Download Circular Motion and Rotation

General Physics
Today:
1. Circular Motion / Rotation
Rotational Kinematics
Definition of position, velocity and acceleration for
rotational motion.
Rotational kinematics
Angular Position
θ (theta)
Linear kinematics
x
Angular Velocity ω (omega)
v
Angular Acceleration α (alpha)
a
Angular Position Θ
Θ=0 : reference line
Θ > 0 : counterclockwise
Θ < 0 clockwise
Units:
degree °
revolution rev (1rev = 360 °)
radian (most convenient unit)
A radian is the angle for which the arc length s on a
circle is equal to the radius r
Æ s=r Θ (s=r for 1 rad) and therefore Θ=s/r [m/m=rad]
1rev=360°= 2 π rad
or 1rad ≡ 180°/π
1rad = 57.3°
Revolution, Period , Frequency and
Angular Velocity
ΔΘ = Θf - Θi (change of angular position)
Revolution rev = 360°
period T (time taken for one revolution) [s]
frequency f (how many revolutions per second) [s-1] = [Hz]
are related by: 1 = T f
[
rad
]
s
Angular velocity ω = ΔΘ/Δt
(average; instantaneous for lim Δt → 0)
Average velocity for one period
ω=
2π
T
Sign of the angular velocity
ω > 0 : counterclockwise
ΔΘ > 0 (change of angular position)
ω < 0 clockwise
ΔΘ < 0 (change of angular position)
Units: rad/s = 1/s
An object moves along a circular path of
radius r; what is its average speed?
total distance rΔθ
⎛ Δθ ⎞
vav =
= r⎜
=
⎟ = rω av
total time
Δt
⎝ Δt ⎠
Linear Kinematics vs Rotational Kinematics
The block falls down with a constant
acceleration G
G
G G
G
a = Δv / Δt or vf = vi + at
The pulley turns right (clockwise) with a
constant angular acceleration
α = Δω / Δt or ωf = ωi + αt
Velocity of
box is equal
the
tangential
velocity of
the pulley
Consider an object moving in a circular path
of radius r at constant speed.
Here, Δv≠0. The direction
of v is changing.
y
v
v
x
v
If Δv≠0, then a≠0. The
net force cannot be zero.
Conclusion: to move in a circular
path, an object must have a
nonzero net force acting on it.
v
Centripetal Acceleration
The velocity of a particle is tangent to its path.
For an object moving in uniform circular
motion, the acceleration is radially inward.
The magnitude of the radial acceleration is:
v2
= r ω 2 = ωv
ar =
r
Centripetal Acceleration of an Rotating
Object
acp= v2/r = (rω)2/r
acp = r ω2 [m/s2]
Tangential acceleration of an accelerating
object
Δv = r Δω
at = Δv / Δt = r Δω / Δt = r α [m/s2]
Total accelerati on
a=
at + a cp
2
Angle φ
φ = tan
−1
a cp
a tan
2
Centripetal Force
According to Newton's second law of
motion, an accelerating object must be
acted upon by an unbalanced force.
This unbalanced force is in the same
direction as the direction of the
acceleration. For objects in uniform
circular motion, the net force and
subsequent acceleration is directed
inwards. It is often said that circular
motion requires an inward (or
"centripetal") force.
Without a centripetal force, an object
cannot travel in circular motion. In
fact, if the forces are balanced, then an
object in motion continues in motion
in a straight line at constant speed.
Example: The rotor is an amusement
park ride where people stand against
the inside of a cylinder. Once the
cylinder is spinning fast enough, the
floor drops out.
(a) What force keeps the people from
falling out the bottom of the cylinder?
y
fs
N
x
Draw an FBD for
a person with
their back to the
wall:
It is the force of static friction.
w
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’t fall out?
Apply Newton’s 2nd Law:
(1) ∑ Fx = N = mar = mω 2 r
(2) ∑ Fy = f s − w = 0
From (2):
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
Example: A coin is placed on a record that is
rotating at 33.3 rpm. If μs = 0.1, how far
from the center of the record can the coin be
y
placed without having it slip off?
Draw an FBD for the coin:
Apply Newton’s
2nd Law:
N
fs
x
(1) ∑ Fx = N = mar = mω 2 r
(2) ∑ Fy = f s − w = 0
From (2)
w
From (1) : f s = mω 2 r
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
∴r = 2 =
= 0.08 m
2
ω
(3.50 rad/s )
(
)
Unbanked and Banked Curves
Example (text problem 5.20): A highway curve has a
radius of 122 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
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
Rewrite (1) and (2):
v2
(1) N sin θ = m
r
(2) N cos θ = mg
Divide (1) by (2):
(
26.8 m/s )
v
=
= 0.6007
tan θ =
2
gr 9.8 m/s (122 m )
θ = 31.0°
2
2
(
)
Circular Orbits
Consider an object of mass m in a
circular orbit about the Earth.
The only force on the satellite is the
force of gravity:
r
Earth
Gms M e
v2
∑ F = Fg = r 2 = ms ar = ms r
Solve for the speed of
the satellite:
Gms M e
v2
= ms
2
r
r
GM e
v=
r
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
From previous slide: v =
r
v=
Also need,
2πr
T
Combine these expressions and solve for r:
⎛ GM e 2 ⎞
r =⎜
T ⎟
2
⎝ 4π
⎠
(
)(
1
3
)
⎛ 6.67 × 10 Nm /kg 5.98 ×10 kg
2⎞
(86400 s ) ⎟⎟
r = ⎜⎜
2
4π
⎝
⎠
= 4.225 ×107 m
−11
2
2
24
r = Re + h ⇒ h = r − Re = 35,000 km
1
3
Example: What is the minimum speed for
the car so that it maintains contact with the
loop when it is in the pictured position?
r
FBD for the car at
the top of the loop:
y
Apply Newton’s 2nd Law:
x
N
w
v2
∑ Fy = − N − w = −mar = −m r
v2
N +w=m
r
v2
The apparent weight at the N + w = m
r
top of loop is:
⎛ v2
⎞
N = m⎜⎜ − g ⎟⎟
⎝ r
⎠
N = 0 when
⎞
⎛ v2
N = m⎜⎜ − g ⎟⎟ = 0
⎠
⎝ r
v = gr
This is the minimum speed needed
to make it around the loop.
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 Apply Newton’s 2nd Law:
bottom of the loop:
v2
∑ Fy = N − w = mac = m r
y
N
v2
N −w=m
r
⎛ v2
⎞
N = m⎜⎜ + g ⎟⎟
x
⎝ r
⎠
w