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Maths internal assesment

β€œusing the shape of top spinner to identify the best method for calculating the volume.”

Rationale Spinning tops are among the oldest toys ever discovered. A spinning top is a toy designed to spin rapidly on the ground, the motion of which causes it to remain precisely balanced on its tip because of its rotational inertia.

Introduction I use wonder how and why there are many different ways for one solution in mathematics, so I thought that my project would be much interesting if comparing different ways to find one solution for my project. When I was thinking about finding volume the first thing came to my mind was a spinning top because I thought top spinners are one of the best example for an irregular object because it has various shape in it.Top spinners are one of the most interesting invention in the world and the most trending toy in 2018, I am going to calculate the volume of the top spin with two different method, and using the volume I am going to find the dimensions of the spinning top using calculus - optimization.

Action plan To obtain the accurate result I am using two methods to calculate volume for better clarity. water displacement. - I will be using measuring cylinder to measure 200 mg of water and the water will be poured in a container with measurements. - I will be dropping the top spinner into the container with water. - The displaced level of water will be noted down. - Subtract the original volume of water (200mg) with the displaced level of water to obtain the volume of the top spinner.

1st method :-mensuration. 30 cm Scale is used to measure the approximate dimensions of the top spinners. 1) Volume of cylinder (tip) = 𝝅 Γ— π’“π’‚π’…π’Šπ’–π’”πŸ Γ— π’‰π’†π’Šπ’ˆπ’‰π’• 2) Volume of Cylinder (bottom) = 𝝅 Γ— π’“π’‚π’…π’Šπ’–π’”πŸ Γ— π’‰π’†π’Šπ’ˆπ’‰π’• 𝝅 3) Volume of frustum( middle part) = Γ— π’‰π’†π’Šπ’ˆπ’‰π’•(π‘Ήπ‘¨π‘«π‘°π‘Όπ‘ΊπŸ + πŸ‘

π’“π’‚π’…π’Šπ’–π’” Γ— 𝑹𝑨𝑫𝑰𝑼𝑺) 4) Volume of frustum (top) =

𝝅 πŸ‘

Γ— π’‰π’†π’Šπ’ˆπ’‰π’•(π‘Ήπ‘¨π‘«π‘°π‘Όπ‘ΊπŸ + π’“π’‚π’…π’Šπ’–π’” Γ—

𝑹𝑨𝑫𝑰𝑼𝑺) 5) volume of cylinder (top) = πœ‹ Γ— π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘  2 Γ— β„Žπ‘’π‘–π‘”β„Žπ‘‘ Note :- after all these calculations add equation 1, 2, 3, 4, 5. To calculate the volume of the top spinner using mensuration in millilitres. 2nd model :- optimization I will be using optimization to find the dimensions of the top spinner without measuring tape.

Maths and calculations

This is the key that relates the shape and the number of the top spinner. This top spinner contains 5 shapes as mentioned in the diagram. I will be using the numbers to denote each shape in mensuration and optimization.

1st model Mensuration :-All the volumes are calculated in millilitres. The dimensions are measured with 30 cm scale. Key:- volume = v Radius = r Height = h Millilitres = mL 1) cylinder r = 0.5 cm h = 0.2 cm 𝑣 = πœ‹ Γ— π‘Ÿ2 Γ— β„Ž 𝑣 = 3.14 Γ— (0.5)2 Γ— 0.2 𝑣 = 0.16 π‘π‘š3 (1 π‘π‘š3 = 1 π‘šπΏ) V = 0.16 mL . 2) frustum h = 0.6 R = 1.8 R = 1.4

𝑣=

𝑣=

πœ‹ Γ— β„Ž Γ— (𝑅 2 + π‘Ÿ 2 + π‘Ÿπ‘… ) 3

3.14 Γ— 0.6 Γ— ((1.8)2 + (1.4)2 + 2.52) 3 𝑣 = 5.22 π‘π‘š3

(1 π‘π‘š3 = 1 π‘šπΏ) V = 5.22 mL

3) frustum h = 3.3 cm R = 2 cm R = 0.4 cm 𝑣=

𝑣=

πœ‹ Γ— β„Ž Γ— (𝑅 2 + π‘Ÿ 2 + π‘Ÿπ‘… ) 3

3.14 Γ— 3.3 Γ— (22 + (0.4)2 + 0.8) 3 𝑣 = 17.14 π‘π‘š3 (1 π‘π‘š3 = 1 π‘šπΏ)

V = 17.2 mL

4) cylinder r = 0.4 cm h = 0.5 cm 𝑣 = πœ‹ Γ— π‘Ÿ2 Γ— β„Ž 𝑣 = 3.14 Γ— (0.4)2 Γ— 0.5 𝑣 = 0.26 π‘π‘š3 (1 π‘π‘š3 = 1 π‘šπΏ)

𝐯 = 𝟎. πŸπŸ” 𝐦𝐋 .

5) cylinder r = 0.2 cm h = 3.2 cm 𝑣 = πœ‹ Γ— π‘Ÿ2 Γ— β„Ž 𝑣 = 3.14 Γ— (0.22 ) Γ— 3.2 𝑣 = 0.41 π‘π‘š3 (1 π‘π‘š3 = 1 π‘šπΏ)

V = 0.41 mL .

Total volume = 23.25 mL

2nd model Optimization :-Using the volume of mensuration the dimensions of the top spinner is identified and hence proved that volume of top spinner matches with the calculated dimensions. ( no scales used ). Key :- v = volume r = radius h = height A = area

1. Cylinder V = 0.16 ml 𝑣 = πœ‹ Γ— π‘Ÿ2 Γ— β„Ž 𝒉=

𝟎.πŸπŸ” π…Γ—π’“πŸ

equation 1 𝐴 = 2 Γ— πœ‹ Γ— π‘Ÿ Γ— (β„Ž + π‘Ÿ) 0.16 )+π‘Ÿ πœ‹ Γ— π‘Ÿ2 𝐴 = 2 Γ— (0.16 + πœ‹ Γ— π‘Ÿ 3 )

𝐴=2Γ—πœ‹Γ—π‘ŸΓ—(

𝐴 = 0.32π‘Ÿ βˆ’1 + 2 Γ— πœ‹ Γ— π‘Ÿ 2 𝑑𝐴 βˆ’0.32 = +4Γ—πœ‹Γ—π‘Ÿ π‘‘π‘Ÿ π‘Ÿ2 𝑑𝐴 =0 π‘‘π‘Ÿ βˆ’0.32 + 4 Γ— πœ‹ Γ— π‘Ÿ 3 =0 π‘Ÿ2 βˆ’0.32 + 4 Γ— πœ‹ Γ— π‘Ÿ 3 = 0 r = 0.3 cm equation 2 substitute equation 2 in equation 1 β„Ž=

0.16 πœ‹ Γ— (𝟎. πŸ‘πŸ )

0.16 0.2826 𝒉 = 𝟎. πŸ“πŸ•

β„Ž=

𝒗 = 𝝅 Γ— π’“πŸ Γ— 𝒉 𝒗 = πŸ‘. πŸπŸ’ Γ— (𝟎. πŸ‘)𝟐 Γ— 𝟎. πŸ“πŸ• 𝑽 = 𝟎. πŸπŸ”π’„π’ŽπŸ‘ πŸπ’„π’ŽπŸ‘ = πŸπ’Žπ’ V = 0.16 mL

2. Frustum V = 5.22 mL h = 0.6 R = 1.8 r = 1.4 Total surface area = A

𝑹= 𝑣=

𝟏. πŸ– 𝒓 𝟏. πŸ’

πœ‹ Γ— β„Ž Γ— (𝑅 2 + π‘Ÿ 2 + (𝑅 Γ— π‘Ÿ)) 3

πœ‹ 9 2 9 5.22 = Γ— β„Ž Γ— (( π‘Ÿ) + π‘Ÿ 2 + ( π‘Ÿ Γ— π‘Ÿ)) 3 7 7 πœ‹ 81 9 Γ— β„Ž Γ— ( + 1 + ) Γ— π‘Ÿ2 3 49 7

Making h the subject β„Ž= 𝒉=

𝟏.πŸπŸ• π’“πŸ

5.22 Γ— 3 Γ— 49 3.14 Γ— 193 Γ— π‘Ÿ 2

equation 1

Total surface area 𝐴 = (𝑅 + π‘Ÿ) Γ— √(𝑅 βˆ’ π‘Ÿ)2 + β„Ž2 2 9 9 1.27 2 𝐴 = ( π‘Ÿ + π‘Ÿ) Γ— √( π‘Ÿ βˆ’ π‘Ÿ) + ( 2 ) 7 7 π‘Ÿ

16 4 2 1.27 2 𝐴= π‘ŸΓ—βˆš π‘Ÿ +( 2 ) 7 49 π‘Ÿ 256 2 4 2 1.61 2 𝐴= π‘Ÿ Γ—( π‘Ÿ +( 2 ) ) 49 49 π‘Ÿ 𝐴 = 0.4π‘Ÿ 4 + 412π‘Ÿ βˆ’2 𝑑𝐴 824 = 1.6π‘Ÿ 3 βˆ’ 3 π‘‘π‘Ÿ π‘Ÿ 𝑑𝐴 =0 π‘‘π‘Ÿ 1.6π‘Ÿ 3 βˆ’ r = 2.83 cm

equation 2

824 =0 π‘Ÿ3

𝑹= 𝑹=

πŸ— 𝒓 πŸ•

πŸ— Γ— 𝟐. πŸ–πŸ‘ πŸ•

𝑹 = πŸ‘. πŸ”πŸ‘ Substitute equation 2 in equation 1 1.27 π‘Ÿ2 1.27 β„Ž= 8 β„Ž=

h = 0.16 cm 𝑉 = πœ‹ Γ— β„Ž Γ— (𝑅 2 + π‘Ÿ 2 + π‘Ÿ Γ— 𝑅 ) 𝝅 𝑽 = Γ— 𝟎. πŸπŸ” Γ— (πŸ– + πŸπŸ‘. πŸπŸ• + 𝟏𝟎. πŸπŸ•) πŸ‘ 𝑽 = πŸ“. πŸπŸ” π’„π’ŽπŸ‘ πŸπ’„π’ŽπŸ‘ = πŸπ’Žπ‘³

V = 5.26 mL

3.frustum V = 17.2 mL R = 2 cm r = 0.4 cm 𝑹=

𝟐 ×𝒓 𝟎. πŸ’

𝑹 = πŸ“π’“ πœ‹ Γ— β„Ž Γ— (𝑅 2 + π‘Ÿ 2 + (𝑅 Γ— π‘Ÿ)) 3 πœ‹ 17.2 = Γ— β„Ž Γ— (25π‘Ÿ 2 + π‘Ÿ 2 + 5π‘Ÿ 2 ) 3 πœ‹ 17.2 = Γ— β„Ž Γ— (31π‘Ÿ 2 ) 3 3 Γ— 17.2 β„Ž= πœ‹ Γ— 31 Γ— π‘Ÿ 2 𝑣=

𝒉=

𝟎.πŸ“πŸ‘ π’“πŸ

equation 1 𝐴 = (𝑅 + π‘Ÿ) Γ— √(𝑅 βˆ’ π‘Ÿ)2 + β„Ž2 2 0.53 𝐴 = (5π‘Ÿ + π‘Ÿ) Γ— √(5π‘Ÿ βˆ’ π‘Ÿ)2 + ( 2 ) π‘Ÿ 2 0.53 𝐴 = 6π‘Ÿ Γ— √16π‘Ÿ 2 + ( 2 ) π‘Ÿ

𝐴 = 6π‘Ÿ Γ— √ 𝐴=

16π‘Ÿ 6 + 0.281 π‘Ÿ4

6 Γ— √16π‘Ÿ 6 + 0.281 π‘Ÿ

36 Γ— (16π‘Ÿ 6 + 0.281) 𝐴= π‘Ÿ2 10.12 𝐴 = 576π‘Ÿ 4 + 2 π‘Ÿ 𝑑𝐴 20.24 = 2304π‘Ÿ 3 βˆ’ 3 π‘‘π‘Ÿ π‘Ÿ 𝑑𝐴 =0 π‘‘π‘Ÿ 20.24 2304π‘Ÿ 3 βˆ’ 3 = 0 π‘Ÿ 2304π‘Ÿ 6 = 20.24 π‘Ÿ6 =

20.24 2304

r = 0.45 r = 0.45 cm

equation 2 𝑹 = πŸ“π’“πŸ 𝑹 = πŸ“ Γ— 𝟎. πŸ’πŸ“ 𝑹 = 𝟐. πŸπŸ“ π’„π’Ž

substitute equation 2 in equation 1 𝒉=

𝟎. πŸ“πŸ‘ 𝟎. πŸ’πŸ“πŸ

𝒉 = 𝟐. πŸ”πŸ π’„π’Ž 𝑽 = 𝝅 Γ— 𝒉 Γ— (π‘ΉπŸ + π’“πŸ + 𝒓 Γ— 𝑹) 𝑽= V = πŸπŸ•. 𝟐 π’„π’ŽπŸ‘

𝝅 Γ— 𝟐. πŸ”πŸ Γ— ([𝟐. πŸπŸ“]𝟐 + (𝟎. πŸ’πŸ“)𝟐 + 𝟏. 𝟎𝟏) πŸ‘

πŸπ’„π’ŽπŸ‘ = πŸπ’Žπ‘³ V = 17.2 mL .

4.Cylinder v = 0.26 mL 𝒗 = 𝝅 Γ— π’“πŸ Γ— 𝒉 𝒉=

𝟎.πŸπŸ” π…Γ—π’“πŸ

equation 1

𝐴 = 2 Γ— πœ‹ Γ— π‘Ÿ Γ— (β„Ž + π‘Ÿ) 𝐴=2Γ—πœ‹Γ—π‘ŸΓ—(

0.26 )+π‘Ÿ πœ‹ Γ— π‘Ÿ2

0.52 + 2 Γ— πœ‹ Γ— π‘Ÿ 3 𝐴= π‘Ÿ 𝑑𝐴 βˆ’0.52 = +4Γ—πœ‹Γ—π‘Ÿ π‘‘π‘Ÿ π‘Ÿ2 𝑑𝐴 =0 π‘‘π‘Ÿ βˆ’0.52 +4Γ—πœ‹Γ—π‘Ÿ =0 π‘Ÿ2 0.52 4Γ—πœ‹Γ—π‘Ÿ = 2 π‘Ÿ 𝒓 = 𝟎. πŸ‘πŸ’πŸ“ π’„π’Ž

equation 2

substitute equation 2 in equation 1

𝒉=

𝟎. πŸπŸ” πŸ‘. πŸπŸ’ Γ— 𝟎. πŸ‘πŸ’πŸ“

h = 0.7 cm 𝒗 = πŸ‘. πŸπŸ’ Γ— (𝟎. πŸ‘πŸ’πŸ“)𝟐 Γ— 𝟎. πŸ• 𝒗 = 𝟎. πŸπŸ” π’„π’ŽπŸ‘ πŸπ’„π’ŽπŸ‘ = πŸπ’Žπ’ v = 0.26 mL .

5.Cylinder v = 0.41 mL 𝒗 = 𝝅 Γ— π’“πŸ Γ— 𝒉 𝒉=

𝟎. πŸ’πŸ 𝝅 Γ— π’“πŸ

𝐴 = 2 Γ— πœ‹ Γ— π‘Ÿ Γ— (β„Ž + π‘Ÿ) 𝐴=2Γ—πœ‹Γ—π‘ŸΓ—( 𝐴=

0.41 )+π‘Ÿ πœ‹ Γ— π‘Ÿ2

0.82 + 2 Γ— πœ‹ Γ— π‘Ÿ2 π‘Ÿ

𝑑𝐴 = βˆ’0.82π‘Ÿ βˆ’2 + 4 Γ— πœ‹ Γ— π‘Ÿ π‘‘π‘Ÿ 𝑑𝐴 =0 π‘‘π‘Ÿ βˆ’0.82π‘Ÿ βˆ’2 + 4 Γ— πœ‹ Γ— π‘Ÿ = 0 βˆ’0.82 +4Γ—πœ‹Γ—π‘Ÿ =0 π‘Ÿ2

π‘Ÿ3 = r = 0.41

0.82 12.56

equation 2

substitute equation 2 in equation 1 𝒉=

𝟎. πŸ’πŸ πŸ‘. πŸπŸ’ Γ— 𝟎. πŸ’πŸπŸ

β„Ž = 0.78 π‘π‘š 𝒗 = πŸ‘. πŸπŸ’ Γ— (𝟎. πŸ’πŸ)𝟐 Γ— 𝟎. πŸ•πŸ– 𝒗 = 𝟎. πŸ’πŸ π’„π’ŽπŸ‘ πŸπ’„π’ŽπŸ‘ = πŸπ’Žπ‘³ v = 0.41 mL .

Conclusion and discussion Firstly starting with mensuration the five shapes I have used is mentioned below.

1) 2) 3) 4) 5)

Volume of the cylinder is 0.16 mL . volume of the frustum is 5.22 mL . volume of the frustum is 17.2 mL . volume of the cylinder is 0.26 mL . volume of the cylinder is 0.41 mL . Total volume is 23.25 mL .

secondly I have used my optimization method to calculate the dimensions without measuring tape, and I have also considered that rate change of surface area of all shapes of the top spinner. 1) Dimensions of cylinder :Radius = 0.3 cm Height = 0.57 cm 2) Dimensions of frustum :Radius 1 (R) = 1.8 cm Radius 2 (r) = 1.4 cm Height = 0.6 cm 3) Dimensions of frustum :Radius 1 (R) 2 cm Radius 2 (r) = 0.4 cm Height = 2.62 cm 4) Dimensions of cylinder :Radius = 0.345 cm Height = 0.7 cm 5) Dimensions of cylinder :Radius = 0.41 cm Height = 0.78 cm Finally I have concluded that the volume I have found in two methods matched and the volume of the top spinner remains the same even if there is change in the surface area.

Limitations Finally to be more accurate I have used water displacement method to check if the volume matches with the volume obtained in mensuration and optimization, but there was some error which is calculated and shown below. Water displacement :-I used 1000 ml measuring cylinder so that my top spinner would fit the container. 1) I filled 200 mL of water 2) Dropped the top spinner in into it and measured the displaced water level. 3) The water displaced to 222.25 mL. 4) π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ π‘£π‘œπ‘™π‘’π‘šπ‘’ – π‘“π‘–π‘›π‘Žπ‘™ π‘£π‘œπ‘™π‘’π‘šπ‘’ 5) 𝟐𝟎𝟎 π‘šπ‘™ βˆ’ 𝟐𝟐𝟐. πŸπŸ“ π‘šπΏ 6) The answer for volume calculated is 22.5 mL.

Error percentage =

23.25βˆ’22.25 22.25

Γ— 100

= 4.45 %

Bibliography 1) https://math.stackexchange.com/questions/74440/optimization-volume-of-a-box 2) https://www.maplesoft.com/support/help/maple/view.aspx?path=MathApps%2FOp timizationVolume. 3) www.ncert.nic.in/ncerts/l/hemh111.pdf 4) https://www.pdfcoke.com/document/341290679/Frustum-Cone-Optimization 5) http://jwilson.coe.uga.edu/EMAT6680Fa2013/Kar/EMAT%206690/Essay%202/Frust um.pdf

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