1. 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 10/19)
For Wednesday: Reading through 1st four sections in Ch. 9 1

2. Chapter8 Reprisal of : Uniform Circular Motion
For an object moving along a curved trajectory with constant speed
a = ar (radial only)
|ar |=
vt2 r ar v

3. Non-uniform Circular Motion
For an object moving along a curved trajectory,
with non-uniform speed
a = ar + aT (radial and tangential) aT
|ar |=
vT2 r ar dt
d| | v
|aT|=

4. Uniform or non-uniform circular motion
|aradial | =vTang2 / r
implies 
and if there is acceleration there MUST be a net force

5. Key steps
Identify forces (i.e., a FBD)
Identify axis of rotation
Apply conditions (position, velocity & acceleration)

6. Consider a person on a swing:
Example The pendulum
axis of rotation
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

7. Example Gravity, Normal Forces etc.mg T q y x
axis of rotation vT mg T
at top of swing vT = 0
Fr = m 02 / r = 0 = T – mg cos q
T = mg cos q
T < mg
at bottom of swing vT is max
Fr = m ac = m vT2 / r = T - mg
T = mg + m vT2 / r
T > mg

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
S Fz = 0 = T cos q – mg so
T = mg / cos q (> mg)
mar = (mg / cos q ) (sin q )
ar = g tan q = vT2/r
 vT = (gr tan q)½ L r
Period:
t = 2p r / vT =2p (r cot q /g)½
= 2p (L cos q /g)½

9. Conical Pendulum (very different)
Swinging a ball on a string of length L around your head
axis of rotation r L
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

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 ?

11. Orbiting satellites vT = (gr)½
Net Force: ma = mg = mvT2 / r gr = vT vT = (gr)½ The only difference is that g is less because you are further from the Earth’s center!

12. Orbiting satellites vT = (gr)½

13. Geostationary orbit

14. Geostationary orbit
The radius of the Earth is ~6000 km but at 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 * )½ m/s = 3000 m/s
At 3000 m/s, period T = 2p r / vT = 2p / sec =
= sec = 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)

15. 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 = (gr)1/2 vT mg

16. 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

17. 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…

18. Recap
Assignment: HW5,
For Wednesday: Finish reading through 1st four sections in Chapter 9 1

19. 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

20. Example Only one force is in the horizontal direction: static friction
x-dir: Fr = mar = -m |vT | 2 / r = Fs = -ms N (at maximum)
y-dir: ma = 0 = N – mg 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 y N x Fs mg
mcar= 1600 kg
mS = 0.5 for tire/road
r = 80 m
g = 10 m/s2

21. 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
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 y x N mg Fs
mpuck= 2 kg
mS = 0.8
r = 0.5 m
g = 10 m/s2

22. 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?

23. 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?

24. Banked Curves q Free Body Diagram for a banked curve.
Use rotated x-y coordinates
Resolve into components parallel and perpendicular to bank y x N
mar Ff mg q
( Note: 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!!)

25. 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 x y N
mar Ff mg q
At this moment you press the accelerator and, because of the frictional force (forward) by the tires on the road you begin to accelerate in that direction.
How does the radial acceleration change?

26. Navigating a hill
Knight concept exercise: A car is rolling over the top of a hill at speed v. At this instant,
n > w.
n = w.
n < w.
We can’t tell about n without knowing v.
At what speed does the car lose contact?
This occurs when the normal force goes to zero or, equivalently, when all the weight is used to achieve circular motion.
Fc = mg = m v2 /r  v = (gr)1/2 ½ (just like an object in orbit)
Note this approach can also be used to estimate the maximum walking speed.