Download - Mechanics - Department of Physics and Astronomy - Home 298 summer...Mechanics Chapter 2 of Essential University Physics, Richard Wolfson, 3rd Edition. Summer 2018 Prof. Sergio B. Mendes

1Prof. Sergio B. MendesSummer 2018

Mechanics

Chapter 2 of Essential University Physics, Richard Wolfson, 3rd Edition


Mechanics

DynamicsKinematics• How something moves ?? • Why something moves ??

• Geometrical description

• Mathematics

• Physical cause

• Physics


Kinematics

• Object in motion: a small particle (a point)

• 1D: along a straight direction

𝑥𝑥, 𝑡𝑡 𝑥𝑥×

𝑥𝑥 = 0

𝓞𝓞


Average Velocity

�̅�𝑣 ≡𝑥𝑥2 − 𝑥𝑥1𝑡𝑡2 − 𝑡𝑡1

Between two well-defined events.

𝑥𝑥1, 𝑡𝑡1 𝑥𝑥2, 𝑡𝑡2

Event “1” Event “2”

Average between WHAT ??

𝑥𝑥×

𝑥𝑥 = 0

𝓞𝓞

Average Velocity


Examples


Displaying the Two Events in a Plot

𝑥𝑥

𝑡𝑡

𝑥𝑥1, 𝑡𝑡1Event “1”


𝑡𝑡2𝑡𝑡1

𝑥𝑥2

𝑥𝑥1𝑡𝑡2 − 𝑡𝑡1

𝑥𝑥2 − 𝑥𝑥1

𝜃𝜃

�̅�𝑣 ≡𝑥𝑥2 − 𝑥𝑥1𝑡𝑡2 − 𝑡𝑡1

= 𝑡𝑡𝑡𝑡𝑡𝑡 𝜃𝜃


Making a Plot to Display the Particle Motion: 𝑥𝑥 𝑡𝑡

𝑡𝑡



𝑡𝑡2𝑡𝑡1

𝑥𝑥2

𝑥𝑥1𝑡𝑡2 − 𝑡𝑡1

𝑥𝑥2 − 𝑥𝑥1

�̅�𝑣 = �̅�𝑣 = �̅�𝑣 = �̅�𝑣Regardless of the different types of motion:

𝑥𝑥 𝑡𝑡


Instantaneous Velocity

𝑡𝑡


Event “2”

𝑡𝑡1

𝑥𝑥1𝜃𝜃1

𝑥𝑥2 = 𝑥𝑥1 + ∆𝑥𝑥, 𝑡𝑡2 = 𝑡𝑡1 + ∆𝑡𝑡

= lim∆𝑡𝑡→0

∆𝑥𝑥∆𝑡𝑡

𝑣𝑣 𝑡𝑡1 ≡ lim∆𝑡𝑡→0

𝑥𝑥2 − 𝑥𝑥1𝑡𝑡2 − 𝑡𝑡1

= 𝑡𝑡𝑡𝑡𝑡𝑡 𝜃𝜃1 =𝑑𝑑𝑥𝑥 𝑡𝑡 = 𝑡𝑡1

𝑑𝑑𝑡𝑡

𝑡𝑡2 − 𝑡𝑡1 = ∆𝑡𝑡

𝑥𝑥2 − 𝑥𝑥1 = ∆𝑥𝑥

𝑥𝑥 𝑡𝑡


Instantaneous Velocity𝑥𝑥 𝑡𝑡

𝑡𝑡

𝑣𝑣 𝑡𝑡 = 𝑡𝑡𝑡𝑡𝑡𝑡 𝜃𝜃 𝑡𝑡 =𝑑𝑑𝑥𝑥 𝑡𝑡𝑑𝑑𝑡𝑡

𝜃𝜃 𝑡𝑡


Evaluating Instantaneous Velocity

𝑡𝑡



𝑡𝑡2𝑡𝑡1

𝑥𝑥1

𝑥𝑥 𝑡𝑡


Acceleration


𝑡𝑡

𝑣𝑣1, 𝑡𝑡1


𝑡𝑡2𝑡𝑡1

𝑣𝑣2

𝑣𝑣1𝑡𝑡2 − 𝑡𝑡1

𝑣𝑣2 − 𝑣𝑣1

𝜙𝜙

= 𝑡𝑡𝑡𝑡𝑡𝑡 𝜙𝜙

Average Acceleration

�𝑡𝑡 ≡𝑣𝑣2 − 𝑣𝑣1𝑡𝑡2 − 𝑡𝑡1

𝑣𝑣 𝑡𝑡


Instantaneous Acceleration𝑣𝑣

𝑡𝑡


𝑡𝑡1

𝑣𝑣1𝜙𝜙1

𝑣𝑣2 = 𝑣𝑣1 + ∆𝑣𝑣, 𝑡𝑡2 = 𝑡𝑡1 + ∆𝑡𝑡

= lim∆𝑡𝑡→0

∆𝑣𝑣∆𝑡𝑡

𝑡𝑡 𝑡𝑡1 ≡ lim∆𝑡𝑡→0

𝑣𝑣2 − 𝑣𝑣1𝑡𝑡2 − 𝑡𝑡1

= 𝑡𝑡𝑡𝑡𝑡𝑡 𝜙𝜙1 =𝑑𝑑𝑣𝑣 𝑡𝑡 = 𝑡𝑡1

𝑑𝑑𝑡𝑡

𝑡𝑡2 − 𝑡𝑡1 = ∆𝑡𝑡

𝑣𝑣2 − 𝑣𝑣1 = ∆𝑣𝑣


Instantaneous Acceleration𝑣𝑣 𝑡𝑡

𝑡𝑡

𝑡𝑡 𝑡𝑡 = 𝑡𝑡𝑡𝑡𝑡𝑡 𝜙𝜙 𝑡𝑡 =𝑑𝑑𝑣𝑣 𝑡𝑡𝑑𝑑𝑡𝑡

𝜙𝜙 𝑡𝑡


𝑣𝑣 𝑡𝑡 =𝑑𝑑𝑥𝑥 𝑡𝑡𝑑𝑑𝑡𝑡

𝑥𝑥 𝑡𝑡

𝑡𝑡 𝑡𝑡 =𝑑𝑑𝑣𝑣 𝑡𝑡𝑑𝑑𝑡𝑡 =

𝑑𝑑𝑑𝑑𝑡𝑡

𝑑𝑑𝑥𝑥 𝑡𝑡𝑑𝑑𝑡𝑡

=𝑑𝑑2𝑥𝑥 𝑡𝑡𝑑𝑑𝑡𝑡2


Constant Acceleration

𝑡𝑡 𝑡𝑡 = 𝑡𝑡

�𝑡𝑡 = 𝑡𝑡 =𝑣𝑣 𝑡𝑡 − 𝑣𝑣𝑜𝑜𝑡𝑡 − 0

�𝑡𝑡 = 𝑡𝑡

𝑣𝑣 𝑡𝑡 = 𝑣𝑣𝑜𝑜 + 𝑡𝑡 𝑡𝑡



𝑣𝑣 𝑡𝑡

𝑡𝑡0

𝑣𝑣𝑜𝑜

�̅�𝑣 = 𝑣𝑣𝑜𝑜 +12𝑡𝑡 𝑡𝑡 =

𝑥𝑥 𝑡𝑡 − 𝑥𝑥𝑜𝑜𝑡𝑡 − 0

𝑣𝑣𝑜𝑜 + 𝑡𝑡 𝑡𝑡

𝑥𝑥 𝑡𝑡 = 𝑥𝑥𝑜𝑜 + 𝑣𝑣𝑜𝑜 𝑡𝑡 +12𝑡𝑡 𝑡𝑡 2



𝑣𝑣𝑜𝑜 = 0





Constant Acceleration

𝑣𝑣2 𝑡𝑡 = 𝑣𝑣𝑜𝑜2 + 2 𝑡𝑡 𝑥𝑥 𝑡𝑡 − 𝑥𝑥𝑜𝑜


0 = 𝑣𝑣𝑜𝑜 + 𝑡𝑡 𝑡𝑡𝑡

𝑡𝑡𝑡 = −𝑣𝑣𝑜𝑜𝑡𝑡

When does the velocity go to zero ?

𝑣𝑣 𝑡𝑡

𝑡𝑡

𝑣𝑣𝑜𝑜

𝑡𝑡𝑡



𝑡𝑡𝑡 = −𝑣𝑣𝑜𝑜𝑡𝑡

𝑥𝑥 𝑡𝑡𝑡 = 𝑥𝑥𝑜𝑜 + 𝑣𝑣𝑜𝑜 𝑡𝑡𝑡 +12𝑡𝑡 𝑡𝑡𝑡2

= 𝑥𝑥𝑜𝑜 −𝑣𝑣𝑜𝑜2

2 𝑡𝑡

𝑥𝑥 𝑡𝑡

𝑡𝑡𝑡𝑡𝑡 = −𝑣𝑣𝑜𝑜𝑡𝑡

𝑥𝑥 𝑡𝑡𝑡 = 𝑥𝑥𝑜𝑜 −𝑣𝑣𝑜𝑜2

2 𝑡𝑡

𝑥𝑥𝑜𝑜

At the time the velocity goes to zero, then


Constant Acceleration due to Gravity𝑥𝑥 → 𝑦𝑦

𝑦𝑦 𝑡𝑡 = 𝑦𝑦𝑜𝑜 + 𝑣𝑣𝑜𝑜 𝑡𝑡 −12𝑔𝑔 𝑡𝑡 2

𝑣𝑣 𝑡𝑡 = 𝑣𝑣𝑜𝑜 − 𝑔𝑔 𝑡𝑡

𝑣𝑣2 𝑡𝑡 = 𝑣𝑣𝑜𝑜2 − 2 𝑔𝑔 𝑦𝑦 𝑡𝑡 − 𝑦𝑦𝑜𝑜

𝑦𝑦

𝑡𝑡 𝑡𝑡 = −𝑔𝑔 𝑔𝑔 ≅ 9.8 𝑚𝑚/𝑠𝑠2


𝑡𝑡𝑡 =𝑣𝑣𝑜𝑜𝑔𝑔 𝑦𝑦 𝑡𝑡𝑡 = 𝑦𝑦𝑜𝑜 +

𝑣𝑣𝑜𝑜2

2 𝑔𝑔

𝑡𝑡𝑡

𝑦𝑦 𝑡𝑡𝑡

𝑦𝑦 𝑡𝑡 = 𝑦𝑦𝑜𝑜 + 𝑣𝑣𝑜𝑜 𝑡𝑡 −12𝑔𝑔 𝑡𝑡 2

𝑣𝑣 𝑡𝑡 = 𝑣𝑣𝑜𝑜 − 𝑔𝑔 𝑡𝑡

𝑡𝑡 𝑡𝑡 = −𝑔𝑔

𝑡𝑡𝑡

𝑦𝑦𝑜𝑜


You toss a ball straight up at 7.3 m/s; it leaves your hand at 1.5 m above the floor.

a) Find its maximum height.

b) Find when it hits the floor.

a) At the maximum height the ball is instantaneously at rest: v = 0

𝑦𝑦𝑜𝑜 = 1.5 𝑚𝑚𝑣𝑣𝑜𝑜 = 7.3 𝑚𝑚/𝑠𝑠

𝑣𝑣2 = 𝑣𝑣𝑜𝑜2 − 2 𝑔𝑔 𝑦𝑦 − 𝑦𝑦𝑜𝑜

What do we know?

𝑔𝑔 ≅ 9.8 𝑚𝑚/𝑠𝑠2

𝑦𝑦 = 𝑦𝑦𝑜𝑜 +𝑣𝑣𝑜𝑜2

2 𝑔𝑔= 4.2 𝑚𝑚

b) When it hits the floor: y = 0

𝑦𝑦 = 𝑦𝑦𝑜𝑜 + 𝑣𝑣𝑜𝑜 𝑡𝑡 −12𝑔𝑔 𝑡𝑡 2 𝑡𝑡 =

−𝑣𝑣𝑜𝑜 ± 𝑣𝑣𝑜𝑜2 + 2𝑦𝑦𝑜𝑜 𝑔𝑔−𝑔𝑔

−0.18 𝑠𝑠

+1.7 𝑠𝑠


𝑑𝑑𝑥𝑥 𝑡𝑡𝑑𝑑𝑡𝑡

= 𝑣𝑣 𝑡𝑡

𝑥𝑥 𝑡𝑡

𝑑𝑑𝑣𝑣 𝑡𝑡𝑑𝑑𝑡𝑡

= 𝑡𝑡 𝑡𝑡

�0

𝑡𝑡𝑣𝑣 𝑡𝑡 𝑑𝑑𝑡𝑡 = 𝑥𝑥 𝑡𝑡 − 𝑥𝑥𝑜𝑜

�0

𝑡𝑡𝑡𝑡 𝑡𝑡 𝑑𝑑𝑡𝑡 = 𝑣𝑣 𝑡𝑡 − 𝑣𝑣𝑜𝑜

Big Picture of Chapter 2:Kinematics in 1D



𝑥𝑥 𝑡𝑡 − 𝑥𝑥𝑜𝑜 = 𝑣𝑣𝑜𝑜 𝑡𝑡 +12𝑡𝑡 𝑡𝑡 2 = �

0

𝑡𝑡𝑣𝑣 𝑡𝑡 𝑑𝑑𝑡𝑡

�0

𝑡𝑡𝑡𝑡 𝑡𝑡 𝑑𝑑𝑡𝑡 = 𝑡𝑡 𝑡𝑡 = 𝑣𝑣 𝑡𝑡 − 𝑣𝑣𝑜𝑜

For example, in the case ofConstant Acceleration

Download - Mechanics - Department of Physics and Astronomy - Home 298 summer...Mechanics Chapter 2 of Essential University Physics, Richard Wolfson, 3rd Edition. Summer 2018 Prof. Sergio B. Mendes

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