Transcript Physics 207: Lecture 2 Notes

Lecture 11

Goals:
 Employ Newton’s Laws in 2D problems with circular motion
 Relate Forces with acceleration
Assignment: HW5, (Chapter 7, 8 and 9 due 10/18)
For Wednesday: Reading through 1st four sections in Ch. 9
Physics 207: Lecture 11, Pg 1
Chapter8
Reprisal of : Uniform Circular Motion
For an object moving along a curved trajectory with constant
speed
a = ar (radial only)
v
vt2
|ar |=
r
ar
Physics 207: Lecture 11, Pg 3
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 11, Pg 4
Uniform or non-uniform circular motion
implies 
|aradial | =vT2 / r
 and if there is acceleration there
MUST be a net force

Physics 207: Lecture 11, Pg 5
Key steps
 Identify
forces (i.e., a FBD)
 Identify
axis of rotation
 Apply
conditions (position, velocity &
acceleration)
Physics 207: Lecture 11, Pg 6
Example
axis of rotation
The pendulum
Consider a person on a swing:
(A)
(B)
(C)
When is the tension equal to the weight of the person + swing?
(A) At the top of the swing (turnaround point)
(B) Somewhere in the middle
(C) At the bottom of the swing
(D) Never, it is always greater than the weight
(E) Never, it is always less than the weight
Physics 207: Lecture 11, Pg 7
Example
Gravity, Normal Forces etc.
axis of rotation
T
T
y
vT
q
mg
x
mg
at top of swing vT = 0
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 11, Pg 8
Conical Pendulum (Not a simple pendulum)

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
L
S Fz = 0 = T cos q – mg
so
T = mg / cos q (> mg)
r
mar = (mg / cos q ) (sin q )
ar = g tan q = vT2/r
 vT = (gr tan q)½
Period:
T= 2p r / vT =2p (r cot q /g)½
= 2p (L cos q /g)½
Physics 207: Lecture 11, Pg 9
Conical Pendulum (very different)

Swinging a ball on a string of length L around your head
axis of rotation
Period:
t = 2p r / vT =2p (r cot q /g)½
= 2p (L cos q / g )½
= 2p (5 cos 5 / 9.8 )½ = 4.38 s
= 2p (5 cos 10 / 9.8 )½ = 4.36 s
= 2p (5 cos 15 / 9.8 )½ = 4.32 s
L
r
Physics 207: Lecture 11, Pg 10
Another example of circular motion
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 11, Pg 11
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 11, Pg 12
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 11, Pg 13
Orbiting satellites vT = (gr)½
Net Force:
ma = mg = mvT2 / r
gr = vT2
vT = (gr)½
The only difference is
that g is less
because you are
further from the
Earth’s center!
Physics 207: Lecture 11, Pg 14
Geostationary orbit
Physics 207: Lecture 11, Pg 15
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 11, Pg 16
Recap
Assignment: HW5,
For Wednesday: Finish reading through 1st four sections in
Chapter 9
Physics 207: Lecture 11, Pg 18