1. Numerical integration
Lecture 5 Chapter 5
Using Newton’s Laws
Friction
Circular motion
Drag Forces
Numerical integration
Misconceptions
1. SHOW CENTRIPETAL FORCE
2. SPINNING TRAY
3.CENTRIPETAL ACCELAERATION
4. BALL ON A STRING IN A TUBE; SHOWS TENSION ON STRING CAUSED BY WEIGHT
5. DROP A CUP OF WATER WITH A HOLE IN IT; APPARENT WEIGHTLESSNESS
6. LOOP THE LOOP : WEIGHTLESSNESS
7. AIR TABLE CIRCULAR MOTION; FRICTIONLESS CIRCULAR MOTION

2. Assignment 2 Chp 4 4-52, 4-67, 4-75 Chp 5 5-30, 5-44
Due Friday 12:00 Mar 11

3. Friction You are standing still, then begin to walk.
What was the external forced that caused
you to accelerate?
Hint: It is very hard to start walking if you
are standing on ice.
The second law states that whenever a body accelerates, an external force must be present to cause the acceleration. It is not always obvious what this force is even in very ordinary circumstances.
In order for you to accelerate forward when you start walking, something must push you forward. What is it? It must be the sidewalk. You push back with your feet, and the sidewalk pushes you forward by an equal amount according to Newton’s third law. This could not happen without friction.
You may think that it is your pushing against the sidewalk that causes you to move forward, but it is really the sidewalk pushing forward on you, otherwise you would not accelerate in that direction.
Same argument for a car accelerating.
What force causes a car to accelerate when
a traffic light turns green?

4. Frictional Forces
Friction is an attractive force between two surfaces that is a
result of the vector sum of many electrical forces between
the surface atoms of the two different bodies.
Only about 10-4 of the surface atoms actually contribute.
Model of dry friction 3 or 4 asperites support top block. Temporarily weld together
DEMONSTRATE STATIC AND KINETIC FRICTION BY PULLING A BLOCK WITH A STRING. PUT A SCALE IN BETWEEN TO MEASURE FORCE.
SHOW BLOCK ON INCLINED PLANE. TILT PLANE UNTIL IT MOVES
Models of friction
See Chabay and Sherwood
Matter and Ineractions
Volume 1
ISBN
The Friction and Lubrication of Solids
F. P. Bowden and D. Tabor, Oxford University Press
1964

5. No motion and no horizontal forces
What is the free body diagram without friction?

6. No Motion
What is the free body diagram with friction but no motion?

7. Still No Motion with larger Force F

8. Increase F more and now you get acceleration
Static friction force decreases to kinetic friction and its value remans constant and now you get acceleration

9. Constant velocity

10. How is the frictional force related to the normal?mg F fs
Fixed block N
The maximum value equals is
where N is the normal force. Above we would have The coefficient of static friction ranges from 0 to 1.2
Kinetic Friction: If we increase F until the block starts to
move, the friction force decreases to

11. Problem Solving with Newton’s 2nd Law involving friction
Vector sum of external forces in x direction = max
Vector sum of external forces in y direction = may
If no acceleration, then set sum equal to 0

12. .
Mass on incline plane at rest with impending motion. Find the coefficient of static friction What is the free body diagram? +y N θ +x h θ d mg θ
2/17/2019
Physics 631

13. Free Body Diagram +y N +x θ mg. θ N mg +x +y
Apply Newton's 2nd law in the x directions and using fs = μs N
x direction
2/17/2019
Physics 631

14. Free Body Diagram +y N +x θ mg. θ N mg +x +y
Apply Newton's 2nd law in the y direction
Apply Newtons 2nd Law
2/17/2019
Physics 631

15. . θ N mg +x +y
2/17/2019
Physics 631

16. Find a and T for the Atwood’s machine with friction between M and surface.mg -y Mg N f +y +x T fk
Demos for inclined planes, pulleys
Normal force and tension
Example problems Atwoods machine, Einstein in an elevator or accelerating scale, blocks accelerating T T

17. Find a and T for the Atwood’s machine with friction between M and surface.mg -y Mg N f
Demos for inclined planes, pulleys
Normal force and tension
Example problems Atwoods machine, Einstein in an elevator or accelerating scale, blocks accelerating

18. Find a and T for the Atwood’s machine with friction between M and surface.fk
Demos for inclined planes, pulleys
Normal force and tension
Example problems Atwoods machine, Einstein in an elevator or accelerating scale, blocks accelerating T T

19. Question: What is the minimum magnitude force required to start the crate moving?φ N fs mg +x +y
Tcosφ
y components
x components

20. Question: What is the initial acceleration if μk=0.35?φ N fs mg +x +y
Tcos φ
Newton's 2nd law
y components
x components

21. Inertial Drag Force and Terminal Velocity
Drag force: Whenever you have a body like a ball moving through a medium that behaves like a fluid, there will be a drag force opposing the motion.
Imagine a falling ball slowed down due to elastic collisions
with air molecules. Simply pushing the air out of the way.
Hand waving argument
air molecules
ball v A dy .
Inertial drag

22. Terminal speeds in air Using Newton’s 2nd law, Stokes-Napier Law
where m is the mass of the falling ball
Solve for v0
Stokes-Napier Law

23. TERMINAL SPEEDS IN AIR Object ` Speed (m/s) Speed (mph)
Feather
Snowflake BB
Mouse
Tennis ball
Baseball
Sky diver
Cannonball
As the above objects fall through the air they accelerate. As their speed increases, so does their air resistance. Eventually the upwards force of air resistance becomes equal to the weight of the object, and there is no net force on the falling body, and it’s speed becomes constant.
The sky diver’s terminal speed of 60 m/s (134 mph) is for a spread-eagle position. If the diver tucks his body into a spherical shape, his terminal speed will roughly double.
So the terminal speed of a falling body depends on its weight and shape. A feather falls slowly partly because it is light, but also because it is spread out and intercepts a great deal of air.
Show demo of falling feather in vacuum

24. How to solve this equation? Two ways
One way is to use a spread sheet in Exel.

25. Use Newtons 2nd Law Initial component of momentum:
Initial force on ball:
Using 2nd Law
Find new p

26. Newtons 2nd Law

27. Go to Excel Spread Sheet
631 Website: Lecture 3 Materials

28. =C16+(g-(b_1/m_1)*C16*C16)*delta_t
=D16+1/2*(C16+C17)*delta_t

31. We can also solve the equation to get the velocity as a function of time before it reaches terminal velocity. Let b = 1/2CrA

32. Solving equation continued

33. This can be integrated, and fixing the constant of integration by the requirement that the velocity be zero at t = 0 , which is the case for free fall we find:
Now show comparison of this solution with numerical integration with Excel.

34. Comparison The curve modeled by velocity squared for terminal velocity
Differs from the true equation due to a large delta t.
When delta t becomes small enough the two curves are
Indistinguishable.

35. Water Resistance and Drag Forces
Whenever you have a body moving through a liquid there will be a drag force opposing the motion. Here the drag force is proportional to - kv. Viscous drag.
Stokes Law: terminal velocity is proportional to mass
A 1000 km boat in the water shuts off its engine
at 90 km/hr. Find the time required to
slow down to 45 km/hr due to a water
drag force equal to -70v, where v is
the speed of the boat. Let k = 70.
ma = - kv
v/v0 = 45/90 =1/2
t = m/k ln 2=1000/70 ln 2 = 9.9 s

36. UNIFORM CIRCULAR MOTION
Centripetal Acceleration: accelerates a body by changing the direction of the body’s velocity without changing the speed. There must be a force also pointing radially inward to make this motion.
Examples:
Ball on a string : show demo: Force is produced by the weight of the mass and transmitted by the tension in the string.
Moon in orbit around the earth: gravitational force
A car making a sharp turn: friction
A carousel; friction and contact forces
Demo: pushing bowling ball with broom in a circle
SHOW HOCKEY PUCK ON AIR TABLE
TENSION PROVIDED BY HANGING MASS
SHOW BALL ON A STRING
TENSION PROVIDED BY YOUR FINGERS
In studying uniform circular motion we will be using the same ideas of kinematics (x, v, a) and Newton’s Laws of motion that we have been studying, but we will apply them to situations involving motion in circles.
A new word in involved: Centripetal. This just means directed towards the center. A centripetal acceleration is one directed towards the center of the circle in which an object is moving.
Roller coasters at amusement parks use circular motion to achieve apparent weightlessness momentarily. This can occur near earth where g is still strong, but the object behaves in some ways as if it were far from any star or planet.
The example we are working towards is the moon, in order to understand how Newton compared the fall of an apple with the motion of the moon, taking his first step towards understanding gravity.

37. Uniform circular motion
Centripetal force is really not a new force like gravity, tension, friction.
Motion of earth around sun – centripetal force is a result of gravity
Rock whirled around on a string – centripetal force is a result of tension
Sometimes it is a result of friction or the normal force

38. CENTRIPETAL ACCELERATION:.
Δv points radially inward
Here we have an object moving in a circle with a constant speed. Why is there any acceleration? Simply because velocity is a vector quantity, and in this case, its magnitude doesn’t change, but its direction does.
Consider our moving object at two times: It has moved from r zero to r. Here I have moved the two r vectors away so we can see them more easily. The change in r is delta r as shown.
Since the position is changing with time, there is a non-zero velocity. The velocity vectors are shown in red. The velocity is perpendicular to the radius vector at all times during uniform circular motion. The velocity is always tangent to the circle describing the motion.
But in that case, the velocity itself is rotating around in a circle just like the radius vector. Here I have moved the two velocity vectors away so we can see them also. Because v is always perpendicular to r, the angle between the two r vectors is the same as that between the two v vectors.
This means that the r triangle is similar to the v triangle. They have the same shape. One thing that means is that the ratios of corresponding sides are equal. We also see that as Δt becomes small, Δv is perpendicular to v just as Δr is perpendicular to r.

39. CENTRIPETAL ACCELERATION.
Here we have an object moving in a circle with a constant speed. Why is there any acceleration? Simply because velocity is a vector quantity, and in this case, its magnitude doesn’t change, but its direction does.
Consider our moving object at two times: It has moved from r zero to r. Here I have moved the two r vectors away so we can see them more easily. The change in r is delta r as shown.
Since the position is changing with time, there is a non-zero velocity. The velocity vectors are shown in red. The velocity is perpendicular to the radius vector at all times during uniform circular motion. The velocity is always tangent to the circle describing the motion.
But in that case, the velocity itself is rotating around in a circle just like the radius vector. Here I have moved the two velocity vectors away so we can see them also. Because v is always perpendicular to r, the angle between the two r vectors is the same as that between the two v vectors.
This means that the r triangle is similar to the v triangle. They have the same shape. One thing that means is that the ratios of corresponding sides are equal. We also see that as Δt becomes small, Δv is perpendicular to v just as Δr is perpendicular to r.
Triangles are similar

40. Centripetal Acceleration
Magnitudes are related by due to similar triangles .
And, so
Magnitude of
From the similarity of the triangles we see that the change in r divided by r is equal to the change in v divided by v during the same small time interval Dt. This allows us to solve for the centripetal acceleration magnitude as shown. This is a very simple and useful result: Whenever an object moves in uniform circular motion, it is undergoing an acceleration equal to v2/r.
But a is a vector also. What direction does it point? Looking at the previous slide we can see that a must be perpendicular to v just as v is perpendicular to r in the limit as theta (delta t) gets very small.
Period of the motion

41. What is the magnitude of ac and its direction for a radius of r = 0
What is the magnitude of ac and its direction for a radius of r = 0.5 m and a period of T= 2 s,
Need to find v
What is the direction of ac ?
SHOW DEMO OF BALL ON STRING IN TUBE Here is a game we have all played: Whirl a ball around on a string. Let us define T to be the period of the motion ie the time for the ball to go around one time. As long as we keep it moving with a constant period and radius, it is exhibiting uniform circular motion. For the given radius and period, what is the magnitude of the centripetal acceleration?
We know the radius, so all we need is the speed. How do we find that? The period is the time for one round trip, ie one time around the circle. How far does the ball go during one period? Just the circumference of the circle. So the speed (magnitude of the velocity) is 2 pi r divided by T which is 1.6 m/s in this case.
Then the centripetal acceleration is just the velocity squared divided by the radius, or 5 m/ss.
Question: What if I speed the ball up so as to cut the period in half. What is the centripetal acceleration now?
INWARD

42. QUALITATIVE QUIZ A ball is being whirled around on a string.
The string breaks. Which path does the
ball take? v a c e d b
Ignore all other forces except that of the string acting on the ball. At a certain moment when the ball is at the position shown in the picture, the string breaks. What path does the ball take immediately after?
What principle can we use to answer this question? What idea is involved?
When the string breaks the ball is moving straight up in the picture, and there are no forces acting on it. So it keeps moving in the same direction in a straight line with constant velocity.

43. See notes on vertical circular motion
Leads to apparent weight and fictitious forces

44. VERTICAL CIRCULAR MOTION
Down is negative, Up is positive
At the top: N
Minimum v for N = 0:
(apparent weightlessness) r
SHOW Wine glass on tray.
SHOW Loop the Loop N
At the bottom:
Apparent weight = N = mv2/r + mg
Weigh more

45. What do we mean by Fictitious Forces Ff = - ma (the fictitious force always acts in the opposite direction of acceleration) T θ ma T T mg Ff mg T θ

46. Example of fictitious force (Ff = - ma)
In a vertically accelerated reference frame, eg. an elevator,
what is your apparent weight? Apparent weight = N
Upwards is positive
Downwards is negative
Upward acceleration
When you are in an elevator and it begins to move upwards, you feel heavier than usual for a moment. As the elevator slows and stops, you feel lighter for a moment. What happens in these examples, is that the normal force the elevator exerts on you increases or decreases as the elevator accelerates. We can understand this using Newton’s second law.
Downward acceleration
Free fall a = g
N = 0 Weightless condition