Chapter 5 and 8 PowerPoint, Mr. Mardit

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Transcript Chapter 5 and 8 PowerPoint, Mr. Mardit

Chapter 5: Circular Motion;
Gravitation
Chapter 8: Rotational Motion
5-1: Kinematics of Circular Motion
β€’ Uniform circular motion: movement in a circle at constant speed
1- Is the object accelerating? YES/NO ? WHY?
HINT: π‘Ž =
βˆ†π‘£
βˆ†π‘‘
2- Direction of π‘Ž ?
HINT: What is the direction of βˆ†π‘£? (What is the direction of 𝑣2 and 𝑣1 ?)
Remember vectors add tip to tail!
𝑣2
3- π‘Žπ‘π‘’π‘›π‘‘π‘Ÿπ‘–π‘π‘’π‘‘π‘Žπ‘™ =
; proof see text page 107
π‘Ÿ
5-1: Kinematics of Circular Motion
β€’ π‘“π‘Ÿπ‘’π‘žπ‘’π‘’π‘›π‘π‘¦ = 𝑓 =
# π‘œπ‘“ π‘Ÿπ‘’π‘£π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›π‘ /𝑐𝑦𝑐𝑙𝑒𝑠
π‘ π‘’π‘π‘œπ‘›π‘‘
β€’ π‘π‘’π‘Ÿπ‘–π‘œπ‘‘ = 𝑇 = time required for one revolution/cycle
‒𝑇 =
1
𝑓
β€’ Examples: Earth, moon, heart rate, waves breaking on the shore,
….Can you name some more….?
5-2: Applying N2L to Uniform Circular Motion
β€’ Remember Newton’s 2nd Law:
β€’
πΉπ‘œπ‘Ÿπ‘π‘’π‘  = 𝐹𝑛𝑒𝑑 = π‘šπ‘Ž
β€’ Let’s apply N2L to an object moving in a circle at constant speed?
β€’ πΉπ‘œπ‘Ÿπ‘π‘’π‘  = 𝐹𝑛𝑒𝑑 = π‘šπ‘Žπ‘π‘’π‘›π‘‘π‘Ÿπ‘–π‘π‘’π‘‘π‘Žπ‘™
β€’ 1-What is the direction of 𝐹𝑛𝑒𝑑 ?
β€’ 2-Brainstorm some possible sources of 𝐹𝑛𝑒𝑑 :
5-6: Newton’s Law of Universal Gravitation
β€’πΉπ‘”π‘Ÿπ‘Žπ‘£π‘–π‘‘π‘¦ =
πΊπ‘š1 π‘š2
π‘Ÿ2
β€’ The above gives only the magnitude of the force of gravity NOT the direction, gravity is always
an attractive force that acts along the line joining π‘š1 and π‘š2
β€’ G=6.67 x 10-11 units??? οƒ What are the units of G?
5-7: Applying N2L to Gravity; g
β€’ Remember Newton’s 2nd Law:
β€’
πΉπ‘œπ‘Ÿπ‘π‘’π‘  = 𝐹𝑛𝑒𝑑 = π‘šπ‘Ž
β€’ Let’s apply N2L to two masses π‘š1 π‘Žπ‘›π‘‘ π‘š2
β€’
πΉπ‘œπ‘Ÿπ‘π‘’π‘  = 𝐹𝑛𝑒𝑑 = πΉπ‘”π‘Ÿπ‘Žπ‘£π‘–π‘‘π‘¦ =
πΊπ‘š1 π‘š2
=π‘š2 π‘Ž
π‘Ÿ2
β€’ Now let’s say π‘š1 is the Earth and π‘š2 is a person standing on the surface of the Earth:
πΊπ‘šπΈπ‘Žπ‘Ÿπ‘‘β„Ž π‘šπ‘π‘’π‘Ÿπ‘ π‘œπ‘›
π‘šπ‘π‘’π‘Ÿπ‘ π‘œπ‘› π‘Ž=
; π‘š2 cancels; substitute
2
π‘Ÿ
πΊπ‘šπΈπ‘Žπ‘Ÿπ‘‘β„Ž
=
π‘Ÿ2
β€’ π‘Ž=
2
Nm
6.67×10βˆ’11 kg2
×(5.98×1024 π‘˜π‘”)
(6.371×106 π‘šπ‘’π‘‘π‘’π‘Ÿπ‘ )
β‰… 9.8 π‘š
𝑠2
≑𝑔
Polar Plot: (r,ΞΈ)
Good news:
We will only be studying rotational motion of CONSTANT radius.
Why do we care?
Imagine you are walking in a circle, riding a bicycle, twisting a wrench,
torqueing a screw driver, rolling a ball, or doing a cart wheel. All of these
consist of motion in both the x and y– which must be resolved into twoβ€”one
dimensional problems before they can be analyzed.
However, using polar coordinatesβ€”when the radius is held constantβ€”
only one variable changes: ΞΈ
YESSSSS! Much Simpler!
2πœ‹ π‘Ÿπ‘Žπ‘‘π‘–π‘Žπ‘›π‘  = 360 π‘‘π‘’π‘”π‘Ÿπ‘’π‘’π‘ 
Example:
The hockey puck spins 185 times between defending zone and
the attacking zone. How many radians? How many degrees?
For every linear (straight line) physical quantity, there is a rotational analogue:
To review the linear physical quantities to date:
x (distance), v, a, m, F, KE, p,
Position οƒ  Velocity οƒ  Acceleration
Ο‰ = angular velocity =
ΞΈ2 βˆ’ΞΈ1
𝑑2 βˆ’π‘‘1
compare with linear velocity=
Ξ± = angular acceleration =
compare with
=
βˆ†ΞΈ
βˆ†π‘‘
π‘₯2 βˆ’π‘₯1
𝑑2 βˆ’π‘‘1
Ο‰2 βˆ’Ο‰1
𝑑2 βˆ’π‘‘1
≑ Ο‰ (omega)
=
βˆ†π‘₯
βˆ†π‘‘
=
≑𝑣
βˆ†Ο‰
βˆ†π‘‘
≑ Ξ± (π‘Žπ‘™π‘β„Žπ‘Ž)
π‘₯2 βˆ’π‘₯1
linear acceleration=
𝑑2 βˆ’π‘‘1
=
βˆ†π‘₯
βˆ†π‘‘
≑𝑣
Examples: ΞΈ, Ο‰, Ξ±
1- A small spider walks an arc length of constant radius
around a sun dial from 1 o’clock to 8 o’clock. What is the βˆ†ΞΈ of the
ant? ( in β€œhours”, in degrees, in radians) The journey takes 2.6
minutes what is the Ο‰ of the spider? (radians/second)
2- Timmy hops on his bicycle and starts pedaling. He
completes 4.2 revolutions of the rear tire in the first 3 seconds.
What is the Ο‰π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ of the tire? One minute later the rear tire
completes 15.6 revolutions in 0.6 seconds. What is the Ο‰π‘“π‘–π‘›π‘Žπ‘™ of the
tire? What is the angular acceleration Ξ± of the tire?
Remember this equation from middle school:
Circumference of a circle = 2PI *r
…well that was just a special case of the following:
Manipulating l=rΞΈ to connect linear and angular quantities; a little bit of calculus (PRODUCT RULE)
𝑑
𝑓 𝑑 ×𝑔 𝑑
𝑑𝑑
𝑑𝑓
𝑑𝑔
=
× π‘” 𝑑 + 𝑓(𝑑) ×
𝑑𝑑
𝑑𝑑
l=r ΞΈ
𝑑
𝑑
𝑙 =
π‘ŸΞΈ
𝑑𝑑
𝑑𝑑
𝑑𝑙
π‘‘π‘Ÿ
𝑑θ
= β€’ΞΈ+π‘Ÿβ€’
𝑑𝑑
𝑑𝑙
𝑑𝑑
𝑑𝑑
𝑑θ
= π‘Ÿ β€’ 𝑑𝑑
v=rω
𝑑𝑑
v=rω
Example: A cyclist with a 24 inch diameter tire
glances down at his speedometer and notices his
speed is 28 miles/hour . What is the angular
velocity of the wheel in radians/second?
Applying the PRODUCT RULE a second time
𝑑
𝑓 𝑑 ×𝑔 𝑑
𝑑𝑑
𝑑𝑓
𝑑𝑔
=
× π‘” 𝑑 + 𝑓(𝑑) ×
𝑑𝑑
𝑑𝑑
v=r Ο‰
𝑑
𝑑
𝑣 =
rω
𝑑𝑑
𝑑𝑑
𝑑𝑣
π‘‘π‘Ÿ
𝑑ω
= β€’Ο‰+π‘Ÿβ€’
𝑑𝑑
𝑑𝑣
𝑑𝑑
𝑑𝑑
= π‘Ÿβ€’
a=rΞ±
𝑑ω
𝑑𝑑
𝑑𝑑
atangetial = r Ξ±
Example: At the Indianapolis 6000 a x-wing
fighter banks a turn around the 4th moon of
Xarnex at 18000 m/𝑠 2 . The radius of curvature is
200 meters. What is the angular acceleration?
β€’ π‘Žπ‘›π‘”π‘’π‘™π‘Žπ‘Ÿ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ Ο‰ = 2Ο€ βˆ— π‘“π‘Ÿπ‘’π‘žπ‘’π‘’π‘›π‘π‘¦ = 2π𝑓 =
β€’ π‘π‘’π‘Ÿπ‘–π‘œπ‘‘ = 𝑇 = time required for one revolution
‒𝑇 =
1
𝑓
# π‘œπ‘“ π‘Ÿπ‘’π‘£π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›π‘ 
2Ο€
π‘ π‘’π‘π‘œπ‘›π‘‘
1 Dimensional kinematic equation also in cases of constant angular acceleration with appropriate change in variable
𝑣𝑓 = 𝑣0 + π‘Žπ‘‘
β†’
ω𝑓 = Ο‰0 + α𝑑
Ξ”π‘₯ = 𝑣0 𝑑 + 1 2 π‘Žπ‘‘ 2
β†’
Δθ = Ο‰0 𝑑 + 1 2 α𝑑 2
𝑣𝑓2 = 𝑣02 + 2π‘ŽΞ”π‘₯
β†’
ω𝑓 = Ο‰0 + α𝑑
𝑣=
𝑣𝑓 + 𝑣0
2
β†’
Ο‰=
ω𝑓 + Ο‰0
2
Example: A cyclist with a 24 inch diameter tire glances down at
his speedometer and notices his speed is 28 miles/hour . What
is the angular velocity of the wheel in radians/second? He
comes to a stop in 6 seconds. How many revolutions does his
wheel turn during this time?
Torque: the rotational analogue of force
suppose we have a rigid, massive body:
οƒ  if a force F is applied a distance
r away from a fixed axis of rotation,
then the rigid body will experience a torque
Ο„
when a force is directed in a straight line the result is a push or a pull; a linear force;
however when that force causes a rotation it is called a torque
It is only when the force is applied perpendicular to the lever arm r, that a torque is generated.
What if the force is at an angle?
Torque Example: Susan is changing her tire and applies a 20
Newton force perpendicular to the end of a wrench that is 20
centimeters long. What is the magnitude of the torque
generated?
𝜏 = π‘Ÿ × πΉ = π‘Ÿ 𝐹 sin πœƒ
Torque examples everyday life:
Center of Mass (cm) or Center of Gravity: average of the masses factored by their distances from a reference
point
If you take out your cell phone and balance it on the tip of your finger why does it balance and not
fall over?
Example:
π‘₯π‘π‘š
π‘š1 π‘₯1 + π‘š2 π‘₯2 + π‘š3 π‘₯3 +
=
=
π‘š1 + π‘š2 + π‘š3 +
π‘šπ‘– π‘₯𝑖
π‘šπ‘– π‘₯𝑖
=
π‘šπ‘–
π‘€π‘‡π‘œπ‘‘π‘Žπ‘™
π‘¦π‘π‘š
π‘š1 𝑦1 + π‘š2 𝑦2 + π‘š3 𝑦3 +
=
=
π‘š1 + π‘š2 + π‘š3 +
π‘šπ‘– 𝑦𝑖
π‘šπ‘– 𝑦𝑖
=
π‘šπ‘–
π‘€π‘‡π‘œπ‘‘π‘Žπ‘™
Continuous Mass Distribution
Moment of Inertia: the rotational analogue of mass
When a force is applied to a mass that then moves in a straight line, we consider as if all of the mass were located
at the center of mass; the balance point of mass distribution and only consider the total mass.
However, when we consider torques we need to know the position of where the mass is located. It takes more
force to rotate a mass around an axis of symmetry when the mass is located 100 meters away then when it is 4
meters away.
Moment of intertia=
2
π‘šπ‘– π‘Ÿπ‘–
Example: Determine the moment of inertia of a system of 4 point masses located at
coordinates (1,2), (-2,3), (-3,4), and (-5,-7) each of mass 2 kilograms when the axis of
rotation is about the z-axis.
Moment of intertia=
π‘šπ‘– π‘Ÿπ‘–2
Continuous mass distributions
Newtons’ 2nd Law:
πΉπ‘œπ‘Ÿπ‘π‘’π‘  = 𝐹𝑛𝑒𝑑 = π‘šπ‘Ž β†’
π‘‡π‘œπ‘Ÿπ‘žπ‘’π‘’π‘  = τ𝑛𝑒𝑑 = 𝐼α
Example: If each of the three rotor blades is 8cm long and has a mass of 95 gram, calculate the amount of
torque that the motor must supply to bring the blades up to a speed of 5 rev/sec in 8.0 seconds.
Linear Kinetic Energy οƒ  Rotational Kinetic Energy
1
2
m𝑣
2
β†’
1
2
Iω
2
Example: A pitcher throws a fastball with all translational kinetic energy (ZERO spin)
at 44 meters/second. If the pitcher throws the same baseball with a curveball pitch
with the same total kinetic energy at 33 meters/second determine how many
rotations/second the baseball spins at. NOTE: mass_baseball=143 grams;
diameter_baseball=75 millimeters;
Angular Momentum
Linear Momentum: p=mv  Angular Momentum: L=Iω
What are the units of angular momentum?
Example: Determine the angular momentum of the baseball in the previous example.
Newton’s 2nd Law revisited:
We initially saw N2L like this:
𝑑
𝑓 𝑑 ×𝑔 𝑑
𝑑𝑑
𝑑𝑓
𝑑𝑔
=
× π‘” 𝑑 + 𝑓(𝑑) ×
𝑑𝑑
𝑑𝑑
𝐹 = π‘šπ‘Ž
βˆ†π‘
𝑑𝑝 𝑑
𝑑
π‘‘π‘š
𝑑𝑣
𝑑𝑣
however, Issac originally wrote it like: 𝐹 = = = (𝑝)= (π‘šπ‘£)= 𝑣 + π‘š =π‘š =π‘šπ‘Ž
βˆ†π‘‘
𝑑𝑑 𝑑𝑑
𝑑𝑑
𝑑𝑑
𝑑𝑑
𝑑𝑑
SO…use this and our new definition of angular momentum to derive N2L for rotational
motion:
𝜏=
βˆ†πΏ
βˆ†π‘‘
=
𝑑𝐿 𝑑
𝑑
𝑑𝐼
= (𝐿)= (πΌπœ”)= πœ”
𝑑𝑑 𝑑𝑑
𝑑𝑑
𝑑𝑑
+
π‘‘πœ”
𝐼
𝑑𝑑
=
π‘‘πœ”
𝐼 =𝐼
𝑑𝑑
𝛼
Conservation of angular momentum: L=Iπœ” = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘
Explain:
Comparison chart:
Translation
Rotation
Connection
x
πœƒ
x=rΞΈ
v
πœ”
v=rω
a
𝛼
a=rΞ±
m
𝐼
I=Ξ£mr2
F
𝜏
Ξ€=rFsinΞΈ
KE=½mv2
KE=½IΟ‰2
p=mv
L=Iω
W=Fd
W=τα
Ξ£F=ma
Στ=IΞ±
Ξ£F=βˆ†π‘/βˆ†π‘‘
Στ=βˆ†πΏ/βˆ†π‘‘
Add example problems after each slide where a new concept is introduced that might be unclear
Comparison chart at the end; with linear and rotational values