1. Lecture 10 Employ Newton’s Laws in 2D problems with circular motion
Goals:
Employ Newton’s Laws in 2D problems with circular motion
Assignment: HW5, (Chapters 8 & 9, due 10/15, Wednesday)
For this Wednesday: Read Chapter 9 1

2. Uniform Circular Motion
For an object moving along a curved trajectory,
with non-uniform speed
a = ar (radial only) v
|ar |=
vT2 r ar
Perspective is important

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. Circular motion
Circular motion implies one thing
|aradial | =vT2 / r

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

6. 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?

7. Example Gravity, Normal Forces etc.
axis of rotation T T y x vT q mg mg
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 (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

9. 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 ?

10. 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

11. 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

12. 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…

13. 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

14. 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

15. 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

16. UCM: Acceleration, Force, Velocity

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?

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?

19. 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
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!!

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 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 accelerate in that direction.
How does the radial acceleration change?

21. Locomotion: how fast can a biped walk?

22. How fast can a biped walk?
What about weight?
A heavier person of equal height and proportions can walk faster than a lighter person
A lighter person of equal height and proportions can walk faster than a heavier person
To first order, size doesn’t matter

23. How fast can a biped walk?
What about height?
A taller person of equal weight and proportions can walk faster than a shorter person
A shorter person of equal weight and proportions can walk faster than a taller person
To first order, height doesn’t matter

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

25. Lecture 10 Employ Newton’s Laws in 2D problems with circular motion
Goals:
Employ Newton’s Laws in 2D problems with circular motion
Assignment: HW5, (Chapters 8 & 9, due 10/15, Wednesday)
For this Wednesday: Read Chapter 9
First exam (mean 71, high 100, low 20) 1