Upload
scienceeducationexpert
View
153
Download
5
Embed Size (px)
DESCRIPTION
Introduction to circular (ballistic) and circular motion, centripetal acceleration and force. Kepler's 3rd law.
Citation preview
Curved motion
Overview
Curved motion 2
The trajectory and velocity of an object is fully determined by… 1. Its initial place 𝑠02. Its initial velocity 𝑣0
3. The present overall force field Σ 𝐹
𝑣0
𝑠0
Σ 𝐹 𝑣0
𝑣0
𝑠0 Σ 𝐹
𝑣0
𝑠0
Σ 𝐹
Uniform Σ 𝐹 ON thesame working line as 𝑣0
= Linear motion
Uniform Σ 𝐹 NOT ON thesame working line as 𝑣0
= Projectile motion
Σ 𝐹 always directed to ONEsingle point + constant 𝑣 .= Circular motion
Velocity change ∆ 𝑣 as a vector
Curved motion 3
𝑣0
Σ 𝐹
𝑣0
𝑣0
Σ 𝐹
𝑣0
Σ 𝐹
Δ 𝑣 = 𝑎Δ𝑡 =1
𝑚Σ 𝐹Δ𝑡 The velocity always changes in the direction of the overall force
Δ 𝑣
𝑣1
Δ 𝑣
𝑣1
𝑣1
𝑣2
Δ 𝑣
Δ 𝑣
Δ 𝑣
𝑣1
Verify in the pictures that always ∆ 𝑣 ∕∕ Σ 𝐹
Difficult because Σ 𝐹changes direction
during Δ𝑡So Δ𝑡 must be small
Projectile (ballistic) motion
Curved motion 4
𝑣0
ℎ1ℎ0
𝑣1
𝑣2
𝐹𝑊
Range
Trajectory
Assumption: Σ 𝐹 = 𝐹𝑊 (only uniform downwards weight, no air resistance)
Projectile motion, scalar treatment - I
Curved motion 5
𝑣0
ℎ1ℎ0
𝑣1
𝑣2
𝐹𝑊
Use the energy conservation law 𝐸0 = 𝐸1 = 𝐸2
Range
Trajectory
𝐸𝑝0 + 𝐸𝑘0 = 𝐸𝑝1 + 𝐸𝑘1 ⇒1
2𝑚𝑣0
2 + 𝑚𝑔ℎ0 =1
2𝑚𝑣1
2 + 𝑚𝑔ℎ1 ⇒1
2𝑣0
2 + 𝑔ℎ0 =1
2𝑣1
2 + 𝑔ℎ1
To calculate the maximum height, apply 𝐸0 = 𝐸1
4 unknowns, so 3 must be known
Projectile motion, scalar treatment - II
Curved motion 6
𝑣0
ℎ1ℎ0
𝑣1
𝑣2
𝐹𝑊
Range
Trajectory
𝐸𝑝0 + 𝐸𝑘0 = 𝐸𝑝2 + 𝐸𝑘2 ⇒1
2𝑚𝑣0
2 + 𝑚𝑔ℎ0 =1
2𝑚𝑣2
2 + 𝑚𝑔 ∙ 0 ⇒1
2𝑣0
2 + 𝑔ℎ0 =1
2𝑣2
2
To calculate the final velocity, apply 𝐸0 = 𝐸2
3 unknowns, so 2 must be known
Projectile motion, scalar treatment - III
Curved motion 7
𝑣0
ℎ1ℎ0
𝑣1
𝑣2
𝐹𝑊
Example: ℎ0 = 10.0𝑚, 𝑣0 = 20.0𝑚𝑠−1, 𝑣1 = 18.0𝑚𝑠−1
Range
Trajectory
Maximum height: 1
2𝑣0
2 + 𝑔ℎ0 =1
2𝑣1
2 + 𝑔ℎ1 ⇒1
220.02 + 9.81 ∙ 10.0 =
1
218.02 + 9.81 ∙ ℎ1 ⇒ ℎ1 = 13.9𝑚
Final velocity: 1
2𝑣0
2 + 𝑔ℎ0 =1
2𝑣2
2 + 𝑔ℎ2 ⇒1
220.02 + 9.81 ∙ 10.0 =
1
2𝑣2
2 ⇒ 𝑣2 = 24.4𝑚𝑠2
Important: Scalar treatment does not solve: time values & angle values & range !
Projectile motion, vector treatment - I
Curved motion 8
𝐹𝑊
𝑣
𝑠
𝑠𝑥
𝑠𝑦 𝑣𝑥
𝑣𝑦
start
origin
Resolve 𝑠, 𝑣, 𝑎, and 𝐹 in separate 𝑥 & 𝑦 direction.1. Make sure 𝑥-axis is horizontal, 𝑦-axis is vertical2. Choose origin below the start point on ℎ = 03. Choose ‘+ directions’: to right and upwards
In that case the equations of motion can be written independently for both directions:
𝑠𝑥 = 𝑠𝑥0 + 𝑣𝑥0 ∙ 𝑡 + 1 2𝑎𝑥 ∙ 𝑡2
𝑣𝑥 = 𝑣𝑥0 + 𝑎𝑥 ∙ 𝑡
𝑠𝑦 = 𝑠𝑦0 + 𝑣𝑦0 ∙ 𝑡 + 1 2𝑎𝑦 ∙ 𝑡2
𝑣𝑦 = 𝑣𝑦0 + 𝑎𝑦 ∙ 𝑡
With the correct initial values this becomes easier…
𝜃
Projectile motion, vector treatment - II
Curved motion 9
𝐹𝑊
𝑣
𝑠
𝑠𝑥
𝑠𝑦 𝑣𝑥
𝑣𝑦
start
origin
Choose:𝑠𝑥0 = 0 & 𝑠𝑦0 = ℎ0
𝑣𝑥0 = 𝑣0𝐶𝑜𝑠𝜃 & 𝑣𝑦0 = 𝑣0𝑆𝑖𝑛𝜃
𝑎𝑥 = 0 & 𝑎𝑦 = −𝑔
Now the equations reduce to:𝑠𝑥 = 𝑣𝑥0 ∙ 𝑡
𝑠𝑦 = ℎ0 + 𝑣𝑦0 ∙ 𝑡 − 1 2𝑔 ∙ 𝑡2
𝑣𝑦 = 𝑣𝑦0 − 𝑔 ∙ 𝑡
And for a horizontal projection (𝜃 = 0)𝑠𝑥 = 𝑣0 ∙ 𝑡
𝑠𝑦 = ℎ0 − 1 2𝑔 ∙ 𝑡2
𝑣𝑦 = −𝑔 ∙ 𝑡
𝜃
Projectile motion, vector treatment - III
Curved motion 10
𝐹𝑊
𝑣 = 20.0𝑚𝑠−1
origin
Exampleℎ0 = 10.0𝑚, 𝜃 = 0.0°, 𝑣0 = 20.0𝑚𝑠−1
𝑠𝑥 = 20.0 ∙ 𝑡𝑠𝑦 = 10 − 4.905 ∙ 𝑡2
𝑣𝑦 = −9.81 ∙ 𝑡
Time at the ground: set 𝑠𝑦 = 0 ⇒ 10 − 4.905 ∙ 𝑡2 = 0 ⇒ 𝑡 = 1.43𝑠
Range: 𝑠𝑥 1.43 = 20 ∙ 1.43 = 28.6𝑚
Final velocity: 𝑣𝑥 1.43 = 20𝑚𝑠−1 𝑣𝑦 1.43 = −9.81 ∙ 1.43 = −14.0𝑚𝑠−1
Magnitude: 𝑣 1.43 = 202 + 142 = 24.4𝑚𝑠−1
Angle: 𝜃 1.43 = 𝑡𝑎𝑛−1 𝑣𝑦 𝑣𝑥 = 𝑡𝑎𝑛−1 −14.0 20.0 = −35.0°
ℎ0
Projectile motion, vector treatment - IV
Curved motion 11
𝐹𝑊
𝑣 = 20.0𝑚𝑠−1
𝑣𝑥
𝑣𝑦
origin
Exampleℎ0 = 10.0𝑚, 𝜃 = 30.0°, 𝑣0 = 20.0𝑚𝑠−1
𝑣0𝑥 = 20𝐶𝑜𝑠 30° = 17.32𝑚𝑠−1
𝑣0𝑦 = 20𝑆𝑖𝑛 30° = 10.0𝑚𝑠−1𝑠𝑥 = 17.32 ∙ 𝑡
𝑠𝑦 = 10 + 10 ∙ 𝑡 − 4.905 ∙ 𝑡2
𝑣𝑦 = 10 − 9.81 ∙ 𝑡
Time at highest point: set 𝑣𝑦 = 0 ⇒ 10 − 9.81 ∙ 𝑡 ⇒ 𝑡 = 1.02𝑠
Height of highest point: 𝑠𝑦 1.02 = 10 + 10 ∙ 1.02 − 4.905 ∙ 1.022 = 15.1𝑚
Time at the ground: set 𝑠𝑦 = 0 ⇒ 10 + 10 ∙ 𝑡 − 4.905 ∙ 𝑡2 = 0
4.905𝑡2 − 10𝑡 − 10 = 0 ⇒ 𝑡 =10 ± 102 − 4 ∙ 10 ∙ −4.905
2 ∙ 4.905= 2.77𝑠
Range: 𝑠𝑥 2.77 = 17.32 ∙ 2.77 = 48.0𝑚Final velocity: 𝑣𝑥 2.77 = 17.32𝑚𝑠−1 𝑣𝑦 2.77 = 10 − 9.81 ∙ 2.77 = −17.2𝑚𝑠−1
Magnitude: 𝑣 2.77 = 17.322 + 17.22 = 24.4𝑚𝑠−1
Angle: 𝜃 2.77 = 𝑡𝑎𝑛−1 𝑣𝑦 𝑣𝑥 = 𝑡𝑎𝑛−1 −17.2 17.32 = −44.8°
ℎ0
30°
Projectile motion, vector treatment - V
Curved motion 12
𝐹𝑊
𝑣 = 20.0𝑚𝑠−1
ℎ0 = 10.0𝑚
30.0°
𝑣 = 24.4𝑚𝑠−1
ℎ1 = 15.1𝑚
𝜃 = −35.0° 𝜃 = −44.8°
28.6𝑚48.0𝑚
RESULTS of the calculation
Acceleration in circular motion
Curved motion 13
𝑣0
𝑣1Δ𝜙
𝑟
𝑟
𝑣0
𝑣1
Δ 𝑣
Δ𝜙 A
BΔ 𝑠
In a circle segment: 𝑎𝑟𝑐 𝐴𝐵 = 𝑟 ∙ Δ𝜙 (𝑟𝑎𝑑)If 𝜙 small: Δ𝑠 = 𝑟 ∙ Δ𝜙 ⇒ Δ𝜙 = ∆𝑠 𝑟
The 2 triangles are similar, so:
∆𝜙(𝑠𝑚𝑎𝑙𝑙) =Δ𝑣
𝑣=
Δ𝑠
𝑟
Divide by Δ𝑡:Δ𝑣
𝑣 ∙ ∆𝑡=
Δ𝑠
𝑟 ∙ ∆𝑡⇒ 𝑟
∆𝑣
∆𝑡= 𝑣
∆𝑠
∆𝑡⇒ 𝑟 ∙ 𝑎 = 𝑣2
𝑎 =𝑣2
𝑟Centripetal acceleration
Force in circular motion
Curved motion 14
𝑣0
𝑣1
A
B
According to Newton’s 2nd law an acceleration 𝑎requires an overall force Σ𝐹 = 𝑚𝑎
In other words:An object that orbits around a point at a distance 𝑟With a velocity 𝑣 requires an overall force:
𝐹𝑐𝑝𝑡 = 𝑚𝑎𝑐𝑝𝑡 =𝑚𝑣2
𝑟Centripetal force
𝐹𝑐𝑝𝑡
Force in circular motion – Example I
Curved motion 15
Moon circles around EarthDistance Earth - Moon 𝑟 = 384.4 ∙ 106𝑚Period 𝑇 = 27.32𝑑Mass 𝑚 = 7.35 ∙ 1022𝑘𝑔
𝑣 =2𝜋𝑟
𝑇=
2𝜋 ∙ 384.4 ∙ 106
27.32 ∙ 24 ∙ 3600= 1023𝑚𝑠−1
𝑎𝑐𝑝𝑡 =𝑣2
𝑟=
10232
384.4 ∙ 106 = 2.724 ∙ 10−3𝑚𝑠−2
𝐹𝑐𝑝𝑡 = 𝑚𝑎𝑐𝑝𝑡 = 2.002 ∙ 1020𝑁
This force is provided by the gravitational pull of the Earth
Force in circular motion – Example IIa
Curved motion 16
Fairground looping – lowest pointRadius 𝑟 = 14𝑚Velocity 𝑣 = 24𝑚𝑠−1
Mass (you) 𝑚 = 65𝑘𝑔
𝑎𝑐𝑝𝑡 =𝑣2
𝑟=
242
14= 41𝑚𝑠−2 𝐹𝑐𝑝𝑡 = 𝑚𝑎𝑐𝑝𝑡 = 2.67 ∙ 103𝑁
Weight 𝐹𝑊 = 0.64 ∙ 103𝑁 works in the wrong direction!
A normal force 𝐹𝑁 = 2.67 ∙ 103 + 0.64 ∙ 103 = 3.3 ∙ 103𝑁is required to provide the necessary 𝐹𝑐𝑝𝑡
“you feel the chair pressing you upwards” 𝐹𝑊
𝐹𝑐𝑝𝑡 = Σ 𝐹
𝐹𝑁
Force in circular motion – Example IIb
Curved motion 17
Fairground looping – highest pointRadius 𝑟 = 14𝑚Velocity 𝑣 = 5.2𝑚𝑠−1
Mass (you) 𝑚 = 65𝑘𝑔
𝑎𝑐𝑝𝑡 =𝑣2
𝑟=
5.22
14= 1.9𝑚𝑠−2 𝐹𝑐𝑝𝑡 = 𝑚𝑎𝑐𝑝𝑡 = 126𝑁
Weight 𝐹𝑊 = 638𝑁 works in the correct direction, but is too much!
An upwards normal force 𝐹𝑁 = 638 − 126 = 5.1 ∙ 102𝑁is required to compensate weight and provide the necessary 𝐹𝑐𝑝𝑡
“you are saved by the braces (belts)”
𝐹𝑊
𝐹𝑐𝑝𝑡 = Σ 𝐹
𝐹𝑁
You are 2 × 14𝑚 higher now and your velocity equals 5.2𝑚𝑠−2
in stead of 24𝑚𝑠−1 . Energy conservation law, check it yourself!
Force in circular motion – Example III
Curved motion 18
Short track skatingRadius 𝑟 = 8.0𝑚Velocity 𝑣 = 10𝑚𝑠−1
Mass 𝑚 = 65𝑘𝑔
𝑎𝑐𝑝𝑡 =𝑣2
𝑟=
102
8.0= 12.5𝑚𝑠−2
𝐹𝑐𝑝𝑡 = 𝑚𝑎𝑐𝑝𝑡 = 813𝑁
Weight 𝐹𝑊 = 638𝑁 works downwards!
Construct a free body diagram to find the contact force by the ice on the skates.
With 1cm ≜ 200𝑁 this yields 𝐹𝑖𝑐𝑒 = 1.0 ∙ 103𝑁Or calculate it algebraically with Pythagoras.
𝑡𝑎𝑛 𝜃 =𝐹𝑊
𝐹𝑐𝑝𝑡=
𝑚𝑔
𝑚𝑣2 𝑟=
𝑔𝑟
𝑣2 = 0.785 ⇒ 𝜃 = 38°
Or measure it in the free body diagram
Vectors must beon scale! 𝐹𝑊
𝐹𝑐𝑝𝑡
𝐹𝑖𝑐𝑒
𝜃
Satellite orbits
Curved motion 19
In case of satellites orbiting a planet (Earth), the 𝐹𝑐𝑝𝑡
Is provided by the gravitational force between to masses:
𝐹𝑔 = 𝐺𝑀 ∙ 𝑚
𝑟2
Newton’s law of gravitation
𝑀Typically large
𝑚Typically small
𝐹
𝑣
𝑟
Gravitational constant 𝐺 = 6.67384 ∙ 10−11𝑚3𝑘𝑔−1𝑠−2
Inverse square law
Kepler’s 3rd law
Curved motion 20
𝐹𝑔 = 𝐹𝑐𝑝𝑡 ⇒ 𝐺𝑀 ∙ 𝑚
𝑟2=
𝑚 ∙ 𝑣2
𝑟⇒ 𝐺 ∙ 𝑀 = 𝑟 ∙ 𝑣2
ℎ
𝑅𝑝
𝑟
Substitute: 𝑣 =2𝜋𝑟
𝑇
𝐺 ∙ 𝑀 = 𝑟 ∙2𝜋𝑟
𝑇
2
= 4𝜋2𝑟3
𝑇2 ⇒𝑟3
𝑇2 =𝐺𝑀
4𝜋2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑇
Attention: 𝑟 = 𝑅𝑝 + ℎ. Always add the planet’s radius to the orbiting altitude to get the correct radius
Kepler’s 3rd law - Example
Curved motion 21
𝑟3
𝑇2 =𝐺𝑀
4𝜋2 =6.67384 ∙ 10−11 ∙ 5.976 ∙ 1024𝑘𝑔
4𝜋2 = 1.010 ∙ 1013𝑚3𝑠−2
Calculate the altitude above the Earth’s surface of geostationary satellites
𝑟3 = 1010 ∙ 1013 ∙ 23.93 ∙ 3600 2 = 7.4975 ∙ 1022
𝑟 =32.355 ∙ 1023 = 42.167 ∙ 106𝑚
ℎ = 𝑟 − 𝑅𝑝 = 42.167 ∙ 106 − 6.378 ∙ 106 = 35.79 ∙ 106 𝑚wikimedia
Earth
Mass 𝑀 = 5.976 ∙ 1024𝑘𝑔
Radius 𝑅𝑝 = 6.378 ∙ 106
Siderial rotation period
𝑇 = 23.93ℎ
END
Curved motion 22
DisclaimerThis document is meant to be apprehended through professional teacher mediation (‘live in class’) together with a physics text book, preferably on IB level.