Transcript Physics 207: Lecture 2 Notes

Lecture 10

Goals:
 Employ Newton’s Laws in 2D problems with circular motion
Assignment: HW5, (Chapters 8 & 9, due 3/4, Wednesday)
For Tuesday: Finish reading Chapter 8, start Chapter 9.
Physics 207: Lecture 10, Pg 1
Uniform Circular Motion
For an object moving along a curved trajectory,
with non-uniform speed
a = ar (radial only)
v
vT2
|ar |=
r
ar
Perspective is important
Physics 207: Lecture 10, Pg 2
Non-uniform Circular Motion
For an object moving along a curved trajectory,
with non-uniform speed
a = ar + at (radial and tangential)
at
vT2
|ar |=
r
ar
|aT |=
d| v |
dt
Physics 207: Lecture 10, Pg 3
Circular motion

Circular motion implies one thing
|aradial | =vT2 / r
Physics 207: Lecture 10, Pg 4
Key steps
 Identify
forces (i.e., a FBD)
 Identify
axis of rotation
 Apply
conditions (position, velocity,
acceleration)
Physics 207: Lecture 10, Pg 5
Example
The pendulum
axis of rotation
Consider a person on a swing:
When is the tension on the rope largest?
And at that point is it :
(A) greater than
(B) the same as
(C) less than
the force due to gravity acting on the person?
Physics 207: Lecture 10, Pg 6
Example
Gravity, Normal Forces etc.
axis of rotation
T
T
y
vT
q
x
mg
at top of swing vT = 0
mg
at bottom of swing vT is max
Fr = m 02 / r = 0 = T – mg cos q
T = mg cos q
Fr = m ac = m vT2 / r = T - mg
T = mg + m vT2 / r
T < mg
T > mg
Physics 207: Lecture 10, Pg 7
Conical Pendulum (very different)

Swinging a ball on a string of length L around your head
(r = L sin q)
axis of rotation
S Fr = mar = T sin q
S Fz = 0 = T cos q – mg
so
T = mg / cos q (> mg)
mar = mg sin q / cos q
ar = g tan q = vT2/r  vT = (gr tan q)½
Period:
t = 2p r / vT =2p (r cot q /g)½
Physics 207: Lecture 10, Pg 8
Loop-the-loop 1
A match box car is going to do a loop-the-loop
of radius r.
What must be its minimum speed vt at the top
so that it can manage the loop successfully ?
Physics 207: Lecture 10, Pg 9
Loop-the-loop 1
To navigate the top of the circle its tangential
velocity vT must be such that its centripetal
acceleration at least equals the force due to
gravity. At this point N, the normal force,
goes to zero (just touching).
Fr = mar = mg = mvT2/r
vT
vT = (gr)1/2
mg
Physics 207: Lecture 10, Pg 10
Loop-the-loop 2
The match box car is going to do a loop-the-loop.
If the speed at the bottom is vB, what is the
normal force, N, at that point?
Hint: The car is constrained to the track.
Fr = mar = mvB2/r = N - mg
N = mvB2/r + mg
N
v
mg
Physics 207: Lecture 10, Pg 11
Loop-the-loop 3
Once again the car is going to execute a loop-the-loop.
What must be its minimum speed at the bottom so
that it can make the loop successfully?
This is a difficult problem to solve using just forces. We
will skip it now and revisit it using energy
considerations later on…
Physics 207: Lecture 10, Pg 12
Example, Circular Motion Forces with Friction
(recall mar = m |vT | 2 / r Ff ≤ ms N )

How fast can the race car go?
(How fast can it round a corner with this radius of curvature?)
mcar= 1600 kg
mS = 0.5 for tire/road
r = 80 m
g = 10 m/s2
r
Physics 207: Lecture 10, Pg 14
Example

Only one force is in the horizontal direction: static friction
x-dir: Fr = mar = -m |vT
y-dir: ma = 0 = N – mg
|2/
y
r = Fs = -ms N (at maximum)
N = mg
vT = (ms m g r / m )1/2
vT = (ms g r )1/2 = (0.5 x 10 x 80)1/2
vT = 20 m/s
N
Fs
mg
mcar= 1600 kg
mS = 0.5 for tire/road
r = 80 m
g = 10 m/s2
Physics 207: Lecture 10, Pg 15
x
Another Example
A horizontal disk is initially at rest and very slowly undergoes
constant angular acceleration. A 2 kg puck is located a point
0.5 m away from the axis. At what angular velocity does it slip
(assuming aT << ar at that time) if ms=0.8 ?
 Only one force is in the horizontal direction: static friction
y
x-dir: Fr = mar = -m |vT | 2 / r = Fs = -ms N (at w)

y-dir: ma = 0 = N – mg
N = mg
vT = (ms m g r / m )1/2
vT = (ms g r )1/2 = (0.8 x 10 x 0.5)1/2
vT = 2 m/s  w = vT / r = 4 rad/s
N
Fs
mg
mpuck= 2 kg
mS = 0.8
r = 0.5 m
g = 10 m/s2
Physics 207: Lecture 10, Pg 16
x
UCM: Acceleration, Force, Velocity
F
a
v
Physics 207: Lecture 10, Pg 17
Zero Gravity Ride
A rider in a “0 gravity ride” finds herself
stuck with her back to the wall.
Which diagram correctly shows the
forces acting on her?
Physics 207: Lecture 10, Pg 18
Banked Curves
In the previous car scenario, we drew the following free
body diagram for a race car going around a curve on a
flat track.
n
Ff
mg
What differs on a banked curve?
Physics 207: Lecture 10, Pg 19
Banked Curves
Free Body Diagram for a banked curve.
Use rotated x-y coordinates
Resolve into components parallel and
perpendicular to bank
N
mar
Ff
q
y
x
mg
For very small banking angles, one can approximate
that Ff is parallel to mar. This is equivalent to the small
angle approximation sin q = tan q, but very effective at
pushing the car toward the center of the curve!!
Physics 207: Lecture 10, Pg 20
Banked Curves, Testing your understanding
Free Body Diagram for a banked curve.
Use rotated x-y coordinates
Resolve into components parallel and
perpendicular to bank
N
mar
Ff
q
x
y
mg
At this moment you press the accelerator and, because
of the frictional force (forward) by the tires on the road
you accelerate in that direction.
How does the radial acceleration change?
Physics 207: Lecture 10, Pg 21
Locomotion: how fast can a biped walk?
Physics 207: Lecture 10, Pg 22
How fast can a biped walk?
What about weight?
(a) A heavier person of equal
height and proportions
can walk faster than a
lighter person
(b) A lighter person of equal
height and proportions
can walk faster than a
heavier person
(c) To first order, size doesn’t
matter
Physics 207: Lecture 10, Pg 23
How fast can a biped walk?
What about height?
(a) A taller person of equal
weight and proportions
can walk faster than a
shorter person
(b) A shorter person of equal
weight and proportions
can walk faster than a
taller person
(c) To first order, height
doesn’t matter
Physics 207: Lecture 10, Pg 24
How fast can a biped walk?
What can we say about the walker’s
acceleration if there is UCM (a
smooth walker) ?
Acceleration is radial !
So where does it, ar, come from?
(i.e., what external forces are on
the walker?)
1. Weight of walker, downwards
2. Friction with the ground, sideways
Physics 207: Lecture 10, Pg 25
How fast can a biped walk?
What can we say about the walker’s
acceleration if there is UCM (a
smooth walker) ?
Acceleration is radial !
So where does it, ar, come from?
(i.e., what external forces are on
the walker?)
1. Weight of walker, downwards
2. Friction with the ground, sideways
Physics 207: Lecture 10, Pg 26
How fast can a biped walk?
What can we say about the walker’s
acceleration if there is UCM (a
smooth walker) ?
Acceleration is radial !
So where does it, ar, come from?
(i.e., what external forces are on
the walker?)
1. Weight of walker, downwards
2. Friction with the ground, sideways
(most likely from back foot, no
tension along the red line)
3. Need a normal force as well
Physics 207: Lecture 10, Pg 27
How fast can a biped walk?
What can we say about the walker’s
acceleration if there is UCM (a
smooth walker) ?
Acceleration is radial !
So where does it, ar, come from?
(i.e., what external forces are on
the walker?)
1. Weight of walker, downwards
2. Friction with the ground, sideways
(Notice the new direction)
Physics 207: Lecture 10, Pg 28
How fast can a biped walk?
Given a model then what does the
physics say?
Choose a position with the simplest
constraints.
If his radial acceleration is greater than g
then he is “in orbit”
Fr = m ar = m v2 / r < mg
Otherwise you will lose contact!
ar = v2 / r  vmax = (gr)½
vmax ~ 3 m/s !
(And it pays to be tall and live on Jupiter)
Physics 207: Lecture 10, Pg 29
Orbiting satellites vT = (gr)½
Physics 207: Lecture 10, Pg 30
Geostationary orbit
Physics 207: Lecture 10, Pg 31
Geostationary orbit

The radius of the Earth is ~6000 km but at 36000 km you are
~42000 km from the center of the earth.

Fgravity is proportional to r-2 and so little g is now ~10 m/s2 / 50

vT = (0.20 * 42000000)½ m/s = 3000 m/s

At 3000 m/s, period T = 2p r / vT = 2p 42000000 / 3000 sec =
= 90000 sec = 90000 s/ 3600 s/hr = 24 hrs

Orbit affected by the moon and also the Earth’s mass is
inhomogeneous (not perfectly geostationary)

Great for communication satellites
(1st pointed out by Arthur C. Clarke)
Physics 207: Lecture 10, Pg 32