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Mathematical Calculus Modeling in Improving the Teaching Performance of Shot Put

Publié en ligne: 15 Jul 2022
Volume & Edition: AHEAD OF PRINT
Pages: -
Reçu: 16 Jan 2022
Accepté: 24 Mar 2022
Détails du magazine
License
Format
Magazine
eISSN
2444-8656
Première parution
01 Jan 2016
Périodicité
2 fois par an
Langues
Anglais
Abstract

The article applies mathematical calculus modeling to analyze the force of shot put. We emphatically analyzed the relationship between the angle of the shot and the initial velocity and calculated the motion trajectory equation of the shot put. The above research results are used to assist in the analysis and help improve the teaching performance of sports shot put. The study found that different forces have different effects on the horizontal displacement of the shot. Under the comprehensive force analysis, the horizontal velocity gradually decreases with time, and under the gravity analysis, the horizontal velocity of the shot is constant. The horizontal displacement of the shot put is related to the shooting angle and initial speed. Under the comprehensive force analysis, the influence of air resistance and buoyancy on the throwing distance is different due to the different combinations of shooting angle and initial speed.

Keywords

MSC 2010

Introduction

We use CNKI journals as the search data source, and the search subject is mechanics. We use shot as a keyword to accurately match papers published in core journals from 2009 to 2020. It can be found that biomechanical methods mainly study 20 papers. We have carried out a detailed analysis of the various physical parameters before the shot, and there are roughly two purposes. The first is to enable readers to understand the biomechanical characteristics of outstanding athletes [1]. The second is to explore the best combination of shooting angle and initial speed. Two papers use a combination of particle dynamics and mathematics but explain the relationship between the shot angle and the initial velocity rather than analyze the force after the shot. Therefore, this article analyzes the movement trajectory of the shot from a new perspective of dynamics and comprehensively discusses its force process from a theoretical perspective.

Research objects and methods

This article takes the whole process from shot put to landing as the research object. We divide the operation process into two stages. The first stage is from the shot to the highest point. When the shot put is at the highest point, its vertical speed is zero. The second stage is from the highest point to landing. This article mainly uses the laws of kinetics, the differential equations of particle motion, and mathematical equations to analyze the entire operation process of the shot. The basic expression form of the formula is: d / dt(mV) = F. We substitute the data measured in the actual game into the differential equation of particle motion to obtain the horizontal displacement of the shot put under two different stress conditions [2]. We use SPSS to test the correlation and difference as the conclusion to provide the basis.

Results and analysis
The trajectory of the shot put under a comprehensive force analysis

In Newton's classical mechanic's system, the mass of an object is constant, and time and space are also constant. However, through modern physics, people have realized that the mass, time, and space of any object change with the object's speed [3]. But when the object's speed is much lower than the speed of light, the change of the object's speed on its quality, time, and space is negligible. (See Figure 1, Figure 2, Figure 3, Table 1, Table 2, Table 3). Since the speed (Vt = Vlight) of the shot put is much smaller than the speed of light when it is moving, the physical expression |Vt ± Vlight|/Vlight ≈ is. Therefore, this article does not consider the impact of changes in the shot speed.

Figure 1

The trajectory of the shot put in the air

The attributes of each symbol in Figure 1

Symbol Attributes Symbol Attributes
V0 Initial shot speed H2 Height of highest point of shot put
α Shot angle h The horizontal distance between the throwing point and the toeboard of the throwing circle
V The speed of the shot put at any moment H1 Point to drop
l The horizontal distance from the shot point to the drop point T The trajectory of the shot put in the air

Figure 2

Force during the shot put

Attributes of each symbol in Figure 2

Symbol Attributes Symbol Attributes
F Air buoyancy G Shot Put Gravity
f1 The vertical component of air resistance f2 The horizontal component of air resistance
f air resistance a Air resistance and y-axis angle

Figure 3

Acceleration during the running of the shot put

The attributes of each symbol in Figure 3

Symbol Attributes Symbol Attributes
g Acceleration of gravity a Air buoyancy acceleration
a1 Resistance horizontal acceleration a2 Drag vertical acceleration

When a particle is under the combined action of several forces (F1, F2, ⋯ Fn, we can get ma = ∑ Fi according to the second law of particle dynamics [4]. The differential equation in vector form is md2x/dt2 = ∑Fi. When the whole space is regarded as a three-dimensional whole, the shot ball moves in the xoy plane (Figure 1). Therefore, it can be concluded that the projection of the force on the shot put on the x axis is md2xdt2=Fix m{{{d^2}x} \over {d{t^2}}} = \sum {{F_{ix}}} , and the projection on the y axis is md2y/dt2 = ∑Fiy (see Table 1 for specific parameters). According to Figure 2, the x axis direction and md2x/dt2 = f2, y axis direction md2y/dt2 = f1 + GF (up phase) and md2y/dt2 = Gf1F (down phase) can be obtained. When f2 = f sin a, we take f = − μV. Resistance and speed are in the opposite direction and μ is the viscous damping coefficient. The picture on the left shows the force in the ascending phase. And the picture on the right shows the force in the descending phase [5]. The specific parameters are shown in Table 2.

Horizontal direction a2 = f2 / m. The vertical direction a1 = f1 / m, a = ρVg / m, V represents the volume (V=34πr3) \left( {V = {3 \over 4}\pi {r^3}} \right) of the shot [6]. The picture on the left is the rising section, and the picture on the right is the falling section. We take the shot position as the origin, and the y axis downward is as shown in Figure 3 (Table 3). The differential equation of motion of the shot during the ascent phase is: md2xdt2=f2=fsina=Vx,md2ydt2=f1+GF=fcosa+mgρVg=Vy+mgρVg. \matrix{ {m{{{d^2}x} \over {d{t^2}}} = {f_2} = - f\,\sin \,a = - {V_x},} \hfill \cr {m{{{d^2}y} \over {d{t^2}}} = {f_1} + G - F = f\,\cos \,a\, + mg - \rho Vg = - {V_y} + mg - \rho Vg} \hfill \cr } . when t = 0 is Vx = V0 cos α, Vy = −V0 sin α. For the above two definite integrals, we can get: V0Vx1VxdVx=0tμmdt,V0Vy1Vy+mgρVg/μdVx=0tμmdt. \int_{{V_0}}^{{V_x}} {{1 \over {{V_x}}}{d_{Vx}} = - } \int_0^t {{\mu \over m}dt} ,\int_{{V_0}}^{{V_y}} {{1 \over { - {V_y} + mg - \rho Vg/\mu }}{d_{{V_x}}} = - } \int_0^t {{\mu \over m}dt} .

In this way, we obtain the law of the change of the shot speed with time as: Vx=e(μt/m)V0cosα {V_x} = {e^{ - \left( {\mu t/m} \right)}}{V_0}\cos \,\alpha Vy=(mgρVgμ)(1eμt/m)eμt/mV0sinα {V_y} = \left( {{{mg - \rho Vg} \over \mu }} \right)\left( {1 - {e^{ - \mu t/m}}} \right) - {e^{ - \mu t/m}}\,{V_0}\,\sin \,\alpha When t = 0 is x = 0, y = 0. Do the definite integral of the change of the shot speed with time in the direction of the x, y axis to obtain 0xdx=0te(μt/m)V0cosαdt,0ydy=0t[(mgρVgμ)(1e(μt/m))e(μt/m)V0sinα]dt. \matrix{ {\int_0^x {dx = } \int_0^t {{e^{ - \left( {\mu t/m} \right)}}\,{V_0}\cos \,\alpha dt} ,} \hfill \cr {\int_0^y {dy = } \int_0^t {\left[ {\left( {{{mg - \rho Vg} \over \mu }} \right)\left( {1 - {e^{ - \left( {\mu t/m} \right)}}} \right) - {e^{^{ - \left( {\mu t/m} \right)}}}{V_0}\,\sin \,\alpha } \right]dt} } \hfill \cr } .

In this way, we obtain the trajectories of the shot put in the direction of the x, y axis as follows: x=V0mμ(1eμt/m)cosα x = {V_0}{m \over \mu }\left( {1 - {e^{ - \mu t/m}}} \right)\cos \,\alpha y=(mgρVgμ)t+mμ(mgρVgμ+V0sinα)eμt/mmμ(mgρVgμ+V0sinα) y = \left( {{{mg - \rho Vg} \over \mu }} \right)t + {m \over \mu }\left( {{{mg - \rho Vg} \over \mu } + {V_0}\,\sin \,\alpha } \right){e^{ - \mu t/m}} - {m \over \mu }\left( {{{mg - \rho Vg} \over \mu } + {V_0}\sin \,\alpha } \right)

The trajectory of the shot is put in the ascending phase in the xoy plane. During the descent phase, the force of the shot put in the x-axis direction does not change, so the shot's trajectory on the x-axis does not change [7]. In the y axis direction, although the direction of the vertical component f1 of resistance changes, it is always opposite to speed. Therefore, the differential equation of the shot on the y-axis is also md2ydt2=Gf1F=mgfcosaρVg=mgμVyρVg \matrix{ {m{{{d^2}y} \over {d{t^2}}}} \hfill & { = G - {f_1} - F} \hfill \cr {} \hfill & { = mg - f\cos a - \rho Vg} \hfill \cr {} \hfill & { = mg - \mu {V_y} - \rho Vg} \hfill \cr } . We regard the descent process of the shot put as a falling body motion that starts at rest. We take the highest point of the shot as the origin of the xoy axis coordinates, as shown in Figure 4. When t′ = 0 is Vy=0 V_y^\prime = 0 . V0Vy1Vy+(mgρVg)/μdVy=0tμmdt \int_{V_0^\prime}^{V_y^\prime} {{1 \over { - V_y^\prime + \left( {mg - \rho Vg} \right)/\mu }}{d_{{V_{y'}}}}} = \int_{0'}^t {{\mu \over m}dt'} is obtained by the definite integral of the above formula [8]. The law of the change of our shot speed with time is t′ = 0, Vy = 0. Do the definite integral of the speed of the shot put in the y axis direction with time to get 0tdy=0t(mgρVgμ)(1eμt/m)dt \int_{0'}^t {dy' = \int_{0'}^t {\left( {{{mg - \rho Vg} \over \mu }} \right)\left( {1 - {e^{ - \mu t'/m}}} \right)} } dt' . We obtain the trajectory of the shot in the y axis direction as: y=mgρVgμt+m(mgρVg)μ2(1+eμt/m) y' = {{mg - \rho Vg} \over \mu }t' + {{m\left( {mg - \rho Vg} \right)} \over {{\mu ^2}}}\left( { - 1 + {e^{ - \mu t'/m}}} \right)

Figure 4

Schematic diagram of the coordinates of the shot drop equation

The trajectory of the shot put under the action of gravity

According to the calculation process in 3.1, we divide the running process of the shot into two stages: rising and falling. In this way, two different coordinates are established respectively [9]. In this way, only the influence of gravity is considered, the horizontal direction is not affected by force, and the vertical direction is affected by gravity (see Figure 4). The law of the shot put speed changing with time in the ascending phase is: Vx=V0cosα,Vy=V0sinα+gt \matrix{ {{V_x} = {V_0}\cos \,\alpha ,} \hfill \cr {{V_y} = - {V_0}\sin \,\alpha + gt} \hfill \cr }

The trajectory equation of the shot can be obtained as: x=V0tcosα x = {V_0}t\,\cos \,\alpha y=V0tsinα+12gt2 y = - {V_0}t\,\sin \,\alpha + {1 \over 2}\,g{t^2}

The law of the change of the shot speed with time in the descent phase is: Vx=V0cosα V_x^\prime = {V_0}\,\cos \,\alpha Vy=gt V_y^\prime = gt'

The trajectory equation of the shot can be obtained as: x=V0tcosα x' = {V_0}t'\,\cos \,\alpha y=12gt2 y' = {1 \over 2}gt{'^2}

Analysis of the trajectory of shot put

From the trajectory equation of the shot put in the x-axis direction, it can be found that the throwing distance of the shot put is positively correlated with the initial velocity (V0) of the shot put and the angle of shot α. It changes with time t. When the shot is at the highest point, the vertical speed is zero [10]. Below is the top eight women's shot puts in the National Games to analyze the influence of air resistance and buoyancy on throwing distance.

Comprehensive force analysis

Considering all the forces in the ascending phase Vy = 0, the time T for the shot to reach the highest point can be obtained according to equation (2). We substitute the time T into (3) and (4) respectively to calculate the horizontal and vertical displacements during the ascent phase (Table 4).

Decline phase data

Horizontal displacement during descent/m Time spent in descending phase/s Vertical displacement during descent/m Total horizontal displacement S1/m
11.489927 1 5.098146 20.533524
10.839931 1 5.098146 19.371923
9.939937 1 5.271457 17.945968
11.369936 1 4.240563 18.881659
9.739939 1 5.035463 17.326696
10.669938 1 4.240783 17.882375
10.339938 1 4.447863 17.719364
9.779941 1 4.751445 17.078781

According to H2 = H1 + y, we substitute H2 into (6) to obtain the time T′ of the falling phase [11]. We can substitute T′ into (3) to obtain the horizontal displacement in the descending stage and thus the total displacement S1 (Table 5).

Rising phase data

Horizontal displacement in ascending phase/m Rising time/s Vertical displacement during ascent stage/m
9.043597 0.787085 3.038146
8.531992 0.787085 3.038146
8.006031 0.805437 3.181457
7.511723 0.660662 2.140563
7.586757 0.778929 2.975463
7.212437 0.675955 2.240783
7.379426 0.713678 2.497863
7.29884 0.746304 2.731445

It can be seen from Table 6 that the athlete's shot angle is lower than 45°, so the speed Vx > Vy after the shot is shot, the acceleration in the vertical direction is much greater than the horizontal acceleration [12]. This makes the horizontal displacement of the shot put greater than the vertical. Whether it is under the full force analysis or the gravity analysis, in the ascending or descending phase, the horizontal displacement shows a decreasing trend. This is consistent with the actual throwing distance. Individual data fluctuates relatively greatly, which is because there are certain errors in the actual measured initial speed and shooting angle and the calculation process.

The shooting speed and angle of the top eight women's shots puts in the National Games

Ranking Grade/m Initial shot speed/ms−1 Shot angle/(°) Shot height/m
1 20.35 13.84 33.89 2.06
2 19.38 13.31 35.48 1.98
3 18.97 12.69 38.49 2.09
4 18.41 13.09 29.72 2.1
5 17.67 12.38 38.15 2.06
6 17.52 12.56 31.85 2
7 17.3 12.48 34.13 1.95
8 17.07 12.22 36.83 2.02
Gravity calculation

Considering that gravity is applied when the ascending phase is Vy = 0, the time T for the shot to reach the highest point can be obtained according to formula (7) (Table 7). We substitute the data T into (8) and (9) to calculate the moving distance in the horizontal and vertical directions during the ascent phase (Table 8). According to H2 = H1 + y, we substitute H2 into (13) to obtain the time T′ of the falling phase. We can substitute T′ into (12) to obtain the horizontal displacement in the descending stage and then calculate the total displacement S2.

Decline phase data

Horizontal displacement during descent/m Time spent in descending phase/s Vertical displacement during descent/m Total horizontal displacement S1/m
11.713451 1.019447 5.097635 20.755531
10.963754 1.011416 5.017635 19.494315
10.304118 1.036631 5.270937 18.308807
10.571418 0.929764 4.240183 18.081877
9.868233 1.013165 5.035005 17.453717
9.920852 0.929789 4.240412 17.132075
9.845927 0.952217 4.447451 17.224113
9.625255 0.984177 4.751009 16.92287

Ascent phase data

Horizontal displacement in ascending phase/m Rising time/s Vertical displacement during ascent stage/m
9.042079 0.786952 3.037635
8.53056 0.786952 3.037635
8.004689 0.8053 3.180937
7.510458 0.66055 2.140183
7.585484 0.778797 2.975005
7.211223 0.67584 2.240412
7.378185 0.713557 2.497451
7.297614 0.746177 2.731009

According to the calculation results of 3.3.1 and 3.3.2, the absolute value |ΔS| = |S1 − S2| of the difference between the comprehensive force analysis and the gravity analysis and the ratio of the absolute value to the actual throwing distance can be obtained (Table 9).

Difference of horizontal displacement under two different forces

Rank S| = |S1 − S2| / m S| / L / %
1 0.222007 1.01
2 0.122392 0.63
3 0.362839 1.91
4 0.799782 4.34
5 0.127021 0.72
6 0.7503 4.28
7 0.495251 2.86
8 0.155911 0.91

It can be seen from Table 10 that the absolute value of the horizontal displacement difference |ΔS| is the ratio 0.6300|ΔS|L4.3400 0.63{0 \over 0} \le {{\left| {\Delta S} \right|} \over L} \le 4.34{0 \over 0} of the absolute value of the horizontal displacement difference to the actual throwing distance within the range of (0.12 ~ 0.80)m. |ΔS| and |ΔS|L {{\left| {\Delta S} \right|} \over L} of different athletes are different. This is related to the initial speed and angle of the athlete's shot. If you only find the difference ΔS of the horizontal displacement, you will find that the direction of the difference ΔS is different [13]. This is because different athletes have different combinations of initial shot speed and angle of shot under full force. Athletes are affected by resistance and buoyancy, which makes the difference between the calculated total time and the total time of gravity different in positive or negative. This makes the ΔS direction of different athletes different. Therefore, the air resistance and buoyancy of different athletes may have a positive or negative effect when the shot put is thrown.

Wilcoxon signed-rank test

Z N Asymptotically significant. (Both sides)
S2-S1 0.84 8 0.401
Conclusion

1) According to the particle dynamics equation of the shot, the horizontal velocity is related to the angle of the shot. Under the comprehensive force analysis, the horizontal velocity gradually decreases with time, and under the gravity analysis, the horizontal velocity of the shot is constant. 2) The horizontal displacement of the shot put is related to the shooting angle and initial speed. Under the comprehensive force analysis, due to the different combinations of shot angle and initial speed, the influence of air resistance and buoyancy on the throwing distance may be positive or negative. 3) The ratio between the absolute value and the actual throwing distance under the comprehensive force analysis and the gravity analysis has relatively large fluctuations. There are certain errors in the shooting angle, initial speed, and the entire calculation process.

Figure 1

The trajectory of the shot put in the air
The trajectory of the shot put in the air

Figure 2

Force during the shot put
Force during the shot put

Figure 3

Acceleration during the running of the shot put
Acceleration during the running of the shot put

Figure 4

Schematic diagram of the coordinates of the shot drop equation
Schematic diagram of the coordinates of the shot drop equation

The attributes of each symbol in Figure 3

Symbol Attributes Symbol Attributes
g Acceleration of gravity a Air buoyancy acceleration
a1 Resistance horizontal acceleration a2 Drag vertical acceleration

The attributes of each symbol in Figure 1

Symbol Attributes Symbol Attributes
V0 Initial shot speed H2 Height of highest point of shot put
α Shot angle h The horizontal distance between the throwing point and the toeboard of the throwing circle
V The speed of the shot put at any moment H1 Point to drop
l The horizontal distance from the shot point to the drop point T The trajectory of the shot put in the air

Rising phase data

Horizontal displacement in ascending phase/m Rising time/s Vertical displacement during ascent stage/m
9.043597 0.787085 3.038146
8.531992 0.787085 3.038146
8.006031 0.805437 3.181457
7.511723 0.660662 2.140563
7.586757 0.778929 2.975463
7.212437 0.675955 2.240783
7.379426 0.713678 2.497863
7.29884 0.746304 2.731445

Difference of horizontal displacement under two different forces

Rank S| = |S1 − S2| / m S| / L / %
1 0.222007 1.01
2 0.122392 0.63
3 0.362839 1.91
4 0.799782 4.34
5 0.127021 0.72
6 0.7503 4.28
7 0.495251 2.86
8 0.155911 0.91

Wilcoxon signed-rank test

Z N Asymptotically significant. (Both sides)
S2-S1 0.84 8 0.401

Ascent phase data

Horizontal displacement in ascending phase/m Rising time/s Vertical displacement during ascent stage/m
9.042079 0.786952 3.037635
8.53056 0.786952 3.037635
8.004689 0.8053 3.180937
7.510458 0.66055 2.140183
7.585484 0.778797 2.975005
7.211223 0.67584 2.240412
7.378185 0.713557 2.497451
7.297614 0.746177 2.731009

Attributes of each symbol in Figure 2

Symbol Attributes Symbol Attributes
F Air buoyancy G Shot Put Gravity
f1 The vertical component of air resistance f2 The horizontal component of air resistance
f air resistance a Air resistance and y-axis angle

The shooting speed and angle of the top eight women's shots puts in the National Games

Ranking Grade/m Initial shot speed/ms−1 Shot angle/(°) Shot height/m
1 20.35 13.84 33.89 2.06
2 19.38 13.31 35.48 1.98
3 18.97 12.69 38.49 2.09
4 18.41 13.09 29.72 2.1
5 17.67 12.38 38.15 2.06
6 17.52 12.56 31.85 2
7 17.3 12.48 34.13 1.95
8 17.07 12.22 36.83 2.02

Decline phase data

Horizontal displacement during descent/m Time spent in descending phase/s Vertical displacement during descent/m Total horizontal displacement S1/m
11.713451 1.019447 5.097635 20.755531
10.963754 1.011416 5.017635 19.494315
10.304118 1.036631 5.270937 18.308807
10.571418 0.929764 4.240183 18.081877
9.868233 1.013165 5.035005 17.453717
9.920852 0.929789 4.240412 17.132075
9.845927 0.952217 4.447451 17.224113
9.625255 0.984177 4.751009 16.92287

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