi-\beta)<\omega^2 r<g\sin(\varphi+\beta)\).

If \(\omega^2 r\) has such a value that the onset of sliding is possible only downward, then, for an infinitely small velocity of fall of the body onto the plane, at the instant of fall the absolute velocity of the body will be equal to zero. Subsequently, owing to the presence of the force of friction, the body will be carried along by the plane and its absolute velocity will begin to increase in magnitude; the direction of this velocity will coincide with the direction of motion of the plane. Since the normal pressure of the body on the plane will be composed of the component due to the weight of the body \(mg\cos\beta\) and of the pressure of the plane on the body \(-m\omega^2 r\sin\alpha\), the friction force will, generally speaking, be variable; its magnitude is determined by the expression (Fig. 3)

Fig. 3.

\[ F=f(mg\cos\beta-m\omega^2 r\sin\alpha) \ldots \ldots \ldots (15) \]

[[unclear: fragment of text from another line/page]] determination of the angle \(\alpha_e\) is performed by drawing a line inclined to the axis of abscissae at the angle \(\gamma_v\), where

\[ Tg\,\gamma_v=g\sin(\varphi+\beta)\frac{m_t}{m_v} \ldots \ldots \ldots (22) \]

In addition, the motion of the body will be influenced by the component \(mg\sin\beta\) of the force of gravity; if the plane moves to the right, then this component will diminish the action of the frictional force, and conversely, if the plane moves to the left, then \(mg\sin\beta\) will be added to the frictional force. In connection with this, the acceleration of the absolute motion will be equal to

\[ J_x=\frac{F\mp mg\sin\beta}{m} = g\,\frac{\sin(\varphi\mp\beta)}{\cos\varphi} -\omega^2 r f\sin\alpha \ldots \tag{16} \]

Here the sign \(+\) refers to the motion of the body up the plane, and the sign \(-\) to the motion of the body down the plane.

The absolute velocity of the body, being directed in the direction of the motion of the plane, will increase in its absolute value and is determined by the expression

\[ W_x=\int_{t_a}^{t} j\,dt = g\,\frac{\sin(\varphi\mp\beta)}{\cos\varphi}(t-t_a) +\omega r f(\cos\alpha-\cos\alpha_a) \ldots \tag{17} \]

Let us consider a concrete example: let \(\omega^2 r=5.13\ \text{m/sec}^2\), \(r=50\ \text{mm}\), \(\beta=12^\circ\); \(f=0.4\); in this case \(\omega=10.42\ \text{rad/sec}\), \(\omega r=0.52\ \text{m/sec}\), \(n=100\ \text{rev/min}\). \(\alpha_0=92^\circ 50'\), \(\alpha_3=311^\circ 50'\).

Let, at the instant of falling, the crank occupy the position determined by the angle \(\alpha_a=20^\circ\). Taking the foregoing into account and, consequently, bearing in mind that the plane moves to the right, while the absolute velocity has a negative sign, we may write the expression for the relative velocity of the body

\[ V_x=W_x-U_x = -g\,\frac{\sin(\varphi-\beta)}{\cos\varphi}(t-t_a) -\omega r f(\cos\alpha-\cos\alpha_a) +\omega r\sin\alpha = \]

\[ = -\frac{1}{\cos\varphi} \left[ -\omega r\sin\varphi\cos\omega t_a + g\sin(\varphi-\beta)(t-t_a) - \omega r\sin(\omega t-\varphi) \right]. \]

The first term of the expression in brackets is a constant quantity, the second term gradually increases, and the third varies according to the sine law; therefore, after a certain interval of time the relative velocity will become equal to zero. The position of the crank corresponding to this instant is found from the condition

\[ V_x=0 \]

or

\[ \omega r\sin\varphi\cos\omega t_a - g\sin(\varphi-\beta)(t-t_a) = -\omega r\sin(\omega t-\varphi). \]

This expression is most simply solved graphically as follows (Fig. 4): we construct the curve \(U_x=-\omega r\sin\omega t\) and, at \(\alpha=\omega t_a-\varphi\), lay off along the ordinate axis the segment \(aa_1\), equal to \(\omega r\sin\varphi\cos\omega t_a\); from

Fig. 4.

lying on the plane. But such a situation can occur only in the case when, for \(t_4<T+t_2\), the following relation (from expression 21) holds:

\[ -\omega r\sin(\omega t_3-\varphi)-g\sin(\varphi-\beta)(T+t_2-t_3)<-\omega r\sin(\omega t_2-\varphi). \]

Whence, after the corresponding transformations, we find the condition ensuring the existence of upward sliding in the form

\[ \omega_2 r< \frac{g\sin(\varphi-\beta)(2\pi+\alpha_2-\alpha_3)} {\sin(\alpha_2-\varphi)-\sin(\alpha_3-\varphi)} \ldots\ldots\ldots\ldots (24) \]

Solving this inequality by trial, it is not difficult to verify that, for example, for \(f=0.4\) and \(\beta=12^\circ\), upward sliding will occur only for

\[ 7.0<\omega_2 r<14.0\ \text{m/sec}^2; \]

In reality, however, the upper limit will be determined by the appearance of separation at \(\omega_2 r=9.6\), and in the presence of separation the conditions of motion will be different and formula (24) will not be valid.

If the corresponding calculations are carried out, for example for \(\omega^2 r=9.0\ \text{m/sec}^2\), we shall find that downward sliding will begin at \(\alpha_3=302^\circ 30'\) and will continue up to \(\alpha_4=2\pi+208^\circ\). Then upward sliding will begin and will continue up to \(\alpha_5=2\pi+222^\circ\); after this the body will share the motion of the plane up to \(\alpha_3\). During downward sliding a relative path \(S_n=200\) mm will be traversed, and during upward sliding the path will be \(S_v=0.1\) mm.

Thus, it follows that for \(\omega^2 r>g\sin(\varphi+\beta)\), despite the possibility of the onset of both upward and downward sliding, the body will nevertheless slide only downward, and only when condition (24) is fulfilled will a slight upward sliding occur.

4. Motion of the body for \(\omega^2 r>g\cos\beta\).

a) In the absence of sliding.

In order to clarify the essence of the process of motion in the presence of separation, let us first consider a simpler case, when at the moment of separation the body is at rest on the plane. Such a situation, for example, will occur in those cases where the surface of the plane is made non-smooth (keys of straw shakers).

It was proved above that the moment of separation is characterized by the position of the crank \(\alpha'\), for which from formula (13) we obtain (Fig. 6)

\[ \sin\alpha'=\frac{g\cos\beta}{\omega^2 r}; \]

\[ \cos\alpha'=\sqrt{1-\left[\frac{g\cos\beta}{\omega^2 r}\right]^2}\ldots (25) \]

Fig. 6.

The relative distance traveled by the body during each period of sliding (or, what is the same thing, during each revolution of the crank) will obviously be equal to

\[ S_n=\int_{t_3}^{t_4} V_n\,dt = -\frac{1}{\cos\varphi} \left\{ r\left[\cos(\omega t_4-\varphi)-\cos(\omega t_3-\varphi)\right] \right. \]

\[ \left. +\omega r\sin(\omega t_3-\varphi)(t_4-t_3) +g\sin(\varphi-\beta)\frac{(t_4-t_3)^2}{2} -r \right\}\ldots \tag{23} \]

In our case \(\alpha_4=544^\circ\), \(t_4=0.910\) sec.

\[ \alpha_3=311^\circ 50,\quad t_3=0.520\ \text{sec.},\quad S_n=140\ \text{mm}. \]

It is possible to determine \(S_n\) graphically. For this it is sufficient to planimeter the area \(F\) in Fig. 3, enclosed between the curve \(U_x\) and the straight line \(3_1—4_1\), multiply it by the scale of distance, and divide by \(\cos\varphi\).

In our case \(F=1970\ \text{mm}^2\), \(m_v=0.01\ \text{m/sec/mm}\), \(m_t=0.00667\ \text{sec/mm}\),

\[ m_s=m_v\,m_t=0.0667\ \text{mm/mm}^2,\quad S_n=\frac{F\cdot m_s}{\cos\varphi}=141\ \text{mm}. \]

The difference, as we see, is insignificant and does not exceed the limits determined by the accuracy of the constructions.

3. Motion of the body for \(g\sin(\varphi+\beta)<\omega^2 r<g\cos\beta\).

The character of the motion of the body in the case when the beginning of sliding both upward and downward is possible is best analyzed on a concrete example.

Let \(\omega^2 r=6.0\ \text{m/sec}^2\), \(\varphi=21^\circ 50'\), \(\beta=12^\circ\), \(r=50\ \text{mm}\). In that case: \(\omega=10.95\), \(\omega r=0.548\), \(\alpha_0=95^\circ 30\), \(\alpha_1=134^\circ 30\), \(\alpha_2=183^\circ 10\), \(\alpha_3=308^\circ\).

At first, after the body falls onto the plane, the character of the motion of the body will be the same as for \(\omega^2 r<g\sin(\varphi+\beta)\), i.e. at the end of the ends the body will begin to move downward along the plane at \(\alpha_3=308^\circ\). The end of the downward sliding will be found at \(\alpha_4=2\pi+190^\circ\) (Fig. 5). But since \(\alpha_4>\alpha_2\), in what follows the body will separate from the motion of the plane up to \(\alpha_3\), and then the phenomenon repeats again. Thus, it follows that in the given case, despite the possibility of the beginning of upward sliding at crank positions from \(\alpha_1\) to \(\alpha_2\), this possibility remains unused only because at these positions of the crank the body is in the process of sliding downward.

Fig. 5.

In order that the possibility of the beginning of upward sliding could be used, it is obviously necessary that \(\alpha_4<\alpha_2\), i.e., in other words, it is necessary that at crank positions from \(\alpha_1\) to \(\alpha_2\) the body should turn out to be at rest.

lying on the plane. But such a position can occur only in the case when, for \(t_4<T+t_2\), the following relation will hold (from expression 21)

\[ -\omega r\sin(\omega t_3-\varphi)-g\sin(\varphi-\beta)(T+t_2-t_3)<-\omega r\sin(\omega t_2-\varphi). \]

Whence, after the corresponding transformations, we find the condition ensuring the presence of upward sliding in the form

\[ \omega^2 r<\frac{g\sin(\varphi-\beta)(2\pi+\alpha_2-\alpha_3)} {\sin(\alpha_2-\varphi)-\sin(\alpha_3-\varphi)}\ . . . . . . . . . . . (24) \]

Solving this inequality by trial, it is not hard to verify that, for example, for \(f=0.4\) and \(\beta=12^\circ\), upward sliding will occur only for

\[ 7.0<\omega^2 r<14.0\ \text{m/sec}^2. \]

In reality, however, the upper limit will be determined by the appearance of the phenomenon of separation at \(\omega^2 r=9.6\), and in the presence of separation the conditions of motion will be different and formula (24) will not be valid.

If the corresponding calculations are carried out, for example, for \(\omega^2 r=9.0\ \text{m/sec}^2\), then we find that downward sliding will begin at \(\alpha_3=302^\circ 30'\) and will continue to \(\alpha_4=2\pi+208^\circ\). Then upward sliding will begin and will continue to \(\alpha_5=2\pi+222^\circ\); after this the body will share the motion of the plane up to \(\alpha_3\). During the downward sliding the relative path traversed will be \(S_n=200\) mm, while during the upward sliding the path will be \(S_v=0.1\) mm.

Thus it follows that for \(\omega^2 r>g\sin(\varphi+\beta)\), despite the possibility of the onset of sliding both upward and downward, the body will nevertheless slide only downward, and only in the case when condition (24) is fulfilled will there be a slight upward sliding.

4. Motion of the body for \(\omega^2 r>g\cos\beta\).

a) In the absence of sliding.

In order to clarify the essence of the process of motion in the presence of separation, let us first consider a simpler case, when at the moment of separation the body is at rest on the plane. Such a position, for example, will occur in those cases when the surface of the plane is made non-smooth (the keys of jackhammers).

It was proved above that the moment of separation is characterized by the position of the crank \(\alpha'\), for which from formula (13) we obtain (Fig. 6)

\[ \sin\alpha'=\frac{g\cos\beta}{\omega^2 r}; \]

\[ \cos\alpha'=\sqrt{1-\left[\frac{g\cos\beta}{\omega^2 r}\right]^2}\ . . (25) \]

Fig. 6.

Before separation, the body shared the motion of the plane; therefore, at the moment of separation the coordinates of the body are

\[ x'=r\cos\alpha'=r\sqrt{1-\left[\frac{g\cos\beta}{\omega^2 r}\right]^2};\quad y'=r\sin\alpha'=\frac{g\cos\beta}{\omega^2}. \tag{26} \]

and the components of the initial velocity are

\[ V'_x=-\omega r\sin\alpha'=-\frac{g\cos\beta}{\omega};\quad V'_y=\omega r\cos\alpha'=\omega r\sqrt{1-\left[\frac{g\cos\beta}{\omega^2 r}\right]^2}. \tag{27} \]

From the moment of separation, only the force of gravity will act on the body; therefore the velocity of the body during flight will be determined by the expressions

\[ V_x=V'_x+g\sin\beta\,(t-t');\quad V_y=V'_y-g\cos\beta\,(t-t'). \tag{28} \]

and the trajectory described by the body by the equations of motion

\[ x=x'+V'_x(t-t')+g\sin\beta\,\frac{(t-t')^2}{2};\quad y=y'+V'_y(t-t')-g\cos\beta\,\frac{(t-t')^2}{2}. \tag{29} \]

Since the velocity along the \(y\)-axis decreases uniformly, the moment of greatest rise \(t'_1\) will be found from the equation

\[ V_y=V'_y-g\cos\beta\,(t'_1-t')=0 \]

Whence the time spent in rising is determined by the expression

\[ t_{\text{rise}}=t'_1-t'=\sqrt{\left[\frac{\omega r}{g\cos\beta}\right]^2-\frac{1}{\omega^2}}. \tag{30} \]

Substituting (30) into (29), we find the coordinates of the body at the moment of greatest rise

\[ x'_1=\frac{g\sin\beta}{2\omega^2}\left[\left(\frac{\omega^2 r}{g\cos\beta}\right)^2-1\right];\quad y'_1=\frac{g\cos\beta}{2\omega^2}\left[\left(\frac{\omega^2 r}{g\cos\beta}\right)^2+1\right]. \tag{31} \]

After the moment of greatest rise, the body, descending, will begin to approach the plane and at some moment \(t'_2\) will fall onto it; at this moment the ordinates of the body and of the plane will be equal, and therefore the position of the crank \(\alpha'_2=\omega t'_2\), corresponding to the moment of fall, will be found from the condition

\[ y'+V'_y(t'_2-t')-g\cos\beta\,\frac{(t'_2-t')^2}{2}=r\sin\omega t'_2 . \tag{32} \]

In an ordinary coordinate grid this equation can be solved if one constructs pointwise the curves corresponding to the right- and left-hand sides and finds their point of intersection (Fig. 7), and the angle \(\alpha'_2=\omega t'_2\) corresponding to this point.

Knowing \(\alpha'_2\) and \(t'_2\), it is not difficult to determine the range of flight by the formula

\[ S=x_{\text{body}}-x_{\text{plane}}=x'+V'_x(t'_2-t')+g\sin\beta\,\frac{(t'_2-t')^2}{2}-r\cos\alpha'_2 . \tag{33} \]

The trajectory of the body’s flight can be constructed pointwise by means of equation (29). Thus all the factors characterizing the flight of the body in the air can be determined.

Let us dwell in more detail on some features of the resulting motion.

Considering in Fig. 8 the flight trajectories for different values of $\omega^2 r$, it is not difficult to see that the range of flight, depending on the value of $\omega^2 r$, will differ both in magnitude and in sign.

Fig. 7. Determination of the instant at which the body falls.

Thus, for example, at $\omega^2 r = 10.5$, $s = +9$ mm; at $\omega^2 r = 15.0$, $s = -23$ mm; at $\omega^2 r \geq 40$, $s$ is positive. In other words, if a plane with circular oscillations is used to move a body upward along an incline, as occurs in key-type straw shakers, then this can be accomplished only for certain values of $\omega^2 r$. In Fig. 9 the flight range is shown as a function $S = f(\omega^2 r)$. From this graph it is evident that upward motion of the body along the incline will occur only at $\omega^2 r = 11.5 \div 38.5\ \mathrm{m/sec^2}$, and the maximum velocity of upward motion corresponds to the value $\omega^2 r = 18 \div 30\ \mathrm{m/sec^2}$. Since an increase in $\omega^2 r$ is associated with an increase in the inertial forces, for key-type straw shakers one can evidently recommend the values

\[ \omega^2 r = 18.0 — 24.0\ \mathrm{m/sec^2}. \]

If $\omega^2 r \geq 38.5\ \mathrm{m/sec}$, then the body will move downward along the incline, and the maximum velocity of motion corresponds to the values

\[ \omega^2 r = 44.0 — 47.0\ \mathrm{m/sec^2}. \]

Fig. 8. Flight trajectories at $\beta = 12^\circ$.

Fig. 9. Flight range of the body at $\beta = 12^\circ$.

If $\omega^2 r$ is increased still further, an even greater maximum velocity can be obtained at $\omega^2 r \cong 80\ \mathrm{m/sec^2}$, but these values can practically not be used because of the very large inertial forces.

It follows from the foregoing that, by a suitable choice of $\omega^2 r$, one can obtain such a motion of the body in which the position of the crank at the instant when the body falls onto the plane will be the same as at the instant of separation, and therefore the body will again be thrown upward. With such a value of $\omega^2 r$, the body will evidently be in a process of continuous throwing.

In those cases when the motion under study is used for cleaning the transported product of large impurities (the principal task of keyboard straw walkers), continuous throwing of the body is highly desirable, since in this case the surface of the key always remains free, and the grain has the possibility of falling through the openings of the key without encountering obstacles from the straw.

Continuous throwing will occur in the case when the time during which the body remains in the air is equal to the time required for one revolution of the crank, i.e., when

\[ \alpha'_2=\alpha'+2\pi \quad \text{or} \quad t'_2-t'=\frac{2\pi}{\omega}=T \ldots \ldots \ldots (34) \]

Substituting these values into expression (32), after the corresponding transformations we find that continuous throwing will occur only in the case when

\[ \omega^2 r=g\cos\beta\sqrt{\pi^2+1}=32{,}5\cos\beta \ldots \ldots \ldots (35) \]

If the straw walker has narrow keys whose cranks are arranged at $180^\circ$ to one another, then it is more correct to assume that straw thrown up by one key will fall onto another, located next to the first. In this case

\[ \alpha'_2=\alpha'+\pi \quad \text{or} \quad t'_2-t'=\frac{\pi}{\omega}=\frac{T}{2} \ldots \ldots \ldots (34a) \]

and from formula (32), analogously to the preceding, we find

\[ \omega^2 r=g\cos\beta\sqrt{\frac{\pi^2}{4}+1}=18{,}2\cos\beta \ldots \ldots \ldots (35a) \]

The operating conditions of keyboard straw walkers are such that the latter case corresponds more closely to the actual situation.

b) With the presence of sliding.

If the coefficient of friction has a finite value and the body can slide along the plane, the process of motion will be more complex than in the absence of sliding.

Let us suppose that the body in one way or another shared the motion of the plane up to $\alpha_3$ (the beginning of the downward sliding zone). At $\alpha_3$ the body will begin to slide downward, and its motion along the $y$ axis will be the same as the motion of the plane, while along the $x$ axis the absolute velocity will be determined by expression (19). If there were no separation, then by the method set forth above one can find that the body would cease sliding at $\alpha_4=2\pi+215^\circ$, and the relative path traversed by the body, graphically represented by the area between $U_x$ and the straight line $3_1—4_1$, would be equal to $s=215$ mm (Fig. 10), for $\omega^2 r=10{,}5$.

However, owing to the possibility of separation, the body will slide only up to \(\alpha' = 2\pi + 66^\circ\) (according to formula 13), and will then break away and perform free flight. In this case the velocity along the \(y\)-axis will be determined by expression (28), and along the \(x\)-axis by the equation

\[ V_x = V'_x + g \sin \beta (t - t'). \tag{36} \]

where \(V'_x\) is found from (19) at \(t = t'\).

Since \(V'_x\) has a positive sign, during flight the body will move away from the origin of coordinates, and its velocity along the \(x\)-axis will be represented in graph No. 10 by the segment of the straight line \(BC\).

From Fig. No. 7 it is seen that the body will fall onto the plane at \(\alpha'_2 = 2\pi + 167^\circ\). The velocity \(V'_{2x}\) of the body along the \(x\)-axis at this moment is found from (36) at \(t = t'_2\).

If it is assumed that the body has no elasticity, then subsequently sliding of the body downward will again take place; in this case its velocity along the \(x\)-axis will be determined by the equation

\[ V_x = V'_{2x} - g \frac{\sin(\varphi - \beta)}{\cos \varphi}(t - t'_2) - \omega r f(\cos \omega t - \cos \omega t'_2) . . \tag{37} \]

Fig. 10.
Motion of the body for:

\[ \omega^2 r = 10.5;\quad r = 0.05;\quad \beta = 12^\circ; \]
\[ f = 0.4;\quad m_v = 0.01; \]
\[ m_t = 0.00482;\quad s = 223\ \mathrm{mm} \]

Having constructed in Fig. 10 the curve \(CD\) by this formula, we find point \(4'\), its intersection with the curve \(U_x\). The position of the crank \(\alpha'_4 = \omega t'_4 = 2\pi + 219^\circ\), corresponding to this point, determines the moment when the absolute velocity of the body becomes equal to the velocity of the plane and sliding ceases.

After this (since \(\alpha'_4 > \alpha_2\)) the body will share the motion of the plane up to \(\alpha_3\); then the phenomenon will repeat again, and so on.

The total relative path traversed by the body along the plane in one revolution is graphically represented by the area between the lines \(V_x\) and \(U_x\), and in our case will be equal to \(s = 223\) mm.

Thus, the difference in path compared with sliding without separation turns out to be insignificant.

Considering Fig. 10, one may conclude that, as \(\omega^2 r\) increases, the points \(A\) and \(D\) will approach one another and, consequently, for a sufficiently large value of \(\omega^2 r\), sliding and flight will follow one another and the body will not share the motion of the plane at all.

Such a situation, obviously, can occur only in the case when the curve \(CD\) passes through point \(A\) or above it; for this, in turn, it is necessary that the condition hold

\[ V'_{2x} - g \frac{\sin(\varphi - \beta)}{\cos \varphi}(t_3 - t'_2) - \omega r f(\cos \omega t_3 - \cos \omega t'_2) \geq -\omega r \sin \omega t_3. \]

Substituting here the value \(V'_{2x}\), and then \(V'_x\), after the corresponding pre-

we obtain, for the motion of the body without stops, the necessary condition in the form

\[ \omega^{2}r \leq g\, \frac{(\alpha'_2-\alpha')\sin\varphi\cos\beta-2\pi\sin(\varphi-\beta)} {(\cos\alpha'-\cos\alpha'_2)\sin\varphi} \ .\ .\ .\ .\ . \tag{38} \]

Thus, for example, at \(\omega^{2}r=15.0\ \mathrm{m/sec^{2}}\), the right-hand side is \(0.83\), and the left-hand side \(1.53\), and therefore, evidently, stops will occur. By selection it is not difficult to verify that sliding—flight without stops at \(f=0.4\) and \(\beta=12^\circ\) can occur only when \(\omega^{2}r \geq 16.8\ \mathrm{m/sec^{2}}\).

Let us consider in more detail the case when condition (38) is satisfied. Let, for example, \(\omega^{2}r=21.8\ \mathrm{m/sec^{2}}\) (Fig. 11). Suppose that the body somehow separates its motion from the plane up to \(\alpha'=26^\circ\). At this moment the body will detach from the plane and fly with an initial velocity directed to the right; without dwelling on the details of the investigation and reasoning analogously to the preceding case, we see that during flight the absolute velocity of the body \(V_x\) will vary along the straight line \(1—2\). At the moment \(\alpha'_2=317^\circ\) the body will fall onto the plane and will then slide upward; in this, the velocity \(V_x\) will increase along the curve \(2—3\). At the moment determined by point 3, the body will stop on the plane and immediately begin to slide downward; in this, the velocity \(V_x\) will vary along the curve \(3—4\). At the moment determined by point 4 (at \(\alpha'=2\pi+26^\circ\)), the body will again detach and fly; the velocity of the body will increase along the straight line \(4—5\). At the moment determined by point 5, the body will fall onto the plane and begin to slide downward; the velocity \(V_x\) will then vary along the curve \(5—6\). Subsequently the process of motion repeats; sliding and flight without stops alternate with one another, and the velocity of the body \(V_x\) in the direction parallel to the plane increases with each revolution. Thus, if condition (38) is satisfied, then the body will move relative to the plane to the left (downward), gradually increasing its speed.

Fig. 11.
Motion of the body for:
\[ \omega^{2}r=21.8;\quad r=0.05; \]
\[ \beta=12^\circ;\quad f=0.4; \]
\[ m_v=0.02;\quad m_t=0.00333 \]

The increase in the body’s velocity \(\Delta V_x\) for each revolution will be a constant quantity, and the average (conditional) acceleration will be determined by the expression

\[ j_{\mathrm{cond}}=\frac{\Delta V_x}{T} = g\left[ f\cos\beta\,\frac{\alpha'_2-\alpha'}{2\pi} - \frac{\sin(\varphi-\beta)}{\cos\varphi} \right] - \frac{\omega^{2}r}{2\pi}f(\cos\alpha'-\cos\alpha'_2) \ .\ . \tag{39} \]

In view of the fact that, as \(\omega^{2}r\) increases, the angle \(\alpha'\) decreases, while the angle \(\alpha'_2\) increases, the points \(2,5,7\ldots\) in Fig. 11 will approach the points \(4,6,8\ldots\). Ultimately, when \(\alpha'_2-\alpha'=2\pi\) and, as was indicated above,

\[ \omega^{2}r=g\cos\beta\sqrt{\pi^{2}+1} \]

these points will merge, sliding will disappear altogether, and the body, continuously thrown up by the plane, will move downward with acceleration

\[ j_{\text{conv}}=g\sin\beta \qquad\qquad\qquad (39a) \]

5. Conclusion.

Summarizing the foregoing, we arrive at the following conclusions:

1. The principal factor determining the character of the motion of the body is the centripetal acceleration \(\omega^{2}r\); depending on the magnitude of this acceleration, different types of motion may occur.

2. If \(\omega^{2}r<g\sin(\varphi-\beta)\), then the body will not slide at all, but, on the contrary, will share the motion of the plane.

3. If

\[ g\sin(\varphi-\beta)<\omega^{2}r<g\sin(\varphi-\beta)\, \frac{2\pi+\alpha_{2}-\alpha_{3}} {\sin(\alpha_{2}-\varphi)-\sin(\alpha_{3}-\varphi)}, \]

then the body will eventually slide along the plane only downward.

1. If

\[ g\sin(\varphi-\beta)\, \frac{2\pi+\alpha_{2}-\alpha_{3}} {\sin(\alpha_{2}-\varphi)-\sin(\alpha_{3}-\varphi)} <\omega^{2}r<g\cos\beta, \]

then the body may slide along the plane both downward and upward; in this case the path during upward sliding will be insignificant, and the overall motion of the body will be directed downward.

1. If

\[ g\cos\beta<\omega^{2}r< g\,\frac{(\alpha'_{2}-\alpha')\sin\varphi\cos\beta-2\pi\sin(\varphi-\beta)} {(\cos\alpha'-\cos\alpha'_{2})\sin\varphi}, \]

then during each revolution of the crank the body will successively pass through the stages: sliding—flight—sliding—stop, with a resultant motion downward.

1. If

\[ g\,\frac{(\alpha'_{2}-\alpha')\sin\varphi\cos\beta-2\pi\sin(\varphi-\beta)} {(\cos\alpha'-\cos\alpha'_{2})\sin\varphi} <\omega^{2}r<g\cos\beta\sqrt{\pi^{2}+1}, \]

then the body will move according to the scheme: sliding—flight without stops. The general

Fig. 12. Relative motion at \(\beta=12^\circ,\ f=0.4\).

the downward motion will then be uniformly accelerated, and the magnitude of the downward acceleration will be equal to

\[ j_{\mathrm{cond}} = g \left[ f \cos \beta \frac{\alpha'_2 - \alpha'}{2\pi} - \frac{\sin(\varphi - \beta)}{\cos \varphi} \right] - \frac{\omega^2 r}{2\pi} f(\cos \alpha' - \cos \alpha'_2). \]

1. If \(\omega^2 r = g \cos \beta \sqrt{\pi^2 + 1}\), then the body, continuously bouncing, will move downward with acceleration \(g \sin \beta\).

2. The path traveled by the body along the plane during each revolution of the crank will increase as \(\omega^2 r\) increases (Fig. 12); in motion with separation, however, this increase is extremely small.

3. In motion according to the sliding–flight scheme, the acceleration increases as \(\omega^2 r\) increases.