1. Lecture 11 Goals Forces in circular motion Introduce concept of work
Problem solving with Newton’s 1st, 2nd and 3rd Laws
Forces in circular motion
Introduce concept of work
Introduce dot product 1

2. Loop-the-loop 1 Fr = mar = mvB2/r = N - mg N = mvB2/r + mg
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 vB mg

3. Another example of circular motion Loop-the-loop 2
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 ?

4. Fr = mar = mg = mvt2/r vt = (gr)1/2 Loop-the-loop 2
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). mg vt
Fr = mar = mg = mvt2/r
vt = (gr)1/2

5. 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…

6. Orbiting satellites
Net Force:
mar = mg = mvt2 / r
gr = vt2
The only difference is that g is less because you are further from the Earth’s center!

7. Example, Circular Motion Forces with Friction (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

8. Navigating a curve
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 (or 45 mph) y N x Fs mg
mcar= 1600 kg
mS = 0.5 for tire/road
r = 80 m
g = 10 m/s2

9. 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?

10. Banked Curves q Free Body Diagram for a banked curve.
(rotated x-y coordinates)
Resolve into components parallel and perpendicular to bank y N x Ff
No speed mg q

11. Banked Curves q q Free Body Diagram for a banked curve.
(rotated x-y coordinates)
Resolve into components parallel and perpendicular to bank x y N mg Ff
mar q
Low speed N mg Ff
mar q
High speed

12. Hanging Pink Fuzzy Dice Problem
You are in a car going around a horizontal curve
of radius 40.0 m and a speed of 10 m/s. There is a
0.10 kg pink fuzzy dice at the end of a 0.10 m string.
What is the angle of the die with respect to vertical?
(Little g is 10 m/s2)
x-dir SFx = mar = -m vT2 / r = -Tx = -0.1 x 2.5= N
y-dir SFy = may = 0 = Ty - mg  Ty = 0.1 x 10= 1.0 N
tan q =  q = 14°
But to you, if you are not paying attention, it may look like there is a mysterious force pushing the die out….a so-called centrifugal (center fleeing) or fictitious force.
Observers in accelerated frames of reference do not see proper physics and so come to erroneous conclusions. T mg

13. Drag forces (forces that oppose motion)
Serway & Jewett describe three models:
The velocity dependent model give a general analytic solution.
Terminal velocity for velocity dependent drag forces occur where drag force equals applied force

14. Drag at high velocities in a viscous medium
With a cross sectional area, A (in m2), D coefficient of drag (0.5 to 2.0),  density, and velocity, v (m/s), the drag force is:
R = ½ D  A v2
Example: Bicycling at 10 m/s (22 m.p.h.), with projected area of 0.5 m2 and drag coefficient 1 exerts a force of ~30 Newtons
At low speeds air drag is proportional to v but at high speeds it is v2
Minimizing drag is often important

15. Energy & Work Work (Force over a distance) describes energy transfer
No “working” definition for energy….yet
Net forces result in acceleration
Velocity can change direction, magnitude or both
Only net force acting along the path changes the speed
Forces acting over a distance  work….energy changes
Forces acting over a time  impulse…..momentum changes

16. Perpendicular Forces
I swing a sling shot over my head. The tension in the rope keeps the shot moving at constant speed in a circle.
Is there a net force?
Does the force act over a distance?
What if there were a tangential force….what would happen?
Only parallel forces change speed v Fc

17. Motion along a line x The only net force is along the horizontal
Start
Finish Fx mg
Fy net = 0
A net force acting along the path, over a distance, induces changes in speed (magnitude of the velocity)
We call this action “work”
The vector “dot” product simplifies the notation

18. Scalar Product (or Dot Product)î A Ax Ay q
Useful for finding parallel components
A  î = Ax
î  î = 1
î  j = 0
Calculation can be made in terms of components.
A  B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz )
Calculation also in terms of magnitudes and relative angles.
A  B ≡ | A | | B | cos q
You choose the way that works best for you!

19. Scalar Product (or Dot Product)
Compare:
A  B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz )
with A as force F, B as displacement Dr
Notice if force is constant:
F  Dr = (Fx )(Dx) + (Fy )(Dz ) + (Fz )(Dz)
Fx Dx +Fy Dy + Fz Dz
So here
Fnet  Dr ≡ Wnet
A Parallel Force acting Over a Distance does “Work”

20. Units: Force x Distance = Work mks cgs Other BTU = 1054 J
Newton x
[M][L] / [T]2
Meter = Joule
[L] [M][L]2 / [T]2
mks
cgs
Other
BTU = 1054 J
calorie = J
foot-lb = J
eV = 1.6x10-19 J
N-m or Joule
Dyne-cm or erg
= 10-7 J

21. Circular Motion
I swing a sling shot over my head. The tension in the rope keeps the shot moving at constant speed in a circle.
How much work is done after the ball makes one full revolution?
(A) W > 0 v
(B) W = 0 Fc
(C) W < 0
(D) need more info

22. Infinitesimal Work vs speed along a linear pathx F
Start
Finish Fx

23. Work vs speed along a linear path
If F is constant
DK is defined to be the change in the kinetic energy

24. Work in 3D….
x, y and z with constant F:

25. Examples of “Net” Work (Wnet)
DK = Wnet
Pushing a box on a smooth floor with a constant force; there is an increase in the kinetic energy
Examples of No “Net” Work
Pushing a box on a rough floor at constant speed
Driving at constant speed in a horizontal circle
Holding a book at constant height
This last statement reflects what we call the “system”
( Dropping a book is more complicated because it involves changes in U and K, U is transferred to K. The answer depends on what we call the system. )

26. Net Work: 1-D Example (constant force)
A force F = 10 N pushes a box across a frictionless floor for a distance x = 5 m. F
q = 0°
Start
Finish x
Net Work is F x = 10 x 5 N m = 50 J
1 Nm ≡ 1 Joule and this is a unit of energy
Work reflects energy transfer

27. Net Work: 1-D 2nd Example (constant force)
A force F = 10 N is opposite the motion of a box across a frictionless floor for a distance x = 5 m.
Start
Finish
q = 180° F x
Net Work is F x = -10 x 5 N m = -50 J
Work again reflects energy transfer

28. Work: “2-D” Example (constant force)
An angled force, F = 10 N, pushes a box across a frictionless floor for a distance x = 5 m and y = 0 m
Start
Finish F
q = -45° Fx x
(Net) Work is Fx x = F cos(-45°) x = 50 x 0.71 Nm = 35 J
Work reflects energy transfer

29. Work and Varying Forces (1D)
Consider a varying force F(x)
Area = Fx Dx
F is increasing
Here W = F · r
becomes dW = F dx Fx x Dx
Start
Finish F F
q = 0° Dx
Work has units of energy and is a scalar!

30. Fini
Read all of Chapter 7