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.