# CHAPTER 7 Circular Motion and the Laws of Gravity by ewghwehws

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CHAPTER 7
Rotational Motion and the Law of Gravity
Angular Speed and Angular Acceleration

s = arc length
Θ = s ; where r = radius
r
Term              Symbol      Units
Θ2 – Θ1
ω=
t2 – t1
∆Θ              Average Angular Speed
ω=              Average rate at which the arc
∆t
angle is changing
(Instantaneous) Angular Speed
ω = lim ∆Θ
The rate at which the arc angle is
∆t0 ∆t changing at a particular instant in time
ω2 – ω1
=
t2 – t1
Average Angular Acceleration
∆ω
=         The average rate at which the angular
∆t               speed is changing

NOTE: You are not required to work with
changing rates of acceleration.
Therefore,  = 
Linear/ Angular Comparison

Term                Linear           Angular

Formulas:       ∆x = ½(vo+v)t       ∆Θ = ½(ωo+ω)t
∆x = vot + ½at2        ∆Θ = ωot + ½ t2
v = vo + at           ω = ωo + t
v2 = vo2 + 2a∆x     ω2 = ωo2 + 2∆Θ
N/A               s = rΘ

NOTE: All points on a rotating disc have the
same values for ω and .
Example Problem (Rotating Disc)
A disc of radius 2.50m accelerates to an angular velocity of
1.35 radians/sec in 12.5 seconds. Calculate (a) the disc’s
angular acceleration; (b) the number of revolutions
completed during the acceleration; (c) the distance traveled
by a point on the edge of the disc.
Identify the variables
ωo = 0.00 rad/sec        = ?
ω = 1.35 rad/sec         t = 12.5 sec       ∆Θ = ?
Choose Appropriate Formulas
ω = ωo + t  = .108 rad/sec2

∆Θ =
[ ]ωo + ω
2
t    ∆Θ = ωot + ½ t2

# rev = 1.34rev                    ∆s = 21.1m
Tangential Speed (vt)     •       arc length of the movement
time to complete the movement
• vt = ∆s
∆t
∆s = r ∆Θ
vt = r ∆Θ
∆t
vt = r ω

Tangential Acceleration (t)         at = r 

NOTE: Not every point on a rotating disc has the
same tangential speed or tangential acceleration.
Centripetal Acceleration (center-seeking)

Question:
The tangential velocity vt illustrated above is
clearly changing. How can you tell it is changing?
Its direction is changing.
a = ∆v = vf – vi = vf + -vi
∆t        ∆t           ∆t
The illustration on the right clearly shows that ∆v
points towards the center of rotation.
Centripetal Acceleration (ac)
∆v = ∆s       from similar triangles
v       r
∆v = v
∆s     and ∆v = ac ∆t
r
v
ac ∆t =      ∆s
r
v ∆s      and ∆s = v
ac =
r ∆t          ∆t
v2
ac =    = Centripetal Acceleration
r
Centripetal Force: • The net force that causes an object
to accelerate towards its center of
rotation (i.e. go in a circle)
• Fc = mac
• Fc is not a new force
• Fc will be either one or a
combination of more than one of the
following forces:
Fg
FN
Ff
Fa (T)
Example Problem (Vertical loop)
A rollercoaster contains a loop 20m in diameter. a) What speed
will cause a passenger to feel “weightless” at the top of the
loop? b) If this same person has a mass of 90kg what weight
will they experience at the bottom of the loop?
Sketch the Force Diagrams

Analysis: The only forces
Analysis: The person is           acting on the person are the
“weightless” if no forces other   normal force and force of
than gravity are acting upon      gravity.
them (i.e. FN=0).                   Fc = Fnet(center) = FN – Fg
Hence, Fg is Fc                     FN = Fc + Fg
mg = mv2                               = mv2 + mg
r                                    r
v = √rg        v = 9.9m/s            FN = 1764 Newtons
Example Problem (Banked Curve)
a) For a car traveling with a speed v around a curve of
radius r, determine the formula for the angle at which
the road should be bank so that no friction is required.
b) what is this angle for an expressway off-ramp of
radius 50m at a design speed of 50km/hr?
Sketch the Force Diagram
Part a:
Analysis
• Since there is no friction,
there is no component of
friction directed to the
center (right).
• Only the component
FNsinΘ of the normal force
is directed to the center.
• FN sinΘ is the net force
towards the center = Fc
FN sinΘ = Fc
FN sinΘ = mv2
r
mg = FN cosΘ
mg · sinΘ = mv2
cosΘ           r
mg tanΘ = mv2
r

Part b:
tanΘ = v2
gr Θ = tan
-1
()
v2
gr

r = 50m
v = 50km 1000m  1hr
= 14m/s
hr   km 3600s
g = 9.8
Θ = 22°
Newton’s Universal Law of Gravitation
Every particle in the universe attracts every other particle
with a force that is directly proportional to the product of
their masses and inversely proportional to the square of
the distance between them.

Fg = G m1 m2
r2

G = universal gravitational
constant
G = 6.67 x 10-11 N·m2/kg2
Force of Earth’s Gravity (on Earth’s Surface)
Fg = G mearth mobject
rearth2

Fg =
[ ]
G mearth
rearth 2
· massobject

(6.67 x10-11 N·m2/kg2)(5.98x1024 kg)
Fg =                                      · massobject
(6.37x10 6 m)2

Fg = 9.8 x massobject

Fg = mobect g
Velocity of an Earth Satellite
Analysis:
What is the only Force acting on the satellite?
Fg
If satellite is in circular orbit, then…
Fg = Fc
mg = mvs2               vs = velocity of satellite
r
vs = √gr        g = GMearth
r2
vs =
√     GMearth · r
r2

√
vs =     GMearth
r
Gravitational Potential Energy
PE = -GMEm         r = distance of object from
r                Earth’s center
• from Calculus
• PE = zero at infinite distance from Earth
• PE = some negative value closer to the Earth
Escape Speed
• Speed at surface of Earth where the object’s KE1 + PE1 is
sufficient to reach infinite distance from Earth (PE2=0J)
with a final speed of 0m/s (KE2=0J)
• From Conservation of Energy:
KE1 + PE1 = KE2 + PE1
mv2 + -GMearthm
=0+0

[      ]
2       Rearth                            2GMearth   1/2
mv2 + -GMearthm               vescape =    Rearth
2       Rearth

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