1. Chapter 5 Newton’s Laws and Friction Circular motion Drag Forces
Apparent forces
Banking curves
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. Friction You are standing still, then begin to walk.
What was the external forced that caused
you to accelerate?
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.
Hint: It is very hard to start walking if you
are standing on ice.
What force causes a car to accelerate when
a traffic light turns green?

3. 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
Friction is still not well understood
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

4. Note that, in general, μs > μk.

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. .
Problem: You have an inclined plane at angle θ with the horizontal with a mass at rest with impending motion. Find the coefficient of static friction μs.
1. Sketch a diagram and label it.
2. Define a coordinate system and origin
3. Write down the forces on your diagram.
4. Draw a free body diagram and resolve the forces
into components.
5. Apply F=ma to each axes.
6. Solve for the unknowns.
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13. 1. Sketch a diagram and label it.θ
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14. 2. Define a coordinate system and originθ +x m h θ d
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15. 3. Write down the forces on your diagram.mg +x +y h d θ N fs θ
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16. 4. Draw a free body diagram for the mass and resolve the forces into componentsθ N mg +x +y
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17. 5. Apply Newton's 2nd law in the x directionθ N mg +x +y
x direction
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18. 5. Apply Newton's 2nd law in the y directionθ N mg +x +y
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19. .
6. Solve for the unknown
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20. If the angle is large enough, then the block accelerates. What is a?θ N mg +x +y
x direction
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21. Problem: Find a and T for the Atwood’s machine with friction between M and surface.mg -y Mg N fk +y +x 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 T

22. 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

23. 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

24. Problem 30 Chap 5 Giancoli

25. Box A and box B, each of mass 2
Box A and box B, each of mass 2.7 kg, are on a surface inclined at 34° to the horizontal. The coefficient of kinetic friction is 0.15 and box is moving down.
(a)What is the acceleration of block A?
(b) What is the smallest value of the coefficient of kinetic friction that will keep the box from acceleration.

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

27. 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

28. Velocity-Dependent Forces: Drag and Terminal Velocity
When an object moves through a fluid, it experiences a drag force that depends on the velocity of the object.
For small velocities, the force is approximately proportional to the velocity like in water
For higher speeds, the force is approximately proportional to the square of the velocity like in air.

29. For slow moving objects in a viscous fluid
If the drag force on a falling object is proportional to its velocity, the object gradually slows until the drag force and the gravitational force are equal. Then it falls with constant velocity, called the terminal velocity.
Figure Caption: (a) Forces acting on an object falling downward. (b) Graph of the velocity of an object falling due to gravity when the air resistance drag force is FD = -bv. Initially, v = 0 and dv/dt = g, and but as time goes on dv/dt (= slope of curve) decreases because of FD. Eventually, v approaches a maximum value, vT, the terminal velocity, which occurs when FD has magnitude equal to mg. ma

30. Drag Force in Water FD=-kv
A 1000 kg 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 kg/s.

32. v = 45 km/hr
v0 =90 km/hr
k=70 kg/s
m=1000 kg

33. For high speed objects in air
For high speed objects moving thru air, (baseball,
skydiver, auto) the drag force due to air resistance is
Solve for v0
Stokes-Napier Law

34. 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.

35. How to solve this equation? Two ways
One way is to use a spread sheet in Exel and discrete coordinates.

37. What is terminal velocity of a baseball

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

39. Newtons 2nd Law

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

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

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

45. Solving equation continued

46. 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.

47. Calculations differ due to a large delta t.
When delta t becomes small enough, the two curves
should be indistinguishable.

48. 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
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.

49. 5-2 Uniform Circular Motion—Kinematics
Uniform circular motion: motion in a circle of constant radius at constant speed
Instantaneous velocity is always tangent to the circle.
Figure Caption: A small object moving in a circle, showing how the velocity changes. At each point, the instantaneous velocity is in a direction tangent to the circular path.

50. Dynamics of Uniform Circular Motion
For an object to be in uniform circular motion, there must be a net force acting on it.
We already know the acceleration, so can immediately write the force:
Figure Caption: A force is required to keep an object moving in a circle. If the speed is constant, the force is directed toward the circle’s center.

51. Dynamics of Uniform Circular Motion
We can see that the force must be inward by thinking about a ball on a string. Strings only pull; they never push.
Figure Caption: Swinging a ball on the end of a string.

52. 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.

53. 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

54. Centripetal Acceleration
Magnitudes are related by due to similar triangles .
And, so
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

55. 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,
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?

56. 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.

57. 5-3 Dynamics of Uniform Circular Motion
Example 5-13: Conical pendulum.
A small ball of mass m, suspended by a cord of length l, revolves in a circle of radius r = l sin θ, where θ is the angle the string makes with the vertical. (a) In what direction is the acceleration of the ball, and what causes the acceleration? (b) Calculate the speed and period (time required for one revolution) of the ball in terms of l, θ, g, and m.
Figure Caption: Example 5–13. Conical pendulum.
Answer: a. The acceleration is towards the center of the circle, and it comes from the horizontal component of the tension in the cord.
b. See text.

58. Dynamics of Uniform Circular Motion
Calculate the speed v
Figure Caption: Example 5–13. Conical pendulum.
Answer: a. The acceleration is towards the center of the circle, and it comes from the horizontal component of the tension in the cord.
b. See text.
Calculate the period

59. 5-4 Highway Curves: Banked and Unbanked
Example 5-14: Skidding on a curve.
A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s (54 km/h). Will the car follow the curve, or will it skid? Assume: (a) the pavement is dry and the coefficient of static friction is μs = 0.60; (b) the pavement is icy and μs = 0.25.
Figure Caption: Example 5–14. Forces on a car rounding a curve on a flat road. (a) Front view, (b) top view.
Solution: The normal force equals the weight, and the centripetal force is provided by the frictional force (if sufficient). The required centripetal force is 4500 N.
The maximum frictional force is 5880 N, so the car follows the curve.
The maximum frictional force is 2450 N, so the car will skid.

60. 5-4 Highway Curves: Banked and Unbanked
Banking the curve can help keep cars from skidding. In fact, for every banked curve, there is one speed at which the entire centripetal force is supplied by the
horizontal component of the normal force, and no friction is required. This occurs when:
Figure Caption: Normal force on a car rounding a banked curve, resolved into its horizontal and vertical components. The centripetal acceleration is horizontal (not parallel to the sloping road). The friction force on the tires, not shown, could point up or down along the slope, depending on the car’s speed. The friction force will be zero for one particular speed.

61. Down is negative, Up is positive
Example 5-12: Revolving ball (vertical circle) What is tension at the top of the arc? Let T=FT1.
A kg ball on the end of a 1.10-m-long cord (negligible mass) is swung in a vertical circle. (a) Determine the minimum speed the ball must have at the top of its arc so that the ball continues moving in a circle.
Down is negative, Up is positive
Figure Caption: Example 5–12. Freebody diagrams for positions 1 and 2.
Solution: See the freebody diagrams.
The minimum speed occurs when the tension is zero at the top; v = m/s.
Substitute to find FT2 = 7.35 N.

62. Down is negative, Up is positive
Example 5-12: Revolving ball (vertical circle). What is tension at the bottom of the arc? Let T=F T2
(b) Calculate the tension in the cord at the bottom of the arc, assuming the ball is moving at twice the speed of part (a).
Down is negative, Up is positive
Figure Caption: Example 5–12. Freebody diagrams for positions 1 and 2.
Solution: See the freebody diagrams.
The minimum speed occurs when the tension is zero at the top; v = m/s.
Substitute to find FT2 = 7.35 N.

63. Example of fictitious force
In a vertically accelerated reference frame,
eg. an elevator, what is your apparent weight?
Apparent weight = N
Upwards is positive
Downwards is negative
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.
Upward acceleration
Downward acceleration

64. VERTICAL CIRCULAR MOTION
Down is negative, Up is positive
At the top:
SHOW Wine glass on tray.
SHOW Loop the Loop N r
Minimum v for N = 0:
(apparent weightlessness) N
At the bottom: