Measurement of kinematic quantities through simple experiments

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]

　　　　　　　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

Fig.2 points recording timer

(2) Using a points recording timer

(ⅱ) marking points on a body periodically at very short time

(ⅰ)measuring time very short as possible

Fig.1 marking points

Fig.3 principle of points recording timer

 iron bar

carbon 　　　　　　　 iron core paper 　　　　　 spring 　　　 coil papertape 　　　　　　 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].

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/s1-2 v={(5.0−2.5)/100}/(1/10)= 0.25m/s2-3 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/s1-2 v={(5.0−2.5)/100}/(1/10)= 0.25m/s2-3 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.

{(7.5–5.0)/100}/(1/10)=0.25m/s

Nr.of points ０ １ ２ ３ ４length(cm) ０        difference(cm)

distance(m)            velocity(m/s)            

Experiment 1 To measure the velocity of linear motion of the hand .Installation 　 points recording timer, paper tape, ruler measureProcedure 　 points timer (1/10)sec

a hand 　 after setting, switching ON, and pulling the

tape

Nr.of points ０ １ ２ ３ ４length(cm) ０ 9.5 20.2    difference(cm)

  9.5

10.7      

distance(m)   0.095

0.107

velocity(m/s)   0.95

1.1      

Experiment 1 To measure the velocity of linear motion of the hand.Installation 　 points recording timer, paper tape, ruler measureProcedure 　 points timer (1/10)sec

a hand 　 after setting, switching ON, and pulling the

tape

Points recording timer

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.

ball paper tape

(3) Using “Be-Spe”

Fig.4 principle       sw1 sw2 two light

switches

Fig.5 Be-Spe

(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

Height 5 cm 10 cm 　 20 cmvelocity 1st m/s

velocity 2nd m/s

velocity 3rd

　 m/s     

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 cmvelocity 1st m/s

0.9

1.3

1.8

velocity 2nd m/s

velocity 3rd

　 m/s     

2 Measurement of force

Dynamics truck wears a four wheels 　whose axle is held in ball bearings, so

wheelrotation is very

smooth.

Rolling friction is 1 / 10 or less

Dynamic friction.   close to the constant motion

(1) 　 dynamics trucks Fig.6

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.

(2)The third law of motion = law of action and reaction Fig.7 action, reactionForce 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.

　　　　　Ａ　　　　Ｂ　　　　　　　　　　　　　　　　　

(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.8

Drawing together   Each force equals

Fig.9 M g 0.98N 100gw

   F=Mg  

100g   gravity force for the body of 100 g ⇒ 0.98 [N]

⇒ roughly equals 1 [N]

3 Measurement of acceleration(1) Acceleration

Acceleration is said the variation of the velocity vector divided by the time at a extremely short time. 　 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

(2) 　 Constant acceleration motionIf the velocity of a body becomes to v1

from v0 in a very short time Δt, the acceleration is 　　　　　　　 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

𝚫𝐭Next, assuming this velocity varies between the time Δt, then the acceleration a in the time is got. 　　　　　

𝑎=𝐯𝟏𝟐−𝐯𝟎𝟏

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 0 　　　 1 　　　 2 　　　 3

length 　ｍ 　　　　　 0      

distance          

velocity m/s          

difference        acceleration m/s2        

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 0 　　　 1 　　　 2 　　　 3

length 　ｍ 　　　　　0

0.105 0.260  

distance   0.105 0.155

velocity m/s   1.05 1.55    

difference   0.55    acceleration m/s2   5.5    

(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 (ⅱ)one truck, 1.0N (ⅲ)two weights, 1.0N

Number 　0　

1 　2 　　 　　

　 3 　0 　　

　 1 　 2 　 3 　 0 1 　 2 　 3

Distance m

　0

      　 0       　 0      

Difference

velocity                              

difference

acceleration

average      

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 　 3 　 0 1 　 2 　 3

Distance m

　 0 0.048

0.107

0.173

　 0 0.045 0.108 0.191

　 0 0.035 0.079

0.132

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 0.44 0.53  

difference

  0.10     0.18 0.20     0.09 0.09  

acceleration

  1.0 0.8     1.8 2.0     0.9 0.9  

average 0.9 1.9 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 truck

    weight

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 　0

　 1 　 2 　 3 　0

　 1 　 2 　 3

Distance 　 0 　 0 　 0

Difference    

Velocity    

Difference  

Acceleration

Average

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 　0

　 1 　 2 　 3 　0

　 1 　 2 　 3

Distance 　 0 0.048 0.107 0.173 　 0 0.045 0.108 0.191 　 0 0.035 0.079 0.132

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 0.44 0.53  

Difference   0.10     0.18 0.20     0.09 0.09  

Acceleration

  1.0 0.8     1.8 2.0     0.9 0.9  

Average 0.9 1.9 0.9

(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  

(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 　⑤Namely, 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

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

small ball

x tube　　

Be-Spe 　　　　 　　　　 v

  　　　 x [m] 　　 v [m/s] 　　 g [m/s2]1st      

2nd      

distance x reached velocity

v ⇒ 　 g = v2 / 2 x

Experiment 9 　 Using Be-Spe

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

small ball

x tube　　

Be-Spe 　　　　 　　　　 v

  　　　 x [m] 　　 v [m/s] 　　 g [m/s2]1st 0.50 3.1 9.6

2nd 0.60 3.4 9.6

distance x reached velocity

v ⇒ 　 g = v2 / 2 x

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 gyrationThe unit is“radian”[rad].Here, if we put θ[rad] for Θ[degree] 　

2π✕Θ / 360 =θ 　　⑥the arc length l for

central angle θ[rad] is following l = rθ 　　⑦

(2) 　 Angular velocityin very short time Δtthe angle of gyration Δθ

the angular velocity refers to below.

Fig.10 radius r　　　 Θ

       60° π/3, 360°( ) 45° ( ), 180°( ) 30° ( ), 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 gyrationThe unit is“radian”[rad].Here, if we put θ[rad] for Θ[degree] 　

2π✕Θ / 360 =θ 　　⑥the arc length l for

central angle θ[rad] is following l = rθ 　　⑦

(2) 　 Angular velocityin very short time Δtthe angle of gyration Δθ

the angular velocity refers to below.

Fig.10 radius r　　　 Θ

       60° /3, 360°( 2 ) 45° ( /4 ), 180°( ) 30° ( /6 ), 90° ( /2 ) 

⑧Fig.11 during Δt

Δθ　

(3) 　 Uniform circular motion the motion with constant

angular velocity ω ⇓uniform circular motion ⇓ the tangential velocity v

　　 v = rω 　　⑨The number of rotation per

second n [c/s], or the number of perminute N [rpm], 　

or one rotation period time T [s], there are relationships such as

　 M 　　　 v Fig.12 tangential velocity

　     r

v　　 ω angular velocity ω

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-Spe ２

pieces, 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

Experiment10 To measure

the velocity of the small balls moving on circular orbit with inertia, and to calculate the angular velocity. 　Installation Be-Spe ２

pieces, 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 PCprocedure 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 =2πAngular 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]

experiment 11 (calculation) To calculate the angular velocity of the rotating top.

installation 　 top, stop watch, movie camera, movie application and PCprocedure 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 =2πAngular 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

(1) 　 Central force Fig.13

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

∆v ωΔt  r r    ωΔ ｔ

With the variation of velocity during time Δt, only direction is changed, and the change is oriented to center.

Δv=v✕ωΔt

= rω2 　⑪

The magnitude of central force of the

body m doing uniform circular motion

f = mrω2 　 ⑫

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

Note that because a weight rotating at high speed it can be dangerous.

measurement

1st 2nd Ave.

Calculation m = 0.010 kg, r = 0.30 m, T = 0.50 s, π = 3.14f = mrω2 ＝ 0.010 × 0.30×( 2 × 3.14 × 1 /

0.50 )2 =( )

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

Note that because a weight rotating at high speed it can be dangerous.

measurement

1st 0.48N 2nd 0.50N Ave. 0.49N

Calculation m = 0.010 kg, r = 0.30 m, T = 0.50 s, π = 3.14f = mrω2 ＝ 0.010 × 0.30×( 2 × 3.14 × 1 /

0.50 )2 =( 0.47 )N

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 forceReceiving a central force,

⇓

Fig.14 　　　 M   

ω 

Mrω2

m

mrω2

(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.

Grasping a centrifugal

force. The sizes are equal.The directions are inverse. 

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? 　

The case is rω2 = g

ω2 = 9.8 / 10 = 0.98 　 ω =

T = 2π / ω = 2 × 3.14 / =( )[ ｓ ]

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　  10m

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? 　

The case is rω2 = g

ω2 = 9.8 / 10 = 0.98 　 ω =

T = 2π / ω = 2 × 3.14 / =( 6.3 )[ ｓ ]

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　  10m

(3) 　 A centrifugal separator

If you impose the large gravitational force by rotating a body.depending on slight differencein 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 differencein 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 volumeIf the direction of the resultant force F is out of the center of gravity, force

acts

Fig.16 (-)  f r   (+) axis of rotation

as rotating the body. Fig.17 a force accelerates C.G.

and rotates whole body around C.G.     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. the direction of moment of

force the direction of angular

velocity ⇓ “right screw direction”⇒all turning or revolving motion

direction f fig.18of 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) 　 0.5 + 1.0 = 1.5 Ｎ　 (ⅱ) 　 1.0 + (-2.0) = ( )N 　

(ⅱ) 　　　 2.0 Ｎ　　　　　

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. the direction of moment of

force the direction of angular

velocity ⇓ “right screw direction”⇒all turning or revolving motion

direction f fig.18of 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) 　 0.5 + 1.0 = 1.5 Ｎ　 (ⅱ) 　 1.0 + (-2.0) = ( −1.0 )N 　

(ⅱ) 　　　 2.0 Ｎ　　　　　

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 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) 50cm 　 50cm    0.5N 1N   (ⅱ) 2N   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

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 ＝ ( 1.0 ) Ｎ , position ＝ length from left end ＝ ( 　1.0 　 ) ｍ

(ⅰ) 50cm 　 50cm    0.5N 1N   (ⅱ) 2N   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

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 =( )Nm 　　　　　　　　　　　　　　　　　　

(ⅱ) F, wheel radius = rEx. 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 )Nm 　　　　　　　　　　　　　　　　　　

(ⅱ) F, wheel radius = rEx. 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

Fig19 tops in free space       

Earth

Fig.20 Kendama Fig.21 Top Fig.22 Boomerang

(2) Altering the rotational motion is "moment of force"Fig.23 when adding the force pulling the head for this side  initial rotation  

torque×Δ t      

rotating axis leans

a moment of force ⇓a rotating body

the angular velocity vector is made change its direction and magnitude. ⇑the gyro effect

case of torque vector

perpendicular in a short time ⇓

only the direction of the

axis of rotation willchange.

(3) Actual examples of changing of rotating body on the Earth

Fig.25 presession or pan-tilt motiondirection 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

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.

(4) The top supported at center of gravity or a gyroscope

Fig.26 gyroscope Even on the Earth, the tops supported at center of gravity are intact because those tops are not subjectedto 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

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?

(ⅱ)(ⅲ)