Kinematics of mass point

Kinematics of mass point
Passing of a train I
Task number 386
A freight train of length L=120m travels at a speed of v=30kmh. How long does it take the train to
pass:
a) a power pole?
b) a platform of length d=50m?
 Hint 1:
Think of what distance the locomotive of the train has to travel so that the whole train passes the pole
or the platform.
Draw a picture for both cases.
o Solution of Hint 1:
a) We suppose that the width of the pole is small compared with the length of the train. The
locomotive has to travel the whole length of the train to pass the pole.
b) The locomotive has to travel the length of the whole train plus the length of the platform in order
to the whole train passes it.
 Hint 2:
You know the distance which the locomotive has to travel and its speed. The desired times can be
found easily.
Make sure that all the assigned quantities are in the base units.
 Complete Solution a)

a) Suppose that the width of the pole is very small compared with the length of the train. Before the
whole train passes the pole, the locomotive has to travel the distance equal to the length of the train,
i.e. L.
It travels this distance in the time:
t1=Lv=1208,3s=14.4s.
 Complete Solution b)
b) We cannot consider the width of the platform insignificant, we have to take it into account. In order
for the whole train pass the platform, the locomotive has to travel the length of the train L (so the
whole train passes the beginning of the platform) and the length of the platform d (the whole train
passes the whole platform).
It travels this distance in the time:
t2=L+dv=120+508.3s=20.4s
 Complete answer
The freight train passes the power pole in 14.4s. The train passes the platform in 20.4s.
 Reference to a similar task
A similar but slightly more difficult task about the passing of two moving trains is Passing of a train
II .
Passing of a train II
Task number 387
A passenger train of length L1=60m travels at speed of v1=80kmh . How long does it take this train to
pass a freight train of length L2=120mtraveling at a speed of v2=30kmh?
Assume the trains travel:
a) in the same direction.
b) in opposite directions.
 Hint 1:
Before you start solving the problem, convert the speed of the trains into base units.

Convert the speed of the trains into base units:
The passenger train: v1=80kmh=˙22.2ms
The freight train: v2=30kmh=˙8.3ms.
 Hint 2:
Draw a picture for both cases.
Imagine that you are the engine driver of the freight train. What is the velocity of the passenger train
relative to you in both cases?
a) The trains travel in the same direction.
We can imagine that from the view of the engine driver of the freight train his train is standing still
and the passenger train travels along at the speed of
va=v1−v2=13.9ms.
b) The trains travel in opposite directions.
We can imagine that from the view of the engine driver of the freight train his train is standing still
and the passenger train travels along at the speed of
vb=v1+v2=30.5ms.
 Hint 3:
What distance does the passenger train have to travel in order to pass the freight train in both cases?
The train has to travel the same distance in both cases,as you can see in the pictures (the grey
passenger train is shown in the place where it starts).
a) the same direction.
b) opposite directions.

In both cases the locomotive of the passenger train has to travel the distance:
L1+L2
 Hint 4:
You know the distance which the trains have to travel to pass each other. And you also know the
speed by which the trains are passing. Therefore you can easily find the time needed by the trains to
pass each other.
We distinguish the two cases when calculating the time. We use the equation for distance traveled
during rectilinear motion: distance = speed ∙ time. From this equation we express the desired
time: time = distance/speed.
a) The trains travel in the same direction:
t1=L1+L2va=60+12013.9s=˙12.9s.
b) The trains travel in opposite directions:
t2=L1+L2vb=60+12030.5s=˙5.9s.
 Solution a) the same direction
First of all we convert the speed of the trains to the base units:
We draw a picture of the situation:
When the trains travel in the same direction the speed at which the trains move relative to each other
is:
va=v1−v2=13.9ms.
The train has to travel the distance which you can see in the picture (the grey passenger train is shown
in the place where it starts).
The locomotive of the passenger train has to travel the distance:
L1+L2

From the determined speed and distance we can now easily find the time needed by the trains to pass
each other.
t1=L1+L2va=60+12013.9s=˙12.9s.
 Solution b) opposite directions
First of all we convert the speed of the trains into base units:
We draw a picture of the situation:
When the trains travel in opposite directions the speed by which the trains move relative to each other
is:
vb=v1+v2=30.5ms.
The train has to travel the distance which you can see in the picture (the grey passenger train is shown
in the place where it starts).
The locomotive of the passenger train has to travel the distance:
L1+L2
From the determined speed and distance we can easily find the time needed by the trains to pass each
other:
t1=L1+L2vb=60+12030.5s=˙5.9s.
 Answer
The passenger train passes the freight train in 12.9s when the trains travel in the same direction.
When the trains travel in opposite directions, it takes the passenger train5.9s.
 Reference to a similar task
A similar but easier task about a train passing a stationary object is the task Passing of a train I .
A lift
Task number 651
Time variation of position x(t) of a lift cage is shown in the picture.
a) Describe verbally the motion of the cage.
b) Draw the time variation of the speed of the lift cage.
c) Draw the time variation of the acceleration of the lift cage.

d) A spring with a ball hangs from the ceiling of the cage. Describe the behaviour of the spring during
the motion of the lift.
Point A: position x = 0 m in the time t = 1 s
Point B: position x = 4 m in the time t = 3 s
Point C: position x = 24 m in the time t = 8 s
Point D: position x = 27 m in the time t = 9 s
 Hint 1 for a): Description of the motion
Divide the motion of the lift cage into separate sections between the origin of coordinates and point A,
between point A and point B, between point B and point C, between point C and point D, and
between point D and the end. You can determine what kind of motion the cage performs in each of
the separate sections based on the shape of the curves corresponding to the time variation of
position x(t) of the lift cage.
o Solution to Hint 1
Between the origin of the coordinates and point A the position of the lift cage doesn’t change (and
equals zero). The cage is thus at rest, it stands in the lower floor.
Between points A and B the time variation of the position has a shape of a parabola, the coordinate
grows quadratically. The lift moves upwards by the motion with a constant acceleration.
Between points B and C the graph of the time variation of the position is a line. The lift moves with a
constant speed,which it reached in point B.
Between points C and D the graph of the time variation of the position is again a part of a parabola.
The coordinate still grows but at a lower and lower rate. It is a motion with a constant negative
acceleration; the cage slows down to a halt.
Between point D and the end the position of the lift cage doesn’t change. The cage is at rest.

 Hint 2 for b) Coordinates of the speed
You know from the previous hint what kind of motion is there in each part, so you can decide if the
magnitude of the speed changes or not and how.
Determination of the magnitude of the speed is easy for the motion of the cage before point A and
after point D.
Between points A and B and between points C and D the lift moves with a constant acceleration. You
know what the initial and the final magnitude of the speed is in both parts. You can draw a shape of
the curve into a graph easily.
You know that the motion maintains constant speed between points B and C. You can determine the
magnitude of the speed of the lift cage from the graph in the assignment.
The function vx(t) is a derivative of the function x(t):
vx(t)=dxdt
Sections of the graph before point A and after point D correspond to the situation when the lift cage is
at rest. A graph of the function x(t) in these sections are lines parallel to the time axis. The angular
coefficient of the tangent (and also the speed of the cage) equals zero.
Between points B and C the cage moves with a constant speed, which we determine as the angular
coefficient of the line BC:
vx=ΔxΔt=(24−4)m(8−3)s=4m.s−1

The speed of the cage changes at the beginning of the motion (part AB) and during the repeated stop
(part CD), i.e. in the time intervals from 1 s to 3 s and from 8 s to 9 s. If we suppose that the lift starts
and stops uniformly, the time variation of the speed will be linear (lines AB and CD).
The time variation of the speed of the cage is depicted in the following graph:
 Note
Consider if it were possible to determine reversely the course of the positionx(t) from the graph vx(t).
A solution to this problem isn’t clear-cut. The graph of the functional vx(t)gives only some
information on the changes of the position (not on the position itself). We can determine changes of
the position from the graph in an arbitrary time interval
∫t2t1vx(t)
which corresponds to the area of the surface under the curve in the graph restricted by the start up and
the end point of the time interval. To determine the position at the beginning and at the end of this
interval we need a next piece of information for example the position in time t = 0 s .
 Hint 3 for c): Coordinates of the acceleration
You already know how the coordinates of the speed of the lift change with the time.
Realise what the acceleration will be like in the sections where the speed doesn’t change.
In sections AD,as the case may be CD, the speed grows linearly, as the case may be declines linearly.
Determine the magnitude of the acceleration from the graph vx(t). Realise what direction the
acceleration will have in these sections.

You know that the function ax(t) is a derivative of the function vx(t):
ax(t)=dvxdt
Sections of the graph before point A and after point D correspond to the situation where the lift cage
is at rest and thus the acceleration equals zero.
The acceleration equals zero again between point B and point C because the cage moves with the
constant speed.
The speed grows up linearly between point A and point B, the acceleration is constant with the
positive value and we determine the magnitude of the acceleration from graph vx(t) as a direction of
line AB:
aAB=ΔvABΔtAB=(4−0)m(3−1)s2=2m.s−2
The speed falls down linearly between point C and point D, the acceleration is constant with the
negative value and we determine the magnitude of the acceleration from graph vx(t) as a direction of
line CD:
aCD=ΔvCDΔtCD=(0−4)m(9−8)s2=−4m.s−2
The time variation of the acceleration of the cage is depicted in the following graph:
 Hint 4 for d): A spring with a ball
Realise which forces affect the ball (from the perspective of an observer who stays on the ground
floor) and what holds for their resultant in the particular sections. The relation of the resultant of the
forces to the acceleration is stated by Newton’s second law.

The ball is affected by the gravitational force of the Earth and by the force of the spring on which the
ball is hung.
Both these forces are in balance and their resultant of forces equals zero in the parts before point A,
between point B and point C and after the point D, where the acceleration of cage equals zero.
However in part AB the spring must pull a bit more to give the ball acceleration upwards – the spring
stretches.
On the contrary in part CD, in which the lift slows down, the acceleration heads downwards against
the direction of the motion. The spring must pull a bit less than when the cage is at rest (or when it
moves with constant speed) so that the resultant of forces would head downwards – the spring shrinks.
 COMPLETE SOLUTION
a)
Between the origin of the coordinates and point A the position of the lift cage doesn’t change (and
equals zero). The cage is thus at rest, it stands in the lower floor.
Between points A and B the time variation of the position has a shape of a parabola, the coordinate
grows quadratically. The lift moves upwards by the motion with a constant acceleration.
Between points B and C the graph of the time variation of the position is a line. The lift moves with a
constant speed,which it reached in point B.
Between points C and D the graph of the time variation of the position is again a part of a parabola.
The coordinate still grows but at a lower and lower rate. It is a motion with a constant negative
acceleration; the cage slows down to a halt.
Between point D and the end the position of the lift cage doesn’t change. The cage is at rest.
b)
The function vx(t) is a derivative of the function x(t):
vx(t)=dxdt
Sections of the graph before point A and after point D correspond to the situation when the lift cage is
at rest. A graph of the function x(t) in these sections are lines parallel to the time axis. The angular
coefficient of the tangent (and also the speed of the cage) equals zero.
Between points B and C the cage moves with a constant speed, which we determine as the angular
coefficient of the line BC:
vx=ΔxΔt=(24−4)m(8−3)s=4m.s−1

The speed of the cage changes at the beginning of the motion (part AB) and during the repeated stop
(part CD), i.e. in the time intervals from 1 s to 3 s and from 8 s to 9 s. If we suppose that the lift starts
and stops uniformly, the time variation of the speed will be linear (lines AB and CD).
The time variation of the speed of the cage is depicted in the following graph:

c)
You know that the function ax(t) is a derivative of the function vx(t):
ax(t)=dvxdt
Sections of the graph before point A and after point D correspond to the situation where the lift cage
is at rest and thus the acceleration equals zero.
The acceleration equals zero again between point B and point C because the cage moves with the
constant speed.
The speed grows up linearly between point A and point B, the acceleration is constant with the
positive value and we determine the magnitude of the acceleration from graph vx(t) as a direction of
line AB:
aAB=ΔvABΔtAB=(4−0)m(3−1)s2=2m.s−2
The speed falls down linearly between point C and point D, the acceleration is constant with the
negative value and we determine the magnitude of the acceleration from graph vx(t) as a direction of
line CD:
aCD=ΔvCDΔtCD=(0−4)m(9−8)s2=−4m.s−2
The time variation of the acceleration of the cage is depicted in the following graph:

d)
The ball is affected by the gravitational force of the Earth and by the force of the spring on which the
ball is hung.
Both these forces are in balance and their resultant of forces equals zero in the parts before point A,
between point B and point C and after the point D, where the acceleration of cage equals zero.
However in part AB the spring must pull a bit more to give the ball acceleration upwards – the spring
stretches.
On the contrary in part CD, in which the lift slows down, the acceleration heads downwards against
the direction of the motion. The spring must pull a bit less than when the cage is at rest (or when it
moves with constant speed) so that the resultant of forces would head downwards – the spring shrinks.
 Answer
a) At first the cage stands downstairs and thus it is at rest – in the graph it is the line between the
origin of coordinates and point A. Then the lift goes up (the positive direction of coordinate axis),
moves with the constant acceleration up to the definite speed – in the graph it is the line between point
A and point B. The cage travels at this constant speed during the definite time (the rectilinear motion
with constant speed) - in the graph it is the line between point B and point C. In the end the cage starts
slowing down - in the graph it is the line between point C and point D. The curve from point D means
that the cage is at rest again.
b) The time variation of the speed of the cage is depicted in the following graph:

c) The time variation of the acceleration of the cage is depicted in the following graph:
d) In these cases, where the cage is at rest (sections before point A and after point D) or when it
travels at constant speed (section between point B and point C), the length of the spring does not
change and stays the same. When the cage starts moving (the section between point A and point B),
unlike in the case when the cage is at rest, the spring starts stretching. On the contrary when the cage
starts slowing down (the section between point C and point D) the spring shrinks in comparison to the
length when the cage is at rest.
A garden hose

Task number 506
We want to splash as high as possible on a vertical wall (which is situated at a horizontal
distance d from the hose’s nozzle) with a jet of water flowing from a garden hose with an initial
speed v0.
a) What elevation angle α do we need to choose?
b) At what height h does the water splash on the wall?
c) At what angle ψ does the water hit the wall? Determine the angle (from the horizontal) of the
instantaneous velocity vector at the moment of the hit.
Disregard the air resistance.
Solve for these values: v0 = 15 m·s-1
, d = 10 m.
 Analysis
We want to determine the elevation angle α at which we have to aim the hose so that the water
splashes as high as possible. First we have to express how the height of the splash h depends on the
angle α. Then we can find the maximum of the function h(α).
Beware of the idea that the maximum height of the splash on the wall will correspond to the
maximum height of the throw for a given angle α. Realize that the distance to the wall is fixed. Draw
the different possible situations.
 Hint 1 for a): picture of the situation, type of motion
Draw a picture of the situation. What kind of motion is it? How do the x-component and y-component
of the speed of the water change with time? How do the x-component and y-component of the
coordinates change?
o Solution to Hint 1 for a): picture of the situation, type of motion
We choose the origin of the coordinate system at the garden hose’s nozzle.

It is an inclined throw, in other words projectile motion.
The x-component and y-component of the speed of the water and the x-component and y-component
of the coordinates change with time in the following way:
vx=v0cosα(1)
vy=v0sinα−gt(2)
x=v0tcosα(3)
y=v0tsinα−12gt2(4)
 Hint 2 for a): time of flight of the water to the wall, height h of the range
How long does it take the water to reach the wall for a fixed angle α? What height h of the water
splash on the wall corresponds to this time?
o Solution to Hint 2 for a): time of flight of the water to the wall, height h of the
range
The time of flight of the water and the height of the splash on the wall follow these relations:
x=d
Then we get from (3):
d=v0tcosα
t=dv0cosα(5)
y=h
h=v0tsinα−12gt2
We substitute for t from (5):
h=dtgα−gd22v20cos2α=dtgα−gd22v20(1+tg2α)(6)
 Hint 3 for a): the elevation angle α
We need to find out for what angle α is the height h of the splash on the wall maximal. We want to
find the maximum of the function h = h(α). How can we do it?
o Solution to Hint 3 for a): the elevation angle α
We find when the derivative of the function h with respect to α equals zero:
dhdα=ddα(dtgα−gd22v20cos2α)
dhdα=dcos2α−gd2sinαv20cos3α=dcos2α(1−gdv20tgα)=0
1−gdv20tgα=0

tgα=v20gd(7)
The function h(α) reaches its maximum for elevation angle α=arctgv20gd.
For the given values:
tgα=1529.81⋅10=2.294
α=arctg2,294=66.4∘
 Hint 4 for b): the maximum height of the water splash
You know the angle for the maximum height of the splash and also the dependence of the height h of
the splash on the wall on the angle α. You can find the maximum height easily.
o Solution to Hint 4 for b): the maximum height of the water splash
You know that the height of the splash on the wall is according to equation (6):
You know that the height h of the splash on the wall will be maximal for the angle α which according
to equation (7) is:
tgα=v20gd(7)
We substitute from equation (7) into equation (6):
h=dv20gd−gd22v20(1+v40g2d2)=v202g−gd22v20
Numerically:
h=(1522⋅9.81−9.81⋅1022⋅152)m
h=9.29m
 Hint 5 for c): deviation ψ at which the water hits the wall
We draw in the picture the vector of the velocity of the water at the moment it hits the wall and its
components in the direction of the x-axis and y-axis. Using that we can easily express the appropriate
angle. We determine both components of the speed at the moment of the hit by equations (1) and (2)
from Hint 1.
o Solution to Hint 5 for c): deviation ψ at which the water hits the wall

We determine the angular deviation ψ from the horizontal direction of the final velocity from the
equation:
tgψ=vyvx
We substitute for vx and vy from equations (1) and (2) and for the time tfrom equation (5):
tgψ=v0sinα−gtv0cosα=tgα−gdv20cos2α
We adjust using equation (7):
tgψ=tgα−1tgαcos2α
tgψ=tgα−1sinαcosα=sin2α−1sinαcosα=−1tgα=−gdv20
Numerically:
tgψ=−9.81⋅10152=−0.436
ψ=−23.6∘
The deviation ψ of the final velocity of the water from the horizontal direction is negative (we subtract
the magnitude in the clockwise direction). See the picture.
Note:
Beware of the idea that at the maximum height of the splash on the wall y-component of the speed
will be zero and thus the deviation ψ will be zero too. In the task the distance to the wall is fixed and
thus, for the given angle, the time of flight of the water is also fixed.
 COMPLETE SOLUTION:

a) The elevation angle
We choose the origin of the coordinate system at the garden hose’s nozzle.
It is an inclined throw, in other words projectile motion.
The x-component and y-componentof the speed of the water and the x-component and y-component of
the coordinates change with time in the following way:
vx=v0cosα(1)
vy=v0sinα−gt(2)
x=v0tcosα(3)
y=v0tsinα−12gt2(4)
The time of flight of the water and the height of the splash on the wall follow these relations:
x=d
d=v0tcosα
t=dv0cosα(5)
y=h
h=v0tsinα−12gt2
We substitute for t from (5):
We need to find out for what angle α is the height h of the splash on the wall maximal. We want to
find the maximum of the function h = h(α).

We find when the derivative of the function h with respect to α equals zero:
dhdα=ddα(dtgα−gd22v20cos2α)
dhdα=dcos2α−gd2sinαv20cos3α=dcos2α(1−gdv20tgα)=0
1−gdv20tgα=0
tgα=v20gd(7)
The function h(α) reaches its maximum for elevation angle α=arctgv20gd.
tgα=1529.81⋅10=2.294
α=arctg2.294=66.4∘
b) The maximum height ofthe water splash
You know that the height of the splash on the wall is according to equation (6):
You know that the height h of the splash on the wall will be maximal for the angle α which according
to equation (7) is:
tgα=v20gd(7)
We substitute from equation (7) into equation (6):
h=dv20gd−gd22v20(1+v40g2d2)=v202g−gd22v20
Numerically:
h=(1522⋅9.81−9.81⋅1022⋅152)m
h=9.29m
c) The deviation ψ ofthe velocity vector ofthe splashing water from the horizontal direction
We mark in the picture the velocity vector of the splashing water,its components and the desired
angle.

We determine the angular deviation ψ from the horizontal direction of the final velocity from the
equation:
tgψ=vyvx
We substitute for vx and vy from equations (1) and (2) and for the time t from equation (5):
tgψ=v0sinα−gtv0cosα=tgα−gdv20cos2α
We adjust using equation (7):
tgψ=tgα−1tgαcos2α
tgψ=tgα−1sinαcosα=sin2α−1sinαcosα=−1tgα=−gdv20
Numerically:
tgψ=−9.81⋅10152=−0.436
ψ=−23.6∘
The deviation ψ of the final velocity of the water from the horizontal direction is negative (we subtract
the magnitude in the clockwise direction). See the picture.
Note:
Beware of the idea that at the maximum height of the splash on the wall the y- component of the
speed will be zero and thus the deviation ψ will be zero too. In the task the distance to the wall is fixed
and thus, for the given angle, the time of flight of the water is also fixed.
 Answer
a) If we want to splash the water at the maximum height on the vertical wall, we have to choose the
elevation angle α for which:
α=arctgv20gd

α=66.4∘
b) The water splashes to the height h, for which this formula holds:
h=v202g−gd22v20
h=9.29m
c) The water hits the wall at the deviation ψ given by:
tgψ=−gdv20
ψ=−23.6∘
An ant on a rod
Task number 384
A slim rod OA of length R rotates with angular velocity ω in a clockwise direction in a plane around
point O. There is an ant crawling along the rod from the point O to the point A with constant
speed v⃗ (t) (measured with regard to the rod). Determine the time dependent location of the ant in the
laboratory reference frame. Assume its position at time t = 0 s was at the centre of the rod.
 Hint 1: A picture
Let’s put point O of the rod at the origin of the coordinate system and let the rod begin at the x-axis at
time t = 0 s. Draw a picture including the starting positions of the rod and the ant and also their
positions at some random time t. Mark the position vector r⃗ of the ant at time t.
o Solution of hint 1
Picture:

 Hint 2: Position vector of an ant
What is the distance the ant crawled in time t and what is the length of the position vector of the ant at
time t? What is the angle the rod has turned in time t?
Using the angle α determine the x and y components of the position vector r⃗ (t).
o Solution of hint 2: Position vector of an ant
Let’s split the motion into two parts; a uniform linear motion with velocityv and a rotary motion with
angular velocity ω:
The ant has crawled a distance along the rod of vt.
The length of the position vector at time t is therefore (since the ant began at the centre of the rod):
r(t)=vt+R2(1)
The rod has reached an angle:
α=ωt
The length of the position vector changes with time according to equation (1):
r(t)=vt+R2
The x component of the position vector therefore equals
x(t)=r(t)cosα=(vt+R2)cosωt
The y component of the position vector equals
y(t)=−r(t)sinα=−(vt+R2)sinωt
Expressing the position vector as the vector sum of its components:
r⃗ (t)=x(t)i⃗ +y(t)j⃗

wherei⃗ , j⃗ are unit vectors in directions x and y
r⃗ (t)=(vt+R2)cos(ωt)i⃗ −(vt+R2)sin(ωt)j⃗
 Complete solution
Picture:
Let’s split the motion into two parts; a uniform linear motion with velocityv and a rotary motion with
angular velocity ω:
The ant has crawled a distance along the rod of vt.
The length of the position vector at time t is therefore (since the ant began at the centre of the rod):
r(t)=vt+R2(1)
The rod has reached an angle:
α=ωt
The length of the position vector changes with time according to equation (1):
r(t)=vt+R2
The x component of the position vector therefore equals
x(t)=r(t)cosα=(vt+R2)cosωt
The y component of the position vector equals
y(t)=−r(t)sinα=−(vt+R2)sinωt

Expressing the position vector as the vector sum of its components:
r⃗ (t)=x(t)i⃗ +y(t)j⃗
wherei⃗ , j⃗ are unit vectors in directions x and y
 Answer
The time dependent position vector of the ant is:
where i⃗ , j⃗ are unit vectors in the corresponding directions x and y.
A ladybug crawling on a rotating cylinder
Task number 385
A cylinder with radius R and height h turns at constant speed clockwise around its own axis. The
cylinder makes one turn per time T. There is a ladybug creeping along the surface of the cylinder. The
speed of the ladybug v is constant and downward (in the axial direction) with regard to the cylinder.
a. Determine the time dependent position vector of the ladybug with respect to its surroundings.
Assume that it starts moving at time t = 0 s from the top of the cylinder.
b. Determine the velocity (magnitude) of the ladybug with respect to its surroundings (not to the
cylinder).
c. Determine the magnitudes of both the tangent and the centripetal accelerations of the ladybug.
d. What is the name of the curve one could use to describe the trajectory of the ladybug?
Determine the radius of curvature of this trajectory.
e. What is the distance the ladybug covered during its motion between the top and bottom
surfaces of the cylinder?
Solve all tasks from the point of view of an observer standing next to the cylinder.
Let’s establish the system of coordinates so the axis z matches the axis of the cylinder and
axes x and y are parts of the cylinder’s bottom.
 Hint 1 (task a): a picture of the situation
Let’s establish the coordinate origin at the centre of the cylinder’s bottom and the ladybug’s starting
point at the edge of the cylinder’s top.
Draw a picture capturing the situation at time t = 0 s and mark in it the ladybug’s position.
Then draw a picture capturing the situation at some random time t and in it the ladybug’s position.
Picture 1:

Picture 2:
 Hint 2 (task a): The time dependent position vector of the ladybug
Let’s split the motion in two; one motion in direction of the z axis and the second motion in
the xy plane. What are these motions for?

Determine the time dependent coordinates x, y, z of the ladybug and after that, using these
coordinates, determine its position vector.
We split the ladybug’s motion into a uniform motion with speed v (in the direction of the z axis) and a
circular motion with angular velocity ω=2πT(in the xy plane):
x(t)=Rcosωt=Rcos2πTt
y(t)=−Rsinωt=−Rsin2πTt
z(t)=h−vt
r⃗ (t)=xi⃗ +yj⃗ +zk⃗
r⃗ (t)=Rcos(2πTt)i⃗ −Rsin(2πTt)j⃗ +(h−vt)k⃗
where i⃗ , j⃗ , k⃗ are the unit vectors codirectional with the x, y and z axes
 Hint 3 (task b): Time dependent velocity of the ladybug
First find the components vx, vy,vz of the velocity. To solve this problem you may use a parametric
expression for the position.
Then find the magnitude of the time dependent velocity of the ladybug.
The components of the beetle’s velocity can be found by taking time derivatives of the components of
the position vector of the ladybug.
vx=dxdt=ddt(Rcosωt=−Rωsinωt=−R2πTsin2πTt
vy=dydt=ddt(−Rsinωt=−Rωcosωt=−R2πTcos2πTt
vz=dzdt=ddt(h−vt)=−v
The time dependent velocity of the ladybug is:
vv=|v⃗ (t)|=v2x+v2y+v2z−−−−−−−−−−√=R2ω2(sin2ωt+cos2ωt)+v2−−−−−−−−−−−−−−−−−−−−−−√=
=v2+R2ω2−−−−−−−−√=v2+R2(2πT)2−−−−−−−−−−−√=4π2R2T2+v2−−−−−−−−−−√
 Hint 4 (task c): The magnitudes of both the tangential and centripetal
accelerations of the beetle
If we know the velocity of the ladybug (if not see hint 3), we may easily calculate its tangential
acceleration.

To find the magnitude of the normal (centripetal) acceleration we would need both the magnitude of
the tangential acceleration and the magnitude of the total acceleration. The overall acceleration could
be also (as in the case of velocity) calculated from its components ax, ay,az.
What is the relation between the total acceleration and its tangential and normal (centripetal) parts?
Express the tangential acceleration at of the ladybug:
at(t)=dvvdt=ddtv2+R2(2πT)2−−−−−−−−−−−√
at=0
The relationship between the magnitudes of the tangential (at),normal (an) and overall (a)
acceleration is:
a=a2t+a2n−−−−−−√
an=a2−a2t−−−−−−√=a
We may now determine the components of the overall acceleration and then also its magnitude:
ax(t)=dvxdt=ddt(−Rωsinωt=−Rω2cosωt=−R4π2T2cos2πTt
ay(t)=dvydt=ddt(−Rωcosωt=Rω2sinωt=R4π2T2sin2πTt
az(t)=dvzdt=ddt(−v)=0
a=a2x+a2y+a2z−−−−−−−−−−√=R2ω4cos2ωt+R2ω4sin2ωt+0−−−−−−−−−−−−−−−−−−−−−−−−−√=
=Rω2=4Rπ2T2
The time dependent normal acceleration an of the ladybug is then:
an=a=Rω2=4Rπ2T2
 Hint 5 (task d): The trajectory of the ladybug, the radius of curvature
First imagine the beetle’s trajectory in the case of its having zero velocity with respect to the rotating
cylinder (i.e. the beetle is sitting on the rotating cylinder). Then add in the ladybug’s motion with
respect to the cylinder. What is the curve swept out by its motion?
Remark: Realize that you must solve this problem from the point of view of an inertial reference
frame, i.e. you watch the ladybug’s motion from a stationary reference frame next to the rotating
cylinder.

What is the relation between the radius of curvature of its trajectory and both its velocity and
centripetal acceleration?
Trajectory of the motion::
The ladybug moves along a helix.
Remark: Finding the radius of curvature at some point of a curve means to fit the nearby points of the
curve to a circle. The radius of such a circle is called the radius of curvature. For circular motion it is
true that:
ac=v2R,
where ac is the centripetal (normal) acceleration, v is the tangential velocity and R is the radius of the
circle.
So, in our case:
The relation between the normal acceleration an,the velocity of the ladybug vv and ρ (the radius of
curvature) is:
an=v2vρ
The radius of curvature is then:
ρ=v2van
ρ=R2ω2+v2Rω2
where ω=2πT
and so:
ρ=R24π2T2+v2R4π2T2=R+v2T24π2R
 Hint 6 (task e): the distance traveled by the ladybug
The distance traveled by the ladybug, i.e. the length of its trajectory, we may easily find using its
velocity and the overall time of its motion. The overall time is the time in which it moves from the top
to the bottom of the cylinder.
The velocity of the ladybug (with respect to its surroundings) is constant and so the distance l could
be expressed as:
l=vvt
where vv=(R2ω2+v2)−−−−−−−−−−√ and t=hv.
l=(hv)(R2ω2+v2)−−−−−−−−−−√,
where ω=2πT.
And so:

l=(hv)4π2R2T2+v2−−−−−−−−−−√
Picture 1:
Picture 2:

a) We split the ladybug’s motion into a uniform motion with speed v (in the direction of the z axis)
and a circular motion with angular velocity ω=2πT (in the xy plane):
x(t)=Rcosωt=Rcos2πTt
y(t)=−Rsinωt=−Rsin2πTt
z(t)=h−vt
r⃗ (t)=xi⃗ +yj⃗ +zk⃗
r⃗ (t)=Rcos(2πTt)i⃗ −Rsin(2πTt)j⃗ +(h−vt)k⃗
where i⃗ , j⃗ , k⃗ are the unit vectors codirectional with the x, y and z axes
b) The components of the beetle’s velocity can be found by taking time derivatives of the components
of the position vector of the ladybug.
vx=dxdt=ddt(Rcosωt=−Rωsinωt=−R2πTsin2πTt
vy=dydt=ddt(−Rsinωt=−Rωcosωt=−R2πTcos2πTt
vz=dzdt=ddt(h−vt)=−v
The time dependent velocity of the ladybug is:
vv=|v⃗ (t)|=v2x+v2y+v2z−−−−−−−−−−√=R2ω2(sin2ωt+cos2ωt)+v2−−−−−−−−−−−−−−−−−−−−−−√=
=v2+R2ω2−−−−−−−−√=v2+R2(2πT)2−−−−−−−−−−−√=4π2R2T2+v2−−−−−−−−−−√

c) Express the tangential acceleration at of the ladybug:
at(t)=dvvdt=ddtv2+R2(2πT)2−−−−−−−−−−−√
at=0
The relationship between the magnitudes of the tangential (at),normal (an) and overall (a)
acceleration is:
a=a2t+a2n−−−−−−√
an=a2−a2t−−−−−−√=a
We may now determine the components of the overall acceleration and then also its magnitude:
ax(t)=dvxdt=ddt(−Rωsinωt=−Rω2cosωt=−R4π2T2cos2πTt
ay(t)=dvydt=ddt(−Rωcosωt=Rω2sinωt=R4π2T2sin2πTt
az(t)=dvzdt=ddt(−v)=0
a=a2x+a2y+a2z−−−−−−−−−−√=R2ω4cos2ωt+R2ω4sin2ωt+0−−−−−−−−−−−−−−−−−−−−−−−−−√=
=Rω2=4Rπ2T2
The time dependent normal acceleration an of the ladybug is then:
an=a=Rω2=4Rπ2T2
d) Trajectory of the motion::
The ladybug moves along a helix.
Remark: Finding the radius of curvature at some point of a curve means to fit the nearby points of the
curve to a circle. The radius of such a circle is called the radius of curvature. For circular motion it is
true that:
ac=v2R,
where ac is the centripetal (normal) acceleration, v is the tangential velocity and R is the radius of the
circle.
So, in our case:
The relation between the normal acceleration an,the velocity of the ladybug vv and ρ (the radius of
curvature) is:
an=v2vρ
The radius of curvature is then:

ρ=v2van
ρ=R2ω2+v2Rω2
where ω=2πT
and so:
ρ=R24π2T2+v2R4π2T2=R+v2T24π2R
e) The velocity of the ladybug (with respect to its surroundings) is constant and so the distance l could
be expressed as:
l=vvt
where vv=(R2ω2+v2−−−−−−−−−√) and t=hv.
l=(hv)(R2ω2+v2)−−−−−−−−−−√,
where ω=2πT.
And so:
l=(hv)4π2R2T2+v2−−−−−−−−−−√
 Answers
a) The time dependent position vector of the ladybug is:
r⃗ (t)=Rcos2πTti⃗ −Rsin2πTtj⃗ +(h−vt)k⃗
where i⃗ , j⃗ , k⃗ are the unit vectors codirectional with the x, y and z axes.
b) The magnitude of the ladybug’s velocity is:
vv=|v⃗ (t)|=v2+R2ω2−−−−−−−−√=4π2R2T2+v2−−−−−−−−−−√
c) The tangential acceleration of the ladybug is:
at=0m⋅s−2
And the magnitude of the normal acceleration an of the ladybug is:
an=a2−a2t−−−−−−√=Rω2=4Rπ2T2

Both the tangential and normal accelerations are constant,i.e. time independent.
d) The ladybug moves along a helix and its radius of curvature is:
ρ=R2ω2+v2Rω2=R+v2T24π2R
e) The distance l is:
l=(hv)4π2R2T2+v2−−−−−−−−−−√
Mouse and Cat
Task number 927
A mouse is moving in a straight line with a constant velocity
v⃗ m=(3,−2)m⋅s−1
from point
A=[0,1]m
and a cat is moving in a straight line with a constant velocity
v⃗ c=(4,1)m⋅s−1
from point
B=[0,−1]m
.
a) Determine the place where their paths will intersect. Are they going to meet there?
b) Determine the time when the cat and mouse are the nearest.
c) Determine the minimum distance between the cat and the mouse.
Note: Parametric equations should be written in the form of e.g.
y=1m−2m⋅s−1⋅t
For simplicity of the notation formulas are unitless.
 Hint 1: Part (a): The Intersection of Cat’s and Mouse’s Trajectories
Mark the initial position of the cat and mouse into coordinate system. Can you draw the lines of their
movements if you know the directions of their velocities? If your drawing was accurate the
intersection of the lines can be easily subtracted from the picture. Intersection point can be also solved
by writing down the parametric equations of the cat’s and mouse’s trajectories. One of its points and
directions is known in each of the lines.
o Solution: Hint 1
Graphical solution:
Picture 1:

A=[0,1]m vm=(3,−2)m⋅s−1
B=[0,−1]m vc=(4,1)m⋅s−1
Numerical Solution:
Parametric equations of the mouse line:
xm=0+3tm
ym=1−2tm
Equation of the mouse trajectory is obtained by eliminating parameter tm.
tm=xm3
Mouse trajectory:
ym=1−2xm3
3ym=3−2xm
Parametric equations of the cat’s line:
xc=0+4tk
yc=−1+tk
Equation of the cat‘s trajectory is obtained by eliminating the parameter tc.
tc=xk4
Cat’s trajectory:
yc=−1+xc4
4yc=xc−4
Intersection point of the trajectories:
xm=xc=x
ym=yc=y

Two equations in two unknowns x and y is solved:
3y=3−2x(1)
4y=x−4(2)
From (2):
x=4y+4.
is substituted to (1):
3y=3−8y−8.
Hence:
y=−511,
x=2411.
Coordinates of the intersection point P:
P=[2411,−511]m.
 Hint 2: Part (a): Cat’s and Mouse’s Meeting
At what time will the mouse and the cat arrive at the point? Is it the same time?
o Solution: Hint 2
Mouse is at the point P
[2411;−511]m
at the time
tm=x3=2411⋅3s=811s.
Cat is at point P
[2411;−511]m
at the time
tc=x4=2411⋅4s=611s.
Arrival times are different, it means that they don’t meet at point P.
 Hint 3: Relationship Between Time and Cat‘s and Mouse’s Distance
Do you know the relationship between time and the cat’s and the mouse’s coordinates? Can you
express the time-dependence of their distance between them? Formula for distance of two points can
be easily derived from the picture 2 or you can look into the mathematical tables.
o Solution: Hint 3
The relationship between time and cat’s and mouse’s distance L(t) can be derivated from the picture 2.
Picture 2:

xc−xm=4t−3t=t
yc−ym=t−1+2t−1=3t−2
By using Pythagorean Theorem:
L(t)=(xc−xm)2+(yc−ym)2−−−−−−−−−−−−−−−−−−−−√=10t2−12t+4−−−−−−−−−−−−√
 Hint 4 Part (b): The Time When the Mouse and the Cat are the Nearest
Now you know the time-dependence of the distance between the cat and the mouse. How do you find
a minimum of this function?
o Solution: Hint 4
To find the time, when the cat and the mouse are the nearest,is to find a minimum of a function L(t).
We’ll find the derivative of the function L(t) with respect to time t and then its extreme:
dLdt=1220tmin−1210t2min−12tmin+4−−−−−−−−−−−−−−−−√=0
20tmin−12=0
tmin=35s
The time when the mouse and the cat are the nearest:
tmin=0.6s.
 Hint 5: Part (c): Minimum Distance Between the Cat and the Mouse
Now you know the time when the mouse and the cat are the nearest and also the time-dependence of
their distance. How do you determinate the minimum distance?
o Solution: Hint 5
tmin=0.6s.
By substituting this value into the function L(t) we get the smallest distance of the mouse and the cat.

Lmin=10t2min−12tmin+4−−−−−−−−−−−−−−−−√
Lmin=10⋅0.62−12⋅0.6+4−−−−−−−−−−−−−−−−√m=0.4−−−√m=0.63m
 Complete Solution
a)
Graphical solution:
Picture 1:
A=[0,1]m vm=(3,−2)m⋅s−1
B=[0,−1]m vc=(4,1)m⋅s−1
Numerical Solution:
Parametric equations of the mouse line:
xm=0+3tm
ym=1−2tm
Equation of the mouse trajectory is obtained by eliminating parameter tm.
tm=xm3
Mouse trajectory:
ym=1−2xm3
3ym=3−2xm
Parametric equations of the cat’s line:
xc=0+4tk
yc=−1+tk
Equation of the cat‘s trajectory is obtained by eliminating the parameter tc.

tc=xk4
Cat’s trajectory:
yc=−1+xc4
4yc=xc−4
Intersection point of the trajectories:
xm=xc=x
ym=yc=y
Two equations in two unknowns x and y is solved:
3y=3−2x(1)
4y=x−4(2)
From (2):
x=4y+4.
is substituted to (1):
3y=3−8y−8.
Hence:
y=−511,
x=2411.
Coordinates of the intersection point P:
P=[2411,−511]m.
Mouse is at the point P
[2411,−511]m
at the time
tm=x3=2411⋅3s=811s.
Cat is at point P
[2411,−511]m
at the time
tc=x4=2411⋅4s=611s.
Arrival times are different, it means that they don’t meet at point P.
b)
The relationship between time and cat’s and mouse’s distance L(t) can be derivated from the picture 2.
Picture 2:

xc−xm=4t−3t=t
yc−ym=t−1+2t−1=3t−2
By using Pythagorean Theorem:
L(t)=(xc−xm)2+(yc−ym)2−−−−−−−−−−−−−−−−−−−−√=10t2−12t+4−−−−−−−−−−−−√
To find the time, when the cat and the mouse are the nearest,is to find a minimum of a function L(t).
We’ll find the derivative of the function L(t) with respect to time t and then its extreme:
dLdt=1220tmin−1210t2min−12tmin+4−−−−−−−−−−−−−−−−√=0
20tmin−12=0
tmin=35s
tmin=0.6s.
c)
tmin=0.6s.
By substituting this value into the function L(t) we get the smallest distance of the mouse and the cat.
Lmin=10t2min−12tmin+4−−−−−−−−−−−−−−−−√
Lmin=10⋅0.62−12⋅0.6+4−−−−−−−−−−−−−−−−√m=0.4−−−√m=0.63m
 Answers
a) Paths of the mouse and the cat will intersect at point:
P=[2411,−511]m.
The mouse and the cat will not meet at point P,because the mouse will arrive at
tm=811s
and the cat will arrive at

tk=611s.
b) The time when and the cat are the nearest:
tmin=0.6s.
c) The minimum distance between the cat and the mouse:
Lmin=10t2min−12tmin+4−−−−−−−−−−−−−−−−√=0.63m.
A Moving Wheel
Task number 1159
A wheel with a radius R moves in a rectilinear motion with a speed of v0. In the coordinate system
related to the ground:
a) find the progress of the position vector of point X on the circumference of the wheel,
b) find the progress of the speed of point X and the progress of its magnitude,
c) using the result a) draw the shape of the trajectory and solve the length of one of its curves,
d) find the length of the travelled distance depending on time.
 Hint 1: The picture of the situation
Choose the appropriate coordinate axes and the origin of the coordinate system.
Draw the situation for the time t = 0 s. Choose the beginning position of point X in the origin of the
coordinate system.
Then draw the situation for the time t and mark the position of point X where the wheel gets in time t.
Mark the position of point X in the figure which the wheelachieved in time t. What is the distance?
We set off the stopwatch (t = 0 s) for example in the moment where point X touches the pavement.
We choose this point on the pavement to be the origin of the coordinate system (denoted by P). We
point the x-axis in the direction of the motion of the wheel along the pavement and y-axis vertically
upward.

The picture shows the situation in time t = 0 s.
The wheel covers the distance v0t in the direction of x-axis and the situation is created showing in the
second picture.
The point X is relocated from the origin P to the left upward. If the wheel doesn´t slip, the arc TX has
the same length as the rolled distance v0t. This angle α (corresponding to the arc TSX) is given:
α=−ωt=−v0tR
 Hint 2 for a): The progress of the position vector
Decompose the motion of point X in the direction of the x-axis and in the rotation around point S.
Try to find three simple vectors by which we can write the position vector PX→.
The position vector PX→=r⃗ (t) is changed quite complicatedly. It stretches right and is in rotation
upward and downward. We can compose this vector from three vectors which behave more easily:
PX→=PT→+TS→+SX→
where PT→ … vector stretches along x-axis with velocity v0: PT→=v0ti→
TS→ … vector is in the direction of y-axis: TS→=Rj→
SX→ … vector rotates equally in the direction clockwise around point S.
SX→=Rcos(−v0tR−π2)i→+Rsin(−v0tR−π2)j→
where (−v0tR−π2) is the angle which the vector SX→ forms to x-axis.
We substitute into the equation:
and we use these equations:
cos(−α−π2)=−sinα
sin(−α−π2)=−cosα
then we get for the motion vector r⃗ (t)=PX→:
PX→=(v0t−Rsinv0tR)i→+(R−Rcosv0tR)j→
alternatively for the components:
x=v0t−Rsinv0tR
y=R−Rcosv0tR
z=0
 Hint 3 for b) The progress of the speed and the progress of its magnitude
How do we get from the progress of the position vector of point X the progress of the speed of point
X? Write them.
How do we find the progress of the magnitude of the speed of point X?
We determine the progress of the speed v⃗ (t) by the derivative of the function r⃗ (t):
r⃗ (t)=PX→=(v0t−Rsinv0tR)i→+(R−Rcosv0tR)j→

v⃗ (t)=d(v0t−Rsinv0tR)dti→+d(R−Rcosv0tR)dtj→
After the derivative we get for the progress of the speed:
v⃗ (t)=(v0−v0cosv0tR)i→+v0sinv0tRj→
The magnitude of the speed is the absolute value of this vector:
v(t)=|v⃗ (t)|=v2x+v2y+v2z−−−−−−−−−−√=(v0−v0cosv0tR)2+(v0sinv0tR)2−−−−−−−−−−−−−−−−−−−
−−−−−−−−−√
v(t)=v20−2v20cos2(v0tR)+v20(cos2(v0tR)+sin2(v0tR))−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
−−−−−−−−−−−√
v(t)=2v20−2v20cos2(v0tR)−−−−−−−−−−−−−−−−−√
v(t)=2v01−cosv0tR2−−−−−−−−−√
Or we use this equation: ∣∣sinα2∣∣=1−cosα2−−−−−−√ and so simply:
v(t)=2v0∣∣sinv02Rt∣∣
 Hint 4 for c): The trajectory of the motion, the length of the arc
Determine the shape of the progress of the speed from the progress of the motion vector of point X.
Draw the picture.
Knowing the progress of the magnitude of the speed how do we calculate the length of a single arc or
the travelled distance for one revolution?
We determine the trajectory in the most comfortable way for example from the components of the
vector PX→:
x=v0t−Rsinv0tR
y=R−Rcosv0tR
z=0
We determine a few points of the trajectory (x, y) for example in these times 0, T6,2T6,3T6 etc.
We assume a facilitation for the numerical calculation R=1m and v=1ms−1.
From these equations:
x=v0t−Rsinv0tR
y=R−Rcosv0tR
z=0
We get then:

x 0 0.18 1.23 3.13 5.05 6.10 6.28
y 0 0.5 1.5 2.0 1.5 0.5 0
The length of the single arc equals the travelled distance of the point during one period and so:
sT=∫T0v(t)dt
where v(t)=2v0∣∣sinv02Rt∣∣.
One period equals T=2πRv0. This expression sinv02Rt assumes (on the whole interval 0 do T) positive
values therefore we don´t write necessarily the absolute value.
sT=∫T0(2v0sinv02Rt)dt=2v02Rv0[−cosv0t2R]T0
sT=4R(−cosv02πR2Rv0+1)=8R
The length of the single arc is equalled this expression: sT=8R.
 Hint 5 for d): The travelled distance as a function of time
We know what distance point X travels for one period.
We can calculate the travelled distance (for the time t < T) by the equation for the progress of the
speed.
How do we express the total travelled distance in the random time t?
We get the total distance for the rolling of much more than one revolution as the sum of the distance
of complete revolutions and the rest of the revolution.
We know the length of one arc is sT=8R.
The distance of the complete revolution equals:
sc=(INTtT)⋅8R
where INTtT is the whole-number of the part quotient tT.
We calculate the rest of the distance from the equation:
sz=∫tz0v(t)dt
values and tz < T therefore we don´t write necessarily the absolute value.
sz=∫tz02v0∣∣sinv02Rt∣∣dt=2v02Rv0[−cosv0t2R]tz0
sz=4R(1−cosv0tz2R)=8Rsin2v0tz4R
The total travelled distance in the time t equals:
s(t)=(INTtT)⋅8R+8Rsin2v0tz4R
The first member presents sums of the completed arcs and the second member presents the last
incomplete arc.
The graph of the function s(t)=(INTtT)⋅8R+8Rsin2v0tz4R looks like (see the figure):

The picture of the situation:
We set off the stopwatch (t = 0 s) for example in the moment where point X touches the pavement.
We choose this point on the pavement to be the origin of the coordinate system (denoted by P). We
point the x-axis axis in the direction of the motion of the wheel along the pavement and y-axis
vertically upward.
The picture shows the situation in time t = 0 s.
The wheel covers the distance v0t in the direction of x-axis and the situation is created showing in the
second picture.
The point X is relocated from the origin P to the left upward. If the wheel doesn´t slip, the arc TX has
the same length as the rolled distance v0t. This angle α (corresponding to the arc TSX) is given:
α=−ωt=−v0tR

a) The position vector
The position vector PX→=r⃗ (t) is changed quite complicatedly. It stretches right and is in rotation
upward and downward. We can compose this vector from three vectors which behave more easily:
where PT→ … vector stretches along x-axis with velocity v0: PT→=v0ti→
TS→ … vector is in the direction of y-axis: TS→=Rj→
SX→ … vector rotates equally in the direction clockwise around point S.
SX→=Rcos(−v0tR−π2)i→+Rsin(−v0tR−π2)j→
where (−v0tR−π2) is the angle which the vector SX→ forms to x-axis.
We substitute into the equation:
and we use these equations:
cos(−α−π2)=−sinα
sin(−α−π2)=−cosα
then we get for the motion vector r⃗ (t)=PX→:
PX→=(v0t−Rsinv0tR)i→+(R−Rcosv0tR)j→
x=v0t−Rsinv0tR
y=R−Rcosv0tR
z=0
b) The progress of the speed and the progress ofits magnitude
We determine the progress of the speed v⃗ (t) by the derivative of the function r⃗ (t):
v⃗ (t)=d(v0t−Rsinv0tR)dti→+d(R−Rcosv0tR)dtj→
After the derivative we get for the progress of the speed:
The magnitude of the speed is the absolute value of this vector:
v(t)=|v⃗ (t)|=v2x+v2y+v2z−−−−−−−−−−√=(v0−v0cosv0tR)2+(v0sinv0tR)2−−−−−−−−−−−−−−−−−−−
−−−−−−−−−√
v(t)=v20−2v20cos2(v0tR)+v20(cos2(v0tR)+sin2(v0tR))−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
−−−−−−−−−−−√
v(t)=2v20−2v20cos2(v0tR)−−−−−−−−−−−−−−−√
v(t)=2v01−cosv0tR2−−−−−−−−−√
Or we use this equation: ∣∣sinα2∣∣=1−cosα2−−−−−−√ and so simply:
v(t)=2v0∣∣sinv02Rt∣∣
c) The trajectory ofthe motion, the length ofthe arc
We determine the trajectory in the most comfortable way for example from the components of the
vector PX→:

x=v0t−Rsinv0tR
y=R−Rcosv0tR
z=0
We determine a few points of the trajectory (x, y) for example in these times 0, T6,2T6,3T6 etc.
We assume a facilitation for the numerical calculation R=1m a v=1ms−1.
From these equations:
x=v0t−Rsinv0tR
y=R−Rcosv0tR
z=0
We get then:
x 0 0.18 1.23 3.13 5.05 6.10 6.28
y 0 0.5 1.5 2.0 1.5 0.5 0
The length of the single arc equals the travelled distance of the point during one period and so:
sT=∫T0v(t)dt
Where v(t)=2v0∣∣sinv02Rt∣∣
values therefore we don´t write necessarily the absolute value.
sT=∫T0(2v0sinv02Rt)dt=2v02Rv0[−cosv0t2R]T0
sT=4R(−cosv02πR2Rv0+1)=8R
The length of the single arc is equalled this expression: sT=8R.
d) The travelled distance as a function oftime
We get the total distance for the rolling of much more than one revolution as the sum of the distance
of complete revolutions and the rest of the revolution.
We know the length of one arc is sT=8R.
The distance of the complete revolution equals:
sc=(INTtT)⋅8R
where INTtT is the whole-number of the part quotient tT.
We calculate the rest of the distance from the equation:

sz=∫tz0v(t)dt
values and tz < T therefore we don´t write necessarily the absolute value.
sz=∫tz02v0∣∣sinv02Rt∣∣dt=2v02Rv0[−cosv0t2R]tz0
sz=4R(1−cosv0tz2R)=8Rsin2v0tz4R
The total travelled distance in the time t equals:
The first member presents sums of the completed arcs and the second member presents the last
incomplete arc.
The graph of the function s(t)=(INTtT)⋅8R+8Rsin2v0tz4R looks like (see the figure):
 Answer
a) It holds for the progress of the position vector of point X (which is on the circumference of the
wheel):
x(t)=v0t−Rsinv0tR
y(t)=R−Rcosv0tR
z(t)=0
b) It holds for the progress of the speed of point X

The magnitude of the speed of point X is:
v(t)=|v⃗ (t)|=v2x+v2y+v2z−−−−−−−−−−√=2v0∣∣sinv02Rt∣∣
c)The shape of the trajectory is:
The length of the single arc is: sT=8R
d) The length of the total arc is:
where INTtT is the whole-number of the part quotient tT and tz=t−(INTtT)T
(The first member presents sums of the completed arcs and the second member presents the last
incomplete arc)
References:---------------- http://physicstasks.eu/uloha.php?uloha=1159

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