Physics 1710 Chapter 4: 2-D Motion—II
Free Fall:
• Kinematics in two (or more) dimensions obeys the
same 1- D equations in each component
independently.
vx =constant
ax = 0
ay = -g
x = xinitial + vx,initial t
y = yinitial – ½ gt 2
Physics 1710 Chapter 4: 2-D Motion—II
Kinematic Equations
x(t) = xinitial + vinitial t + 1/2 ao t 2
v(t) = dx/dt = vinitial + ao t
a= dv/dt = ao
Observe Air Track
REVIEW
Physics 1710 Chapter 4: 2-D Motion—II
1′ Lecture:
• The velocity and acceleration of a body in a moving
(and accelerating) frame of reference (FoR ) is equal to
that of a stationary FoR minus the velocity or acceleration
of the moving FoR.
 v ′ = v – vframe of reference
 a ′ = a – aframe of reference
• Motion in a circle at a constant speed is due to an
acceleration toward the center of the circle, a centripetal
acceleration of
 a = - ω2 r and |a| = v 2/ |r| toward the center.
Physics 1710 Chapter 4: 2-D Motion—II
Frame of Reference
Fly
v = v′ + vframe of reference
v′
and
v′ = v - vframe of reference
v
v′
vframe of reference
Physics 1710 Chapter 4: 2-D Motion—II
Relative Motion
and the Galilean Transformation:
r ′ = r – vframe of reference t
d r ′/dt = d r/dt – vframe of reference
v ′ = v – vframe of reference
d v ′/dt = d v/dt – d vframe of reference/dt
a ′ = a – aframe of reference
Physics 1710 Chapter 4: 2-D Motion—II
Relative Motion
in a Free Falling Frame of Reference
In Lab
Frame
In Moving
Frame
a ′ = a – aframe of reference
Physics 1710 Chapter 4: 2-D Motion—II
Everyday Physics:
• When
a bird is flying at the same
velocity(same speed and direction) as a car
what is its relative velocity v′ to the car?
•When you brake an automobile, which
direction is the acceleration on the vehicle?
Which direction do the passengers sense as
the acceleration relative to the frame of
reference of the car?
Physics 1710 Chapter 4: 2-D Motion—II
Relative Motion
and the Galilean Transformation:
v ′ = v – vframe of reference
a ′ = a – aframe of reference
Physics 1710 Chapter 4: 2-D Motion—II
Uniform Circular Motion
d θ/dt = ω = constant
x = R sin θ
y = R cos θ
v
θ
r
R2 = x2 + y2
r = (x,y) = x i + y j
vx
= R cos θ d θ/dt
= Rω cos θ
vy
= - R sin θ d θ/dt
= - Rω sin θ
Physics 1710 Chapter 4: 2-D Motion—II
Uniform Circular Motion
v
θ
r
vx
= R cos θ d θ/dt
= R ω cos θ
vy
= - R sin θ d θ/dt
= - R ω sin θ
ax
= - R ω2 sin θ
ay
= - R ω2 cos θ
a
= - ω2 r
|a| = v2/R, toward center
Physics 1710 Chapter 4: 2-D Motion—II
Uniform Circular Motion—in review
⃒a⃒ = a,
a constant value always pointing toward the center of the
circle: Centripetal acceleration.
a = ax i+ a y j = - ω2 r
where
ax = a sin θ = - (v2/R) sin θ = - ω2 R sin θ
ay = a cos θ = - (v2/R) cos θ = - ω2 R cos θ
Physics 1710 Chapter 4: 2-D Motion—II
Uniform Circular Motion
|a| = v 2 /R, toward center
The Centripetal acceleration, where v is the tangential
speed and R is the radius of the circle.
v = ω R = 2πR / T ,
Where T is the “period” or time to make one revolution.
|a |= 4π 2R/ T 2 = ω2 R
Physics 1710 Chapter 4: 2-D Motion—II
Uniform Circular Motion
Little Johnny on the Farm—part II
v
-g
.
a
?
.
a
v
-g
Physics 1710 Chapter 4: 2-D Motion—II
Uniform Circular Motion
aFoR
-g
In Frame of
Reference of Bucket
a′ = -g – aFoR
If
|aFoR | ≽ |g|
In same (down)
direction,
a′ is up!
Physics 1710 Chapter 4: 2-D Motion—II
Uniform Circular Motion
a
g
a = (2π/T)2 R,
T ≈1. sec
R ≈1. m
2π ≈ 6.
Do you believe
this?
a ≈ (6./1.sec)2 (1.m)
a≈ 36. m/sec 2
a> 3 g
Physics 1710 Chapter 4: 2-D Motion—II
80/20 Summary:
• In a moving or accelerating Frame of Reference
• v ′ = v – vframe of reference
• a ′ = a – aframe of reference
• The Centripetal acceleration is
• a = - ω2 r
or |a| = v 2/ |r|, toward the center.