Transcript 212 Lecture 10

Lecture 10
Employ Newton’s Laws in 2D problems
with circular motion
Uniform Circular Motion
For an object moving along a curved trajectory,
with non-uniform speed
a = ar (radial only)
v
v2T
|ar |=
r
ar
Perspective is important
Non-uniform Circular Motion
For an object moving along a curved trajectory,
with non-uniform speed
a = ar + at (radial and tangential)
at
v2T
|ar |=
r
ar
d| v |
|aT | =
dt
Circular motion

Circular motion implies one thing
|aradial | =v2T / r
Key steps

Identify forces (i.e., a FBD)

Identify axis of rotation

Apply conditions (position, velocity, acceleration)
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?
Example
Gravity, Normal Forces etc.
axis of rotation
T
T
y
vT
q
mg
x
at top of swing vT = 0
Fr = m 02 / r = 0 = T – mg cos
q
T = mg cos q
T < mg
mg
at bottom of swing vT is max
Fr = m ar = m vT2 / r = T - mg
T = mg + m vT2 / r
T > mg
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 = vT
2/r
 vT = (gr tan
q)½
Period:
t = 2p r / vT =2p (r cot q /g)½
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 ?
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
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 = mv2B/r = N - mg
N = mvB2/r + mg
N
v
mg
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…
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
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
x
Acceleration, Force, Velocity
F
a
v
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?
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?
Banked Curves
Free Body Diagram for a banked curve.
Use rotated x-y coordinates
Resolve into components parallel and perpendicular toy
bank
N
mar
Ff
q
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!!
x
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?