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Measurement of kinematic quantities
through simple experiments
 1 Measurement of velocity
 2 Measurement of force
 3 Measurement of acceleration
 4 Measurement of angular velocity of a body
moving on circular orbit
 5 Measurement of the central force and the
moment of force, or torque
 6 Expansion=the motion of rotating body,
especially “gyroscope”
1 Measurement of velocity
(1) Using a stop watch and a measure
a body moving distance L [m], time taken t [s], the velocity v [m/s]
Ｌ
ｖ= ｔ
①
Exercise1 L=10m, required time t=0.50s, ask the velocity of the vehicle.
V=( 10 )/( 0.50 )=( 20 )m/s
t : large ⇒ v : average
t : small ⇒ v : instant
(2) Using a points recording timer
(ⅰ)measuring time very short as possible
(ⅱ) marking points on a body
periodically at very short time
Fig.1 marking points
Fig.2 points recording timer
Fig.3 principle of points recording timer
iron bar
iron core
carbon
paper
spring
paper
tape
Using 50Hz
AC supply
coil
AC
⇒(ⅰ) number of vibration of the iron bar per second is 50.
⇒(ⅱ) periodic pointing number per second is 50.
⇒(ⅲ) the time between adjacent points becomes (1 / 50) [s].
Now, the time duration between points can be set (1 / 50), or (1 / 10)
seconds .
Exercise 2 period time is (1/10)seconds.
paper tape
Nr.Point
0 1 2
3 4
5
Length cm 0 2.5 5.0 7.5 1 0.0 12.5
Asking for the velocity of the body.
0-1 v={(2.5−0.0)/100}/(1/10)= 0.25m/s
1-2 v={(5.0−2.5)/100}/(1/10)= 0.25m/s
2-3 v=
Caution for this equipment.
(ⅰ) To set the side of the paper tape chemicals painted upwards.
(ⅱ) Discharge electrode may be corrupted by putting a tape from the other side.
(ⅲ) Turn on the power switch after all have been set.
Now, the time duration between points can be set (1 / 50), or (1 / 10)
seconds .
Exercise 2 period time is (1/10)seconds.
paper tape
Nr.Point
0 1 2
3 4
5
Length cm 0 2.5 5.0 7.5 1 0.0 12.5
Asking for the velocity of the body.
0-1 v={(2.5−0.0)/100}/(1/10)= 0.25m/s
1-2 v={(5.0−2.5)/100}/(1/10)= 0.25m/s
2-3 v= {(7.5–5.0)/100}/(1/10)=0.25m/s
Caution for this equipment.
(ⅰ) To set the side of the paper tape chemicals painted upwards.
(ⅱ) Discharge electrode may be corrupted by putting a tape from the other side.
(ⅲ) Turn on the power switch after all have been set.
Experiment 1 To measure the velocity of linear motion of the hand .
Installation points recording timer, paper tape, ruler measure
Procedure
points timer (1/10)sec
a hand
after setting, switching ON, and pulling the tape
Nr.of points
０
length(cm)
difference(cm)
distance(m)
velocity(m/s)
１
０
２
３
４
Experiment 1 To measure the velocity of linear motion of the hand.
Installation points recording timer, paper tape, ruler measure
Procedure
points timer (1/10)sec
a hand
after setting, switching ON, and pulling the tape
Nr.of points
０
length(cm)
difference(cm)
distance(m)
velocity(m/s)
１
２
9.5
０
9.5
0.095
0.95
３
20.2
10.7
0.107
1.1
４
Experiment 2
Motion of hovering soccer ball.
rotating fan
Points on paper tape
air
lined up almost evenly.
⇓
constant velocity linear motion
or, uniform motion.
Points recording timer
ball
paper tape
layer
(3) Using “Be-Spe”
Fig.5 Be-Spe
Fig.4 principle
sw1
sw2
two light switches
(ⅰ) When the body has interrupted one⇒the timer switch is ON
(ⅱ) next it interrupts the other one ⇒ the timer switch is OFF
(ⅲ) Internal computer calculates and shows the value of the
velocity.
Experiment 3 To measure the velocity of small steel balls
moving inside the tube.
Installation
“Be-Spe”, transparency tube, small steel ball
Be-Spe
Height
velocity 1st
m/s
velocity 2nd
m/s
velocity 3rd
m/s
5 cm
height
10 cm
20 cm
Experiment 3 To measure the velocity of small steel balls
moving inside the tube.
Installation
“Be-Spe”, transparency tube, small steel ball
Be-Spe
Height
velocity 1st
m/s
velocity 2nd
m/s
velocity 3rd
m/s
height
5 cm
10 cm
0.9
20 cm
1.3
1.8
2 Measurement of force
(1)
dynamics trucks
Dynamics truck
wears a four wheels
whose axle is held in
ball bearings, so wheel
rotation is very smooth.
Rolling friction is
1 / 10 or less
Dynamic friction.
close to the
constant motion
Fig.6
(2)The third law of motion = law of action and reaction
Force is that a body operates to other body. Ａ
a body A operates a force to a body B,
⇔ the body B operates a force to the body A.
Fig.7 action, reaction
Ｂ
Experiment4 To depart two trucks. Making two trucks confront each
other.
Releasing the coil spring ⇒ Two trucks detach with the same speed.
Correctly speaking, detach
with same acceleration.
Naturally, it causes in the
case mass of trucks are equal.
(3) To measure forces by a spring balance
In many cases we use a scale or a spring balance.
Especially a spring balance is often used. For example they
are used when checking action reaction law.
Fig.9
M
g
Fig.8
0.98N
Drawing together
100gw
F=Mg
Each force equals
gravity force for the body of 100 g ⇒ 0.98 [N]
⇒ roughly equals 1 [N]
100g
3 Measurement of acceleration
(1) Acceleration
Acceleration is said the variation of the velocity vector divided by
the time at a extremely short time. ａ =
𝚫ｖ
𝚫ｔ
ｖ𝟐 −ｖ𝟏
=
𝚫ｔ
Experiment5 (demonstration) An experiment of acceleration display
Installation acceleration display, plane board
Both downward and upward
(ⅰ) instant value of acceleration
(ⅱ) direction of acceleration
downward
②
(2)
Constant acceleration motion
If the velocity of a body becomes to v1 from v0
in a very short time Δt, the acceleration is
𝑎=
𝚫𝐯
𝚫𝐭
=
𝐯𝟏 −𝐯𝟎
𝚫𝐭
③
If you can measure xi, body position, at extremely short cycle
time Δt each, you can calculate the velocity. For example, if
you can measure x0, x1, x2 ,
𝐯𝟏𝟐 =
𝐱𝟐 − 𝐱𝟏
𝚫𝐭
,
𝐯𝟎𝟏 =
𝐱𝟏 − 𝐱𝟎
𝚫𝐭
Next, assuming this velocity varies between the time Δt,
then the acceleration a in the time is got.
𝐯𝟏𝟐 − 𝐯𝟎𝟏
𝑎=
𝚫𝐭
④
Experiment6
To do the same construction as Exp.1, and
pull the paper tape at a accelerated velocity. Provided that
period time is (1/10) s.
nr.of points
length
ｍ
distance
velocity m/s
difference
acceleration m/s2
0
1
0
2
3
Experiment6
To do the same construction as Exp.1, and
pull the paper tape at a accelerated velocity. Provided that
period time is (1/10) s.
nr.of points
length
ｍ
0
1
0
distance
0.105
velocity m/s
1.05
2
0.105
0.260
0.155
1.55
difference
0.55
acceleration m/s2
5.5
3
(3) To check out the relation between force and acceleration=the
second law
The acceleration is proportional to the force and inversely
proportional to mass.
Experiment7-1 pulling twice of power? and also, make truck mass be twice?
installation dynamics truck(0.50kg), spring balance, plane board,
weight(0.25kg✕2), points timer (period time is 1/10 s)
spring balance
truck
points timer
Continue pulling the truck by ways of 3 type following.
(ⅰ) Pull the truck by the balance with the dial at 0.50[N].
(ⅱ) Pull the truck by the balance with the dial at 1.0[N].
(ⅲ) Put 2 weights upon the truck, and pull with the dial at 1.0[N].
Though you pull the trucks by the balance with the dial constantly,
of course, as the trucks will be accelerated, you should make the
balance move the same movement as the trucks.
From the paper tape to calculate velocity and acceleration.
(ⅰ)one truck, 0.50N
Number
0
Distance
m
Difference
0
velocity
difference
accelerati
on
average
1
2
3
(ⅱ)one truck, 1.0N
0
0
1
2
(ⅲ)two weights, 1.0N
3
0
0
1
2
3
Though you pull the trucks by the balance with the dial constantly,
of course, as the trucks will be accelerated, you should make the
balance move the same movement as the trucks.
From the paper tape to calculate velocity and acceleration.
(ⅰ)one truck, 0.50N
1
(ⅱ)one truck, 1.0N
(ⅲ)two weights, 1.0N
Number
0
Distance
m
Difference
0
0.048 0.059 0.066
0.045 0.063 0.083
0.035 0.044
0.053
velocity
0.48
0.45
0.35
0.53
2
3
0.048 0.107 0.173
0.58
difference
0.10
accelerati
on
average
1.0
0.9
0.66
0
1
2
3
0
0
0.045
0.108
0.191
0
0.63
0.83
0.18
0.20
1.8
2.0
0.8
1.9
1
2
0.035
3
0.079 0.132
0.44
0.09
0.09
0.9
0.9
0.9
Experiment7-2 By gravity imposed on weight, to pull trucks.
It is hard pulling with constant force.
making gravitational force of weight pull a truck.
But approximately and provisionally proportional.
Installation points timer(period time 1/10 s), 50g-weight, pulley,
plane board, strap(fishing line),
timer
pulley
weight
truck
Then make the weight pull the truck in the three types below.
(ⅰ) Pull with 50g-weight
(ⅱ) Pull with 50g-weight✕2
(ⅲ) Pull a truck and 2 weights on it with 50g-weight✕2
From the paper tape to calculate velocity and acceleration.
(ⅰ) 50g-weight
Number
0
Distance
0
Difference
Velocity
Difference
Acceleratio
n
Average
1
2
(ⅱ) 100g-weight
3
0
0
1
2
(ⅲ) 2 weights 100g
3
0
0
1
2
3
Then make the weight pull the truck in the three types below.
(ⅰ) Pull with 50g-weight
(ⅱ) Pull with 50g-weight✕2
(ⅲ) Pull a truck and 2 weights on it with 50g-weight✕2
From the paper tape to calculate velocity and acceleration.
(ⅰ) 50g-weight
(ⅱ) 100g-weight
(ⅲ) 2 weights 100g
Number
0
Distance
0
Difference
0.048
0.059
0.066
0.045
0.063
0.083
0.035 0.044
0.053
Velocity
0.48
0.58
0.66
0.45
0.63
0.83
0.35
0.53
1
3
0.048 0.107
Difference
0.10
Acceleratio
n
1.0
Average
2
0.9
0.8
0.173
0
1
2
0.045
0
0.108
0.18
0.20
1.8
2.0
1.9
3
0.191
0
0
1
0.035
2
0.079
0.44
0.09
0.09
0.9
0.9
0.9
3
0.132
(4) To determine the value of the gravitational acceleration
The value can be obtained by doing the following way.
Though we can ask the value by easier method in the Exp.9.
Experiment8(demonstration) To make a
weight attached a paper tape free fall,
and to measure the distances of points.
(ⅰ) Set a paper tape through a points
timer (period time 1/10 s).
(ⅱ) Attach the paper tape end to a
weight.
(ⅲ) Fall the weight free.
(ⅳ) Calculation.
timer
weight
(5)
To determine the gravitational acceleration by fall
distance and velocity
If you free fall at the field of gravitational acceleration g,
as a = g, v0 = 0
v2 = 2 g x exists,
Therefore g = v2 / 2 x ⑤
Namely, at a point x [m] fallen
if you measure the velocity v [m / s],
g can be obtained by a calculation.
Experiment 9 Using Be-Spe
small ball
To measure the velocity of
a steel ball at the point
where the ball have fallen
a certain distance.
Installation Be-Spe,
transparency tube, small ball
x
tube
distance x
reached velocity v
⇒ g = v2 / 2 x
x [m]
1st
2nd
Be-Spe
v
v [m/s]
g [m/s2]
Experiment 9 Using Be-Spe
small ball
To measure the velocity of
a steel ball at the point
where the ball have fallen
a certain distance.
Installation Be-Spe,
transparency tube, small ball
x
tube
distance x
reached velocity v
⇒ g = v2 / 2 x
x [m]
Be-Spe
v
v [m/s]
g [m/s2]
1st
0.50
3.1
9.6
2nd
0.60
3.4
9.6
4 Measurement of angular velocity of a body moving on circular orbit
(1) The rotational or revolutionary motion a round a certain center.
central angle ⇒ angle of gyration
Fig.10 radius r
The unit is“radian”[rad].
Here, if we put θ[rad] for
Θ[degree]
2π✕Θ / 360 =θ
Θ
⑥
the arc length l for central
angle θ[rad] is following
l = rθ
⑦
60° π/3,
Angular velocity
45° ( ),
in very short time Δt
30° ( ),
the angle of gyration Δθ
the angular velocity refers to below.
(2)
360°(
180°(
90° (
)
)
)
4 Measurement of angular velocity of a body moving on circular orbit
(1) The rotational or revolutionary motion a round a certain center.
central angle ⇒ angle of gyration
Fig.10 radius r
The unit is“radian”[rad].
Here, if we put θ[rad] for
Θ[degree]
2π✕Θ / 360 =θ
Θ
⑥
the arc length l for central
angle θ[rad] is following
l = rθ
⑦
60° /3,
Angular velocity
45° ( /4 ),
in very short time Δt
30° ( /6 ),
the angle of gyration Δθ
the angular velocity refers to below.
(2)
360°( 2 )
180°( )
90° ( /2 )
Fig.11
𝚫𝛉
𝛚＝
⑧
𝚫𝐭
(3) Uniform circular motion
the motion with constant
angular velocity ω
⇓
uniform circular motion
⇓
the tangential velocity v
v = rω
during Δt
Δθ
M
v
Fig.12 tangential velocity
⑨
The number of rotation per second
n [c/s], or the number of per
minute N [rpm],
r
ω angular velocity ω
v
or one rotation period time T [s], there are relationships such as
n = 1 / T, ω = 2πn= 2π/T, As well, n = N/60
⑩
Experiment10 To measure
the velocity of the small
balls moving on circular
orbit with inertia, and to
calculate the angular
velocity.
Installation Be-Spe２pieces,
transparency tube, small
ball of steel or glass
r=0.50, the tube edge to
10 cm height.
ω= v/r =(average velocity)/0.50= 2.0✕(average velocity)
Be-Spe 1
1st m/s
2nd m/s
Be-Spe 2
difference
average
angular velocity [rad/s]
Experiment10 To measure
the velocity of the small
balls moving on circular
orbit with inertia, and to
calculate the angular
velocity.
Installation Be-Spe２pieces,
transparency tube, small
ball of steel or glass
r=0.50, the tube edge to
10 cm height.
ω= v/r =(average velocity)/0.50= 2.0✕(average velocity)
Be-Spe 1
Be-Spe 2
difference
average
angular velocity [rad/s]
1st m/s
1.3
1.1
0.2
1.2
2.4
2nd m/s
1.4
1.3
0.1
1.35
2.7
experiment 11 (calculation) To calculate the angular velocity of the
rotating top.
installation top, stop watch, movie camera, movie application and PC
procedure Taking photograph of the spinning top with the
stopwatch,and make it slow-motion replay, and measure the time
and angle of rotation,
Then, calculate the angular velocity.
Δt = 2m 01s 49 - 2m 01s 39= 0.10 s, θ =2π
Angular velocity of rotating top ω is ω= 2π/0.10 =(
Number of revolution per second ｎ is
n = 1 / T =１ / 0.10 =(
)[rad/s]
)[s-1]
experiment 11 (calculation) To calculate the angular velocity of the
rotating top.
installation top, stop watch, movie camera, movie application and PC
procedure Taking photograph of the spinning top with the
stopwatch,and make it slow-motion replay, and measure the time
and angle of rotation,
Then, calculate the angular velocity.
Δt = 2m 01s 49 - 2m 01s 39= 0.10 s, θ =2π
Angular velocity of rotating top ω is ω= 2π/0.10 =(62.8)[rad/s]
Number of revolution per second ｎ is
n = 1 / T =１ / 0.10 =( 10 )[s-1]
Experiment12 To messure the angular velocity of the rotating top
another way.
Installation top, points recording timer,
paper tape, Be-spe, transparency tape
(ⅰ) Rotating the top
attached with a tape.
(ⅱ) Rotating the top
with a transparency
tape.
points timer
5 Measurement of the central force and the moment of force, or torque
Fig.13
(1) Central force
Of the body, which doing free
movement without force in space,
the motion of the center of gravity
and the motion of rotation around
a fixed point are saved.
Simply put, those remain constant.
The value of the acceleration
𝐚=
𝐯 𝛚 𝚫𝐭
𝚫𝐭
= 𝐯𝛚 = rω2
⑫
ωΔt
r
r
ωΔｔ
⑪
The magnitude of central force of the
body m doing uniform circular motion
f = mrω2
∆v
With the
variation of
velocity during
time Δt, only
direction is
changed,
and the change is
oriented to center.
Δv=v✕ωΔt
Experiment 13 To ask
for a central force.
Installation
slim cylinder like a
Ball-pointpen barrel
(polyvinyl chloride)
Rotating the weight
(period time 0.50
seconds)
⇓
Measuring the scale
value.
Note that because a
weight rotating at high
speed it can be dangerous.
strap(length 30 cm)
10g
cylinder
2 rev. per 1sec.
balance
measurement
1st
2nd
Ave.
Calculation m = 0.010 kg, r = 0.30 m, T = 0.50 s,
π = 3.14
f = mrω2 ＝ 0.010 × 0.30×( 2 × 3.14 × 1 / 0.50 )2
=(
)
Note that because a weight
rotating at high speed it can be
dangerous.
Then instead of pulling by a
balance, you set down 50 g
weight. How you make the
rotation for weight be still in
the air?
(
)
strap(length 30 cm)
10g
cylinder
a weight 50g
measurement
1st
0.48N
2nd 0.50N
Ave. 0.49N
Calculation m = 0.010 kg, r = 0.30 m, T = 0.50 s,
π = 3.14
f = mrω2 ＝ 0.010 × 0.30×( 2 × 3.14 × 1 / 0.50 )2
=( 0.47 )N
Note that because a weight
rotating at high speed it can be
dangerous.
Then instead of pulling by a
balance, you set down 50 g
weight. How you make the
rotation for weight be still in
the air?
( approx. 2rev. per 1 sec.
)
strap(length 30 cm)
10g
cylinder
a weight 50g
(2)Think of centrifugal Fig.14
force
Receiving a central
force,
M
m
ω
⇓
Grasping a centrifugal
force.
Mrω2
mrω2
The sizes are equal.
The directions are
inverse.
(centrfugal force)
=(magnification)✕(gravity force at ground level)
m r ω2 = x×m g, therefore x =r ω2 / g
⑬
Then, by this value, you can express the centrifugal
force is x times g.
Exercise 3
To get a gravity by rotation
as large as the Earth surface
gravity.
The spinning donut-shaped
space ship of 20 m in diameter
is rotating.
10m
Gravity is similar with on
Earth. the rotating period?
The case is rω2 = g
ω2 = 9.8 / 10 = 0.98
ω = ０. ９８
T = 2π / ω = 2 × 3.14 / ０. ９８ =(
)[ｓ]
Exercise 3
To get a gravity by rotation
as large as the Earth surface
gravity.
The spinning donut-shaped
space ship of 20 m in diameter
is rotating.
10m
Gravity is similar with on
Earth. the rotating period?
The case is rω2 = g
ω2 = 9.8 / 10 = 0.98
ω = ０. ９８
T = 2π / ω = 2 × 3.14 / ０. ９８ =( 6.3
)[ｓ]
(3) A centrifugal separator
If you impose the large
gravitational force by rotating a
body.depending on slight difference
in density you can layer in fluid
each powder or fluid.
Fig.15 centrifugal separation
ω
Exercise4 (i) Putting milk for domestic use centrifugal separator and
rotating about one hour, and fat little isolating is said. Ask for the value g
taken for the milk using a 10 cm radius of rotation and rotating number
1500 rpm equipment.
x = rω2 / g =(
)
(ii) You used a centrifugal extractor of household washing machine as a
substitute for a centrifugal separator. It is said the rotating number of
the extractor is 1200 rpm, and radius of rotation is 30 cm. Ask the value g
in this rotation.
x = rω2 / g =(
)
(iii) Stuffing small jar with milk, attached 1 m strap, and we rotated it
with a rotating number of 2 rotations per second. Ask g taken for this
milk.
x = rω2 / g =(
)
(3) A centrifugal separator
If you impose the large
gravitational force by rotating a
body.depending on slight difference
in density you can layer in fluid
each powder or fluid.
Fig.15 centrifugal separation
ω
Exercise4 (i) Putting milk for domestic use centrifugal separator and
rotating about one hour, and fat little isolating is said. Ask for the value g
taken for the milk using a 10 cm radius of rotation and rotating number
1500 rpm equipment.
x = rω2 / g =( 252 )
(ii) You used a centrifugal extractor of household washing machine as a
substitute for a centrifugal separator. It is said the rotating number of
the extractor is 1200 rpm, and radius of rotation is 30 cm. Ask the value g
in this rotation.
x = rω2 / g =( 483 )
(iii) Stuffing small jar with milk, attached 1 m strap, and we rotated it
with a rotating number of 2 rotations per second. Ask g taken for this
milk.
x = rω2 / g =( 16 )
(4) Measurement of the moment of force=for the body with certain volume
If the direction of the resultant force F
is out of the center of gravity, force acts
as rotating the body.
Fig.17 a force accelerates C.G.
and rotates whole body
Fig.16
(-)
around C.G.
f
r
(+) axis of rotation
force F, Δt sec
a physical quantity of starting rotation ⇒ a moment of
force “N”.
In many cases, this is calculated around the center of gravity or
the fulcrum.
N = f1 r1 + f2 r2 + f3 r3 + ⑭ (units are [Nm] (Newton meter))
Moment of force makes vector.
direction f
the direction of moment of force of N
the direction of angular velocity
⇓
“right screw direction”⇒
angular
all turning or revolving motion
fig.18
velocity
Exercise5 To sum up moments of force.
(ⅰ) 0.5m CG 0.5m
Result force is obtainable by adding vectorially.
To put downward on fig. positive.
0.5N
1.0N
(ⅰ) 0.5 + 1.0 = 1.5 Ｎ
(ⅱ) 1.0 + (-2.0) = (
(ⅱ)
)N
2.0Ｎ
0.5m CG 0.5m
The moment of force around the point G is
obtained.
To put counterclockwise on fig. positive.
1.0N
(ⅰ) 0.5✕0.5 – 1.0✕0.5 = -0.25 Nm
(ⅱ) 1.0✕0.5 + 2.0✕0.0 =(
)Nm
Moment of force makes vector.
direction f
the direction of moment of force of N
the direction of angular velocity
⇓
“right screw direction”⇒
angular
all turning or revolving motion
fig.18
velocity
Exercise5 To sum up moments of force.
(ⅰ) 0.5m CG 0.5m
Result force is obtainable by adding vectorially.
To put downward on fig. positive.
0.5N
1.0N
(ⅰ) 0.5 + 1.0 = 1.5 Ｎ
(ⅱ) 1.0 + (-2.0) = ( −1.0
(ⅱ)
)N
2.0Ｎ
0.5m CG 0.5m
The moment of force around the point G is
obtained.
To put counterclockwise on fig. positive.
1.0N
(ⅰ) 0.5✕0.5 – 1.0✕0.5 = -0.25 Nm
(ⅱ) 1.0✕0.5 + 2.0✕0.0 =( 0.5 )Nm
Making a ruler be body, pulling by spring balances, we look actual example.
Experiment14 Acting two forces for a body in the certain size, and
making balance by third
(ⅰ)
50cm
50cm
force.
Pulling two balances
⇓
0.5N
1N
The ruler moving
⇓
Clips
For the ruler being still
(ⅱ)
2N
⇓
By the third balance
Pulling where?
1N
(the value indicated by the third balance)=(the deductive sum of two
forces)
(around every position the sum of the moment of force) = 0
(ⅰ) magnitude＝(
(ⅱ) magnitude＝(
)Ｎ, position＝length from left end＝(
)Ｎ, position＝length from left end＝(
)ｍ
)ｍ
Making a ruler be body, pulling by spring balances, we look actual example.
Experiment14 Acting two forces for a body in the certain size, and
making balance by third
(ⅰ)
50cm
50cm
force.
Pulling two balances
⇓
0.5N
1N
The ruler moving
⇓
Clips
For the ruler being still
(ⅱ)
2N
⇓
By the third balance
Pulling where?
1N
(the value indicated by the third balance)=(the deductive sum of two
forces)
(around every position the sum of the moment of force) = 0
(ⅰ) magnitude＝( 1.5 )Ｎ, position＝length from left end＝( 0.67 )ｍ
(ⅱ) magnitude＝( 1.0 )Ｎ, position＝length from left end＝( 1.0
)ｍ
Usually "moment of force" ⇔ "torque". In the case,
Symmetric around the center of gravity, or rotating axis fixed.
Exercise6
(ⅰ) wrench, or torque meter
arm length = r , force=F
F
Ex. force=20N, arm=20cm, How is torque?
torque=F× r =(
(ⅱ)
)Nm
F, wheel radius = r
Ex. torque at wheel shaft = 3000Nm,
r=0.3m, How is driving force？
torque = F×r , 3000=F×0.3
F=(
)N
Usually "moment of force" ⇔ "torque". In the case,
Symmetric around the center of gravity, or rotating axis fixed.
Exercise6
(ⅰ) wrench, or torque meter
arm length = r , force=F
F
Ex. force=20N, arm=20cm, How is torque?
torque=F× r =( 4.0
(ⅱ)
)Nm
F, wheel radius = r
Ex. torque at wheel shaft = 3000Nm,
r=0.3m, How is driving force？
torque = F×r , 3000=F×0.3
F=( 10000 )N
6 Expansion=the motion of rotating body, especially “gyroscope”
(1) The motion of rotating body is
held constant
Witheout outer forces,
i.e. in free state,
A body rotating keeps
the rotation intact.
the direction of axis
the number of revolution
Fig.20 Kendama
Fig.21 Top
Fig19 tops in free space
Earth
Fig.22 Boomerang
(2) Altering the rotational motion is "moment of force"
Fig.23 when adding the force pulling
a moment of force
the head for this side
⇓
a rotating body
initial rotation
the angular velocity
vector is made change its
direction and magnitude.
⇑
the gyro effect
torque×Δ t
case of torque vector
perpendicular
in a short time
⇓
only the direction of the
axis of rotation will
change.
rotating axis leans
(3) Actual examples of changing of rotating body on the Earth
In the case of a top rotating Fig.25 presession or pan-tilt motion
on Earth, gravity is working to direction of gravity varies over time
the direction down vertical.
At a contact point
the normal force is acting.
Fig.24 when push the axis
a top rotating clockwise
if pulling
if pushing
Moment of force acts to the direction for
this side of the paper. Angular velocity
vector varies and it leans the other side of
paper.
(4) The top supported at center of gravity or a gyroscope
Even on the Earth, the tops Fig.26 gyroscope
supported at center of gravity
are intact because those tops
are not subjected
to the moment of force.
This is the principle of
“gyroscope”
or“gyrocompass”.
Its axis of rotation is
permanently constant, so it
points the relative changing
of direction of the bodies
nearby, for example
latitude and longitude,
and a position
of an airplane and a robot.
Experiment15 To operate “space top” to make
sure the pan-tilt motion and the gyro effect.
Spacetop or Chikyuu-koma is the equipment
that is so much simplified from a gyroscope.
(ⅰ) Rotating the space top, applying force
to the axis of rotation and checking pantilt motion.
(ⅱ) Rotating the top, holding the circle part
with two fingers like the Fig., and tilting
the gimbal, then you will receive the
force perpendicular to the action original.
(ⅲ)If you can support (ⅲ)
the circle part with
bearing, fulcrum, or
swivel, the top will
be “gyroscope”.
How do you realize it?
(ⅱ)