Download Circular Motion

Chapter 5: Circular Motion
•Uniform Circular Motion
•Radial
Radial Acceleration
•Banked and Unbanked Curves
•Circular Orbits
•Nonuniform Circular Motion
•Tangential and Angular Acceleration
•Artificial Gravity
1
Uniform Circular Motion
y
θ is the angular position.
θf
Δθ
θi
Angular displacement:
x
Δθ = θ f − θ i
Note: angles measured CW are negative and angles
measured CCW are p
positive. θ is measured in radians.
2π radians = 360° = 1 revolution
2
The average and instantaneous angular velocities are:
Δθ
Δθ
ωav =
and ω = lim
Δt →0 Δt
Δt
ω is measured in rads/sec.
3
y
arclength = s = rΔθ
θf
r
Δθ
θi
x
s
Δθ =
r
Δθ is a ratio of two lengths; it is
a dimensionless ratio!
4
y
An object moves along a
circular path of radius r; what
is its average speed?
θf
r
Δθ
θi
x
total distance rΔθ
⎛ Δθ ⎞
vav =
=
= r⎜
⎟ = rω av
total time
Δt
⎝ Δt ⎠
Also,
v = rω
(instantaneous values).
5
The time it takes to go one time around a closed path is
called the period (T).
totall distance
di
2πr
=
vav =
total time
T
2π
Comparing to v = rω: ω =
= 2πf
T
f is called the frequency, the number of revolutions (or
cycles) per second.
6
Centripetal Acceleration
y
Consider an object moving
in a circular path of radius r
at constant speed
speed.
v
v
x
Here, Δv ≠ 0. The
direction of v is changing.
v
If Δv ≠ 0, then a ≠ 0. The
net force cannot be zero
zero.
v
7
Conclusion: to move in a circular path, an object must have
a nonzero net force acting on itit.
It is still true that ΣF = ma, but what acceleration do we use?
8
Th velocity
The
l i off a particle
i l iis tangent to iits path.
h
For an object moving in uniform circular motion
motion, the
acceleration is radially inward.
9
The magnitude of the centripedal (or radial)
acceleration is:
2
v
2
ar =
= rω = ωv
r
10
Uniform Circular Motion
Slide 6-13
Examples
The disk in a hard drive in a desktop computer rotates at 7200
rpm. The disk has a diameter of 5.1 in (13 cm.) What is the
angular speed of the disk?
The hard drive disk in the previous example rotates at 7200
rpm. The disk has a diameter of 5.1 in (13 cm.) What is the
speed
d off a point
i t 6.0
6 0 cm ffrom th
the center
t axle?
l ? Wh
Whatt iis th
the
acceleration of this point on the disk?
Slide 6-14
Quiz
1 For uniform circular motion
1.
motion, the acceleration
A. is parallel to the velocity.
B. is directed toward the center of the circle.
C is larger for a larger orbit at the same speed
C.
speed.
D. is always due to gravity.
E. is always negative.
Slide 6-2
Answer
1 For uniform circular motion
1.
motion, the acceleration
B. is directed toward the center of the circle.
Slide 6-3
Understanding
When a ball on the end of a string is swung in a vertical circle:
We know that the ball is accelerating because
A. the speed is changing.
B. the direction is changing.
C. the speed and the direction are changing.
Slide 6-9
Answer
When a ball on the end of a string is swung in a vertical circle:
We know that the ball is accelerating because
B. the direction is changing.
Slide 6-10
Understanding
When a ball on the end of a string is swung in a vertical circle:
What is the direction of the acceleration of the ball?
A. Tangent to the circle, in the direction of the ball’s
motion
B. Toward the center of the circle
Slide 6-11
Answer
When a ball on the end of a string is swung in a vertical circle:
What is the direction of the acceleration of the ball?
B. Toward the center of the circle
Slide 6-12
Circular Motion Dynamics
When the ball reaches the break in the circle,
circle which path will it
follow?
Slide 6-19
Answer
When the ball reaches the break in the circle,
circle which path will it
follow?
Slide 6-20
Forces in Circular Motion
v = ωr
v2
a = — = ω2 r
r
→
→
Fnett = ma =
{
}
mv2
—, toward center of circle
r
Slide 6-21
Example
A level curve on a country road
has a radius of 150 m. What is
the maximum speed at which
this curve can be safely
negotiated on a rainy day when
the coefficient of friction between
th tires
the
ti
on a car and
d th
the road
d iis
0.40?
Slide 6-24
Driving over a Rise
A car of mass 1500 kg goes over a
hill at a speed of 20 m/s. The shape
of the hill is approximately circular,
with a radius of 60 m, as in the figure
at right. When the car is at the
highest point of the hill,
a What is the force of gravity on
a.
the car?
b. What is the normal force of the
road on the car at this point?
Slide 6-26
Example: The rotor is an amusement park ride where people
stand
t d against
i t the
th inside
i id off a cylinder.
li d
O
Once th
the cylinder
li d iis
spinning fast enough, the floor drops out.
(a) What force keeps the people from falling out the bottom of
the cylinder?
y
fs
Draw an FBD for a person
with
ith th
their
i b
back
k tto th
the wall:
ll
N
x
w
It is the force of static friction.
24
Example continued:
(b) If μs = 0.40 and the cylinder has r = 2.5 m, what is the
minimum angular speed of the cylinder so that the people
don’tt fall out?
don
Apply
pp y Newton’s 2nd Law:
From (2):
(1) ∑ Fx = N = mar = mω 2 r
(2) ∑ Fy = f s − w = 0
From (1)
fs = w
μ s N = μ s (mω 2 r ) = mg
ω=
9.8 m/s 2
=
= 3.13 rad/s
(0.40)(2.5 m )
μs r
g
25
Example (text problem 5.79): A coin is placed on a record
that is rotating at 33
33.3
3 rpm
rpm. If μs = 0.1,
0 1 how
ho far from the
center of the record can the coin be placed without having it
slip off?
y
Draw an FBD for the coin:
N
fs
x
Apply Newton’s 2nd Law:
(1) ∑ Fx = f s = mar = mω
(2) ∑ Fy = N − w = 0
2
w
r
26
Example continued:
From (1) : f s = mω 2 r
From (2)
f s = μ s N = μ s (mg ) = mω 2 r
μs g
Solving for r: r = 2
ω
What is ω?
rev ⎛ 2π rad ⎞⎛ 1 min ⎞
ω = 33.3
⎟⎜
⎟ = 3.5 rad/s
⎜
min ⎝ 1 rev ⎠⎝ 60 sec ⎠
μ s g (0.1)(9.8 m/s 2 )
= 0.08 m
r= 2 =
2
ω
(3.50 rad/s )
27
Unbanked and Banked Curves
Example (text problem 5.20): A highway curve has a radius
of 825 m. At what angle should the road be banked so that
a car traveling at 26.8 m/s has no tendency to skid sideways
on the road? (Hint: No tendency to skid means the frictional
force is zero.))
Take the car’s motion
to be into the page.
θ
28
Example continued:
y
FBD for the car:
θ
N
x
w
Apply Newton’s Second Law:
v2
(1) ∑ Fx = N sin θ = mar = m
r
(2) ∑ Fy = N cosθ − w = 0
29
Example continued:
Rewrite (1) and (2):
v2
(1) N sin θ = m
r
(2) N cosθ = mg
Divide (1) by (2):
2
(
)
26
.
8
m/s
v
tan θ =
=
= 0.089
2
gr (9.8 m/s )(825 m )
2
θ = 5.1°
30
The Force of Gravity
Slide 6-31
Circular Orbits
r
Earth
Consider an object of mass m in a
circular orbit about the Earth.
The only force on the satellite is the force
of gravity:
Gms M e
v2
∑ F = Fg = r 2 = ms ar = ms r
Solve for the speed
p
of the satellite:
Gms M e
v2
= ms
2
r
r
GM e
v=
r
32
Example: How high above the surface of the Earth does a
satellite need to be so that it has an orbit period of 24 hours?
GM e
v
=
From previous slide:
r
Also need
need,
Combine these expressions and solve for r:
(
)(
)
2πr
v=
T
⎛ GM e 2 ⎞
r =⎜
T ⎟
2
⎠
⎝ 4π
⎛ 6.67 ×10 Nm /kg 5.98 ×10 kg
2⎞
⎟⎟
(
)
r = ⎜⎜
86400
s
2
4π
⎝
⎠
= 4.225 ×107 m
−11
2
2
24
1
1
3
3
r = Re + h ⇒ h = r − Re = 35,000 km
k
33
⎛ GM e 2 ⎞
r =⎜
T ⎟
2
⎝ 4π
⎠
1
3
is Kepler’s Third Law.
It can be generalized to:
⎛ GM 2 ⎞
r =⎜ 2 T ⎟
⎝ 4π
⎠
1
3
Where M is the mass of the central body.
y For example,
p , it
would be Msun if speaking of the planets in the solar system.
34
Example: What is the minimum speed for the car so that it
maintains
i t i contact
t t with
ith the
th loop
l
when
h it iis iin th
the pictured
i t d
position?
FBD ffor the
th car att
the top of the loop:
r
y
Apply Newton’s 2nd Law:
x
N
w
v2
∑ Fy = − N − w = −mar = −m r
v2
N +w=m
r
35
Example continued:
The apparent weight at the top of loop is:
N = 0 when
v2
N +w=m
r
⎛ v2
⎞
N = m⎜⎜ − g ⎟⎟
⎝ r
⎠
⎛ v2
⎞
N = m⎜⎜ − g ⎟⎟ = 0
⎝ r
⎠
v = gr
This is the minimum speed needed to make it around the
loop.
36
Example continued:
Consider the car at the bottom of the loop; how does the
apparent weight compare to the true weight?
FBD for the car at the
bottom of the loop:
p
y
N
x
w
Apply Newton’s 2nd Law:
v2
∑ Fy = N − w = mac = m r
v2
N −w=m
r
⎛ v2
⎞
N = m⎜⎜ + g ⎟⎟
⎝ r
⎠
Here, N > mg
37
Nonuniform Circular Motion
Here, the speed is not constant.
a
at
There is now an acceleration
tangent
g
to the p
path of the p
particle.
ar
v
The net acceleration of the body is a = ar + at
2
2
38
a
ar
at
at changes
g the magnitude
g
of v.
ar changes the direction of v.
Can write:
∑ F = ma
∑ F = ma
r
r
t
t
39
Artificial Gravity
A large rotating cylinder in
deep space (g≈0).
40
FBD for person at
th bottom
the
b tt
position
iti
FBD for person at
the top position
y
y
N
x
x
N
Apply Newton’s 2nd Law to each:
2
F
=
N
=
ma
=
m
ω
r
∑ y
r
2
F
=
−
N
=
−
ma
=
−
m
ω
r
∑ y
r
41
Example (text problem 5.56): A space station is shaped like
a ring
i and
d rotates
t t tto simulate
i l t gravity.
it If th
the radius
di off the
th
space station is 120m, at what frequency must it rotate so
that it simulates Earth’s gravity?
g
y
Using the result from the previous slide:
2
F
=
N
=
ma
=
m
ω
r
∑ y
r
N
mg
ω=
=
=
mr
mr
g
= 0.28 rad/sec
r
The frequency is f = (ω/2π) = 0.045 Hz (or 2.7 rpm).
42
Tangential and Angular
A
Acceleration
l
i
The average and instantaneous angular acceleration are:
Δω
Δω
α av =
and α = lim
Δt →0 Δt
Δt
α is measured in rads/sec2.
43
Recalling that the tangential velocity is vt = rω means the
tangential acceleration is
Δvt
Δω
αt =
=r
= rα
Δt
Δt
44
The kinematic equations:
Linear
Angular
v = v0 + aΔt
ω = ω0 + αΔt
1
2
x = x0 + v0 Δt + aΔt
2
v 2 = v02 + 2aΔx
1
2
θ = θ 0 + ω0 Δt + αΔt
2
ω 2 = ω02 + 2αΔθ
With
vt = rω and at = rα
45
Example (text problem 5.66): A high speed dental drill is
rotating
t ti att 3.14×10
3 14 104 rads/sec.
d /
Th
Through
hh
how many d
degrees
does the drill rotate in 1.00 sec?
Given: ω = 3.14×104 rads/sec; Δt = 1 sec; α = 0
Want Δθ.
1
θ = θ 0 + ω0 Δt + αΔt 2
2
θ = θ 0 + ω0 Δt
(
)
Δθ = ω0 Δt = 3.14 ×10 rads/sec (1.0 sec )
4
= 3.14 ×10 rads = 1.80 ×10 degrees
g
4
6
46