Physicsguy51 ha scritto: ↑27 ago 2023, 8:34
The middle band of a horizontal table is formed by a (infinite) rubber strip of width
moving at a constant speed
and exactly fitting to the resting surface of the tabletop. A solid rubber ball is launched perpendicular to the edge of the conveyor belt, rolled smoothly, with an initial speed
. We find that when the ball rolls smoothly again after leaving the conveyor belt, it moves parallel to its original direction of movement. By how much distance
did the trajectory of the ball shift in the direction parallel to the conveyor belt? (The coefficient of friction between the horizontal surfaces and the ball is large, so the duration of sliding and rolling movements is much smaller than
).
Figure:
https://ibb.co/jg1bP8V
This problem is much less difficult than the "
inhomogeneous Rubik's Cube" problem, but the obscure text makes it fascinating.
Consider an appropriate
reference, with
vertical axis perpendicular to the horizontal plane,
horizontal axis having the same direction of motion as the conveyor belt, and
having the same motion direction as the ball on the static part of the table. The ball is thrown onto the table with initial velocity
along the
direction, without an initial angular velocity
imparted to it along the same direction. Therefore:
.
Since the ball rolls smoothly along the static surface of the table, it carries out a pure rolling motion with constant initial velocity
, without friction between the object and plane affecting the motion. So the ball, in the absence of frictional forces, initially does not assume initial translation velocity of the centre of mass
along
; in the direction of the horizontal axis, it possesses only initial rotation velocity
around an axis parallel to
. Therefore:
.
To recap:
.
Since in the beginning, the rubber ball performs a constant rotation, when it arrives in the vicinity of the conveyor belt, the coordinates of the angular velocity and translational velocity (of the centre of mass) are the same as those at the starting point: the velocity of the centre of mass has only component along
, the angular speed has only horizontal component along
. Assuming the start of motion (
) at the entrance on the mat, we therefore have that:
The ball arrives in the vicinity of the treadmill with the above velocities and begins a rototranslational motion. It does not begin to roll from the very beginning, but there is a settling period in which, before rolling, the ball slides across the table. So, it first carries out a sliding, then begins a rolling motion, in a small amount of time
. Since the coefficient of friction, and hence the frictional force, between the horizontal surfaces and the ball is assumed to be very large, the pure rolling motion begins (after the sliding and rotational motion is finished) much sooner than the time
it takes for the ball to move to the other side of the conveyor belt. Therefore, it must be assumed that the pure rolling motion begins when the ball is already on the conveyor belt, before it reaches the other side of the rubber strip, that is, near the point of contact between the mat and the static surface of the table.
Three forces are acting on the ball: the weight force
, the normal reaction
and the dynamic friction force
. The friction will be a dynamic one - and not a static one - because, when the ball enters the conveyor belt, it is subjected to the constant speed
of the rubber strip along
, which promotes sliding at the point of contact: the ball does not immediately enter rolling, but begins to slide, thus promoting the initiation of dynamic friction, which acts in the same direction as the
velocity of the mat along the positive
and tends to slow down the same
velocity of the contact point. Project the forces on the 2nd Newton's Law, along the horizontal
and vertical
axes: along
, the normal reaction
counterbalances the weight
; on the horizontal, along the
axis of the ball's motion, the friction is counter motive - in that it wants to slow the
velocity of the point of contact - and produces the deceleration of the centre of mass.
With no rotation, the only force that changes the velocity is the force of friction, constant in modulus, intensity and direction,
, with
dynamic friction coefficient between ball and horizontal surface and
mass of the ball. Substituting in the second equation of the system, we obtain that the deceleration of the centre of mass, a fraction of the acceleration of gravity according to the coefficient of dynamic friction, is equal to:
Since the ball has entered the conveyor belt, it has acquired an initial velocity along
equal to the
velocity of the moving part of the table. The constant force causes a linear decrease in the translation velocity of the centre of mass, which starts at
, gradually decreasing due to deceleration:
.
Let us analyze the rotational analog of Newton's Law, calculating the torques with respect to the centre of mass. Since the weight force
arises from the centre of mass and the normal reaction
passes through it, their respective distance from the reference pole will be zero and, having zero "arm", they will have torques equal to
:
.
The only force contributing to the motion is dynamic friction, whose "arm" will be equal to the radius of the wheel, being the force applied at the point of contact and being the "arm" a distance from that point to the centre of the ball. The frictional force will therefore have torque:
.
Using Newton's second law in angular form, we obtain an equation of motion for angular acceleration:
, da cui:
. Integrating, we obtain:
. In this case, the initial angular velocity
will be absent, since, in the launch, only the translational part was deliberately aroused without initiating rotation. So, for
, we have:
.
To compare the linear velocity of translation with the angular velocity, it is possible to transform the angular velocity
into the
velocity of the peripheral points of the ball. When the ball rolls,
the instantaneous velocity of the point tangent to the surface on which it rolls is zero. Consequently, its rotational velocity
and the translational velocity of its centre of rotation
) are related by
, with
radius of the sphere. If the centre of rotation of the ball moves faster than
, the rotation cannot "keep up" and the object slides on the surface: this type of motion is called sliding. Because of friction, the ball undergoing a sliding motion generally slows down rapidly to
, at which point it rolls without sliding. Therefore:
. Substituting
, we have:
.
The rotational velocity
increases over time, while the translational velocity
decreases with time: the frictional force
, while trying to decrease the translational velocity, also increases the rotational velocity, dispossessing the translational kinetic energy to give it to the rotation, to bring itself into an equilibrium in which the two velocities are those of pure rolling motion. Summing up, then, these two velocities are given by the equations:
.
The sliding and rolling motion is combined by translation and rotation. Initially, at the moment of launch, there is pure sliding; during the motion, friction tries to decrease the
and increase the
, until, at the point of contact, the two velocities equalize and the rolling regime is set off. Thus, friction plays a dual role, decreasing translation and increasing rotation until it can cancel the relative motion of one relative to the other, minimizing the energy losses in sliding, which are very frequent. Consider the graph below:
.
Translational velocity decreases progressively, while rotational velocity increases linearly: they will equal each other at some point corresponding to an instant of time
, from which, thereafter, there will be only pure rolling. Therefore, to find the time
at which the ball enters a pure rolling motion, it is sufficient to equate
with
, whence:
.
The ball, in this case, is assimilated to a homogeneous solid sphere, so its moment of inertia is equal to
. Substituting that value of
into
:
Note that the time
, until complete rolling motion is achieved, decreases inversely with the coefficient of friction: for a coefficient of friction tending to infinity
, the time for the ball to slip and roll tends to
,
, so it should be assumed that the pure rolling motion begins when the ball enters the conveyor belt. After this time
, friction stops playing its role: from dynamic, it disappears, since it no longer has any reason to exist, having succeeded in achieving the desired equality. As soon as the sphere rolls, its surface no longer moves relative to that on which it is rolling, since its instantaneous velocity at the point of contact is zero. As a result, the frictional force is very small and the sphere can roll for a great distance before it stops. From this instant
onward, the two velocities do not reverse, but proceed together, and, once rolling is established, it is very difficult for it to be interrupted: this is a very stable situation, preserved for a long time, in which the energies of translation and rotation travel together at the same rate. Therefore, the ball proceeds at a constant rolling speed over time.
With this preliminary analysis completed, we need to consider which physical model to use for the determination of
.
On the conveyor belt, along
the ball performs a vertical displacement
(equal to the width of the rubber strip) with velocity
; in the same interval of time
, along
the ball carries out a horizontal displacement
with rolling velocity relative to the table equal to
, corresponding to the maximum value (constant from
until the exit from the strip) of the velocity
assumed by the ball on the mat during the roll-translation. Finally, it is expected that this velocity
will cancel upon exiting the conveyor belt, from which the ball is no longer subject to the velocity
of the same and can restart its pure rolling motion with only constant
velocity along
. Therefore, in conclusion, we have:
, whence:
To calculate
, as can be seen, an evaluation of
is required.
Describe the phenomenon not in a frame of reference
fixed to the table, but in a frame of reference K' that moves with the the conveyor belt. Assume the table as the fixed laboratory frame of reference
(see Figure 1), and the conveyor belt as the moving system
(see Figure 2).
Figure 1:
Figure 2:
When the two velocities
and
come together, the ball will continue its motion at a common velocity corresponding to the rolling velocity
, which is obtained by substituting the required time
for one of the two velocities:
.
The final rolling speed
is independent of the friction force, but depends instead on the moment of inertia of the ball.
Now, assume the table as the fixed
frame of reference of the laboratory, and the conveyor belt as the moving
frame of reference. For the Galilean composition of velocities, along
we have:
, where
is the absolute velocity, measured by the observer located in the fixed frame of reference
of the laboratory;
is the relative velocity, measured by the observer located in the moving frame of reference
;
is the drag speed, which is the (constant) speed of the moving frame of reference
relative to the fixed frame of reference
. In this case, the relative velocity
is the
rolling velocity assumed by the ball at a time
, while the drag velocity
is the constant
velocity of the conveyor belt. Therefore:
From the beginning of the sliding and rotational motion to the beginning of pure rolling (i.e., from
to
), the frictional force exerts a constant action on the body in the direction of positive
(where
is directed). Since friction has a decelerating stopping action on the centre of mass, it always acts in the direction opposite to the velocity of the contact point relative to the moving system (i.e., the conveyor belt), which at time
is equal to
. So, this velocity is always directed in the
direction, that is, toward the negative
. Therefore:
The previous equation in vector form can be written as:
. Substituting
, we have:
Substituting in
, we have:
Substituting numerical values: