1. 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”

2. 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]
　　　　　　　　　　　ｖ= Ｌ ｔ 　　　　　　　①
Exercise1 L=10m, required time t=0.50s, ask the velocity of the vehicle. V=( 10 )/( 0.50 )=( 20 )m/s

3. 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.2 points recording timer
Fig.1 marking points

4. Fig.3 principle of points recording timer
iron bar
carbon 　　　　　　　iron core
paper
　　　　　 spring 　　　 coil
tape 　　　　　　 AC
Using 50Hz ⇒(ⅰ) number of vibration of the iron bar per second is 50.
AC supply ⇒(ⅱ) periodic pointing number per second is 50.
　　　　　　　 ⇒(ⅲ) the time between adjacent points becomes (1 / 50) [s].

5. Exercise 2 period time is (1/10)seconds. paper tape
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
Length cm
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.

6. 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
Length cm
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.

7. after setting, switching ON, and pulling the tape
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 １ ２ ３ ４
length(cm)
difference(cm)
distance(m)
velocity(m/s)

8. after setting, switching ON, and pulling the tape
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 １ ２ ３ ４
length(cm)
9.5
20.2
difference(cm)
10.7
distance(m)
0.095
0.107
velocity(m/s)
0.95
1.1

9. Experiment 2 Motion of hovering soccer ball. rotating fan
Points on paper tape
air layer
lined up almost evenly. ⇓
constant velocity linear motion
　　　　　
or, uniform motion.
Points recording timer
ball paper tape

10. (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.

11. 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
Height
5 cm
10 cm
　 20 cm
velocity 1st
m/s
velocity 2nd
velocity 3rd
　 m/s

12. 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
Height
5 cm
10 cm
　 20 cm
velocity 1st
m/s
0.9
1.3
1.8
velocity 2nd
velocity 3rd
　 m/s

13. 2 Measurement of force (1) dynamics trucks Rolling friction is
Fig.6
(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

14. Force is that a body operates to other body.
(2)The third law of motion = law of action and reaction Fig.7 action, reaction
　Ａ　　　　Ｂ
Force is that a body operates to other body.
a body A operates a force to a body B,
⇔ the body B operates a force to the body A.
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.

15. (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
0.98N gw
F=Mg
100g
Fig.8
Drawing together
Each force equals
gravity force for the body of 100 g ⇒ 0.98 [N]
⇒ roughly equals 1 [N]

16. the time at a extremely short time. ａ= 𝚫ｖ 𝚫ｔ = ｖ 𝟐 − ｖ 𝟏 𝚫ｔ ②
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 = ｖ 𝟐 − ｖ 𝟏 𝚫ｔ 　 ②
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

17. 𝑎= 𝚫𝐯 𝚫𝐭 = 𝐯 𝟏 − 𝐯 𝟎 𝚫𝐭 ③ 𝐯 𝟏𝟐 = 𝐱 𝟐 − 𝐱 𝟏 𝚫𝐭 , 𝐯 𝟎𝟏 = 𝐱 𝟏 − 𝐱 𝟎 𝚫𝐭
(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 = 𝐯 𝟏 − 𝐯 𝟎 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 ,
𝐯 𝟏𝟐 = 𝐱 𝟐 − 𝐱 𝟏 𝚫𝐭 ,　𝐯 𝟎𝟏 = 𝐱 𝟏 − 𝐱 𝟎 𝚫𝐭
Next, assuming this velocity varies between the time Δt, then the acceleration a in the time is got.　　　　　
𝑎= 𝐯 𝟏𝟐 − 𝐯 𝟎𝟏 𝚫𝐭 ④　

18. Experiment6 To do the same construction as Exp
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
　　　1
　　　2
　　　3
length　ｍ
　　　　　0
distance
velocity m/s
difference
acceleration m/s2

19. Experiment6 To do the same construction as Exp
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
　　　1
　　　2
　　　3
length　ｍ
　　　　　0
0.105
0.260
distance
0.155
velocity m/s
1.05
1.55
difference
0.55
acceleration m/s2
5.5

20. (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].

21. From the paper tape to calculate velocity and acceleration.
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
(ⅱ)one truck, 1.0N
(ⅲ)two weights, 1.0N
Number
　0　 1
　2　　　　 　3
　0　　 　1 　2 　0
Distance m
Difference
velocity
difference
acceleration
average

22. From the paper tape to calculate velocity and acceleration.
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
(ⅱ)one truck, 1.0N
(ⅲ)two weights, 1.0N
Number
　0　 1
　2　　　　 　3
　0　　 　1 　2 　0
Distance m
0.048
0.107
0.173
0.045
0.108
0.191
0.035
0.079
0.132
Difference
0.059
0.066
0.063
0.083
0.044
0.053
velocity
0.48
0.58
0.66
0.45
0.63
0.83
0.35
0.44
0.53
difference
0.10
0.18
0.20
0.09
acceleration
1.0
0.8
1.8
2.0
0.9
average
1.9

23. 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 truck
weight

24. 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 　1 　2 　3
Distance
Difference
Velocity
Acceleration
Average

25. 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 　1 　2 　3
Distance
0.048
0.107
0.173
0.045
0.108
0.191
0.035
0.079
0.132
Difference
0.059
0.066
0.063
0.083
0.044
0.053
Velocity
0.48
0.58
0.66
0.45
0.63
0.83
0.35
0.44
0.53
0.10
0.18
0.20
0.09
Acceleration
1.0
0.8
1.8
2.0
0.9
Average
1.9

26. (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 timer
(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.
　　　　　　　　　　　　　　　　　　　　　　　　　weight

27. as a = g, v0 = 0 Therefore g = v2 / 2 x ⑤ v2 = 2 g x exists,
(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.

28. distance x reached velocity v ⇒ g = v2 / 2 x Experiment 9 Using Be-Spe
small ball x
tube
Be-Spe　　
　　 　　　　v
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
distance x
reached velocity v
⇒　g = v2 / 2 x
　　　x [m]
　　v [m/s]
　　g [m/s2]
1st
2nd

29. distance x reached velocity v ⇒ g = v2 / 2 x Experiment 9 Using Be-Spe
small ball x
tube
Be-Spe　　
　　 　　　　v
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
distance x
reached velocity v
⇒　g = v2 / 2 x
　　　x [m]
　　v [m/s]
　　g [m/s2]
1st
0.50
3.1
9.6
2nd
0.60
3.4

30. 2π✕Θ / 360 =θ ⑥ l = rθ ⑦ central angle ⇒ angle of gyration
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
The unit is“radian”[rad].
Fig.10 radius r
　　　 Θ
60° π/3, °( )
45° ( ), °( )
30° ( ), ° ( )
Here, if we put θ[rad] for
Θ[degree]　
2π✕Θ / 360 =θ　　⑥
the arc length l for central angle θ[rad] is following
l = rθ　　⑦
(2)　Angular velocity
in very short time Δt
the angle of gyration Δθ
the angular velocity refers to below.

31. 2π✕Θ / 360 =θ ⑥ l = rθ ⑦ central angle ⇒ angle of gyration
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
The unit is“radian”[rad].
Fig.10 radius r
　　　 Θ
60°  /3, °( 2  )
45° ( /4 ), °(  )
30° ( /6 ), ° ( /2 )
Here, if we put θ[rad] for
Θ[degree]　
2π✕Θ / 360 =θ　　⑥
the arc length l for central angle θ[rad] is following
l = rθ　　⑦
(2)　Angular velocity
in very short time Δt
the angle of gyration Δθ
the angular velocity refers to below.

32. 𝛚 ＝ 𝚫𝛉 𝚫𝐭 ⑧ v = rω ⑨ n = 1 / T, ω = 2πn= 2π/T, As well, n = N/60 ⑩
Fig during Δt Δθ
𝛚 ＝ 𝚫𝛉 𝚫𝐭 ⑧
(3)　Uniform circular motion
the motion with constant
angular velocity ω ⇓
uniform circular motion
the tangential velocity v
　 M　　　v Fig.12 tangential velocity
r v
　　ω angular velocity ω
　　 v = rω　　⑨
The number of rotation per second
n [c/s], or the number of per
minute N [rpm], 　
or one rotation period time T [s], there are relationships such as
n = 1 / T, ω = 2πn= 2π/T, As well, n = N/ ⑩

33. ω= v/r =(average velocity)/0.50= 2.0✕(average velocity)
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
2nd m/s

34. ω= v/r =(average velocity)/0.50= 2.0✕(average velocity)
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
0.1
1.35
2.7

35. 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 m 01s 39= 0.10 s, θ =2π
Angular velocity of rotating top ω is ω= 2π/0.10 =(　 　)[rad/s]
Number of revolution per second ｎ is　
n = 1 / T =１ / 0.10 =(　　　)[s-1]

36. 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 m 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]

37. (ⅰ) Rotating the top attached with a tape. (ⅱ) Rotating the top
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

38. 𝐚= 𝐯 𝛚 𝚫𝐭 𝚫𝐭 =𝐯𝛚 = rω2 ⑪ f = mrω2 ⑫
5 Measurement of the central force and the moment of force, or torque
Fig.13
(1)　Central force ∆v
ωΔt
r r
ωΔｔ
With the
variation of velocity during time Δt, only direction is changed, and the change is oriented to center.
Δv=v✕ωΔt
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　⑪
The magnitude of central force of the
body m doing uniform circular motion
f = mrω2　 ⑫

39. 2 rev. per 1sec. Experiment 13 To ask for a central force. 10g
　　　　　　strap(length 30 cm)
　10g
　　　　　　 cylinder
2 rev. per 1sec.　　
　　　　　　　 balance
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.

40. Calculation m = 0.010 kg, r = 0.30 m, T = 0.50 s, π = 3.14
measurement
1st
2nd
Ave.
Calculation m = kg, r = 0.30 m, T = 0.50 s,
π = 3.14
f = mrω2 ＝ × 0.30×( 2 × 3.14 × 1 / 0.50 )2
=( )
Note that because a weight rotating at high speed it can be dangerous.
　　　strap(length 30 cm)
10g
　　 cylinder
a weight 50g
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?
( )

41. Calculation m = 0.010 kg, r = 0.30 m, T = 0.50 s, π = 3.14
measurement
1st N
2nd N
Ave N
Calculation m = kg, r = 0.30 m, T = 0.50 s,
π = 3.14
f = mrω2 ＝ × 0.30×( 2 × 3.14 × 1 / 0.50 )2
=( )N
Note that because a weight rotating at high speed it can be dangerous.
　　　strap(length 30 cm)
10g
　　 cylinder
a weight 50g
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 )

42. m r ω2 = x×m g, therefore x =r ω2 / g ⑬
(2)Think of centrifugal
force
Fig 　　　M ω
Mrω2 ⇓ m
Receiving a central
force,
Grasping a centrifugal
force.
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.

43. 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.
Gravity is similar with on Earth. the rotating period?　
10m
The case is rω2 = g
ω2 = 9.8 / 10 = 0.98　ω = ０.９８
T = 2π / ω = 2 × 3.14 / ０.９８ =( )[ｓ]

44. 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.
Gravity is similar with on Earth. the rotating period?　
10m
The case is rω2 = g
ω2 = 9.8 / 10 = 0.98　ω = ０.９８
T = 2π / ω = 2 × 3.14 / ０.９８ =( )[ｓ]

45. in density you can layer in fluid each powder or fluid.
(3)　A centrifugal separator
Fig.15 centrifugal separation
If you impose the large gravitational force by rotating a body.depending on slight difference
　　　 ω
in density you can layer in fluid
each powder or fluid.
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 =( )

46. in density you can layer in fluid each powder or fluid.
(3)　A centrifugal separator
Fig.15 centrifugal separation
If you impose the large gravitational force by rotating a body.depending on slight difference
　　　 ω
in density you can layer in fluid
each powder or fluid.
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 =( )

47. a physical quantity of starting rotation ⇒ a moment of force “N”.
(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
around C.G.
force F, Δt sec
Fig (-) f r
(+) axis of rotation
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))

48. (ⅱ) 1.0 + (-2.0) = ( )N Moment of force makes vector.
the direction of moment of force
the direction of angular velocity ⇓
“right screw direction”⇒
all turning or revolving motion
direction f fig.18
of N
angular 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)　 = 1.5 Ｎ　
(ⅱ)　1.0 + (-2.0) = ( )N　
(ⅱ)　　　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 = Nm
　 　　 　　　　(ⅱ)　1.0✕ ✕0.0 =( )Nm　　　　

49. (ⅱ) 1.0 + (-2.0) = ( −1.0 )N Moment of force makes vector.
the direction of moment of force
the direction of angular velocity ⇓
“right screw direction”⇒
all turning or revolving motion
direction f fig.18
of N
angular 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)　 = 1.5 Ｎ　
(ⅱ)　1.0 + (-2.0) = ( − )N　
(ⅱ)　　　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 = Nm
　 　　 　　　　(ⅱ)　1.0✕ ✕0.0 =( )Nm　　　　

50. 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
force.
Pulling two balances
⇓ 　
The ruler moving ⇓
Clips For the ruler being still
By the third balance
Pulling where?
(ⅰ) magnitude＝( 　)Ｎ, position＝length from left end＝(　 　)ｍ
(ⅱ) magnitude＝( )Ｎ, position＝length from left end＝(　 　)ｍ
(ⅰ) cm　 cm
0.5N N
(ⅱ) N 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

51. 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
force.
Pulling two balances
⇓ 　
The ruler moving ⇓
Clips For the ruler being still
By the third balance
Pulling where?
(ⅰ) magnitude＝( 1.5　)Ｎ, position＝length from left end＝(　0.67　)ｍ
(ⅱ) magnitude＝( )Ｎ, position＝length from left end＝(　1.0 　)ｍ
(ⅰ) cm　 cm
0.5N N
(ⅱ) N 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

52. Usually "moment of force" ⇔ "torque". In the case,
Symmetric around the center of gravity, or rotating axis fixed.
Exercise (ⅰ) wrench, or torque meter　　　　　
arm length = r , force=F
　　　 F Ex. force=20N, arm=20cm, How is torque?
torque=F× r =( )Nm　　　　　　　　　　　　　　　　　　
(ⅱ) F, wheel radius = r
Ex. torque at wheel shaft = 3000Nm,
r=0.3m, How is driving force？
torque = F×r , =F×0.3
F=( )N 　　　　　　　　　　　　　　　　　　　

53. Usually "moment of force" ⇔ "torque". In the case,
Symmetric around the center of gravity, or rotating axis fixed.
Exercise (ⅰ) wrench, or torque meter　　　　　
arm length = r , force=F
　　　 F Ex. force=20N, arm=20cm, How is torque?
torque=F× r =( )Nm　　　　　　　　　　　　　　　　　　
(ⅱ) F, wheel radius = r
Ex. torque at wheel shaft = 3000Nm,
r=0.3m, How is driving force？
torque = F×r , =F×0.3
F=( )N 　　　　　　　　　　　　　　　　　　　

54. The motion of rotating body is held constant Witheout outer forces,
6 Expansion=the motion of rotating body, especially “gyroscope”
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
Fig19 tops in free space
Earth
Fig.20 Kendama
Fig.21 Top
Fig.22 Boomerang

55. (2) Altering the rotational motion is "moment of force"
a moment of force ⇓
a rotating body
the angular velocity vector is made change its direction and magnitude. ⇑
the gyro effect
Fig.23 when adding the force pulling
the head for this side
initial rotation
torque×Δ t
rotating axis leans
case of torque vector
perpendicular
in a short time ⇓
only the direction of the
axis of rotation will
change.

56. (3) Actual examples of changing of rotating body on the Earth
In the case of a top rotating on Earth, gravity is working to the direction down vertical.
At a contact point
the normal force is acting.
Fig.25 presession or pan-tilt motion
direction of gravity varies over time　　　　　　　　　　　　　　　
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.
Fig.24 when push the axis
a top rotating clockwise
if pulling if pushing

57. (4) The top supported at center of gravity or a gyroscope
Even on the Earth, the tops 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.
Fig.26 gyroscope

58. 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 pan-
tilt 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
the circle part with
bearing, fulcrum, or
swivel, the top will
be “gyroscope”.
How do you realize it?
(ⅲ)
(ⅱ)