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] 　　　　　　　　　　　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

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

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].

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 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.

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)

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

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

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

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

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.

(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 100gw   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]

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

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　

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

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

(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].

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

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

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 Distance Difference   Velocity 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 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

(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  

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.

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

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

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, 360°( ) 45° ( ), 180°( ) 30° ( ), 90° ( ) 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.

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, 360°( 2  ) 45° ( /4 ), 180°(  ) 30° ( /6 ), 90° ( /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.

𝛚 ＝ 𝚫𝛉 𝚫𝐭 ⑧ v = rω ⑨ n = 1 / T, ω = 2πn= 2π/T, As well, n = N/60 ⑩ Fig.11 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/60 ⑩

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

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

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 =(　 　)[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 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]

(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

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

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.

Calculation m = 0.010 kg, r = 0.30 m, T = 0.50 s, π = 3.14 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. 　　　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? ( )

Calculation m = 0.010 kg, r = 0.30 m, T = 0.50 s, π = 3.14 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. 　　　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. )

m r ω2 = x×m g, therefore x =r ω2 / g ⑬ (2)Think of centrifugal force Fig.14 　　　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.

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]

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 =( 6.3 )[ｓ]

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 =( )

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 =( 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 )

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.16 (-)   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))

(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)　0.5 + 1.0 = 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 = -0.25 Nm 　 　　 　　　　(ⅱ)　1.0✕0.5 + 2.0✕0.0 =( )Nm　　　　  

(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)　0.5 + 1.0 = 1.5 Ｎ　 (ⅱ)　1.0 + (-2.0) = ( −1.0 )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 = -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 = 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 )Nm　　　　　　　　　　　　　　　　　　 (ⅱ) 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 　　　　　　　　　　　　　　　　　　　  

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

(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.

(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

(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

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)