1. REVIEW Phys-151 Lectures 1-11 Displacement, velocity, acceleration
Vectors
Newton’s laws of motion
Friction, pulleys

2. What is the mass of a single carbon atom ?
See text : 1-3
Atomic Density
In dealing with macroscopic numbers of atoms (and similar small particles) we often use a convenient quantity called Avogadro’s Number, NA = 6.02 x 1023.
Molar Mass and Atomic Mass are nearly equal
1. Molar Mass = mass in grams of one mole of the substance.
2. Atomic Mass = mass in u (a.m.u.) of one atom of a substance, is approximately the number of protons and neutrons in one atom of that substance.
Molar Mass and Atomic Mass are other units for density.
What is the mass of a single carbon atom ?
= 2 x g/atom

3. Displacement, Velocity, Acceleration
If the position x is known as a function of time, then we can find both velocity v and acceleration a as a function of time! x t v t a t

4. 1-D Motion with constant acceleration
High-school calculus:
Also recall that
Since a is constant, we can integrate this using the above rule to find:
Similarly, since we can integrate again to get:

5. Derivation:
Solving for t:
Plugging in for t:

6. Lecture 2, ACT 4
Alice and Bill are standing at the top of a cliff of height H. Both throw a ball with initial speed v0, Alice straight down and Bill straight up. The speed of the balls when they hit the ground are vA and vB respectively.
Which of the following is true:
(a) vA < vB (b) vA = vB
(c) vA > vB v0
Bill
Alice H vA vB

7. Converting Coordinate Systems
See text: 3-1
Converting Coordinate Systems
In circular coordinates the vector R = (r,q)
In Cartesian the vector R = (rx,ry) = (x,y)
We can convert between the two as follows: y x
(x,y)  r ry rx
rx = x = r cos 
ry = y = r sin 
arctan( y / x )
In cylindrical coordinates, r is the same as the magnitude of the vector

8. Unit Vectors: A Unit Vector is a vector having length 1 and no units.
See text: 3-4
Unit Vectors:
A Unit Vector is a vector having length 1 and no units.
It is used to specify a direction.
Unit vector u points in the direction of U.
Often denoted with a “hat”: u = û U û x y z i j k
Useful examples are the cartesian unit vectors [ i, j, k ]
point in the direction of the x, y and z axes.
R = rxi + ryj + rzk

9. Multiplication of vectors / Recap
There are two common ways to multiply vectors
“scalar product”: result is a scalar
A B = |A| |B| cos(q)
A B = 0 q
“vector product”: result is a vector
A B = 0 q
|A B| = |A| |B| sin(q)

10. Problem 1:
We need to find how high the ball is at a distance of 113m away from where it starts.
Animation  v h D yo

11. Problem 1
Variables
vo = 36.5 m/s
yo = 1 m
h = 3 m
qo = 30º
D = 113 m
a = (0,ay)  ay = -g
t = unknown,
Yf – height of ball when x=113m, unknown,
our target

12. Problem 1 For projectile motion, Equations of motion are:
vx = v0x vy = v0y - g t
x = vx t y = y0 + v0y t - 1/ 2 g t2 y x g  v
v0x
v0y y0
And, use geometry to find vox and voy
Find v0x = |v| cos .
and v0y = |v| sin .

13. Problem 3 (correlated motion of 2 objects in 3-D)
Suppose a projectile is aimed at a target at rest somewhere above the ground as shown in Fig. below. At the same time that the projectile leaves the cannon the target falls toward ground.
Would the projectile now miss or hit the target ?  y x v0
t = 0
TARGET
PROJECTILE
t = t1

14. What is Uniform Circular Motion (UCM) ?
See text: 4-4
What is Uniform Circular Motion (UCM) ?
Motion in a circle with:
Constant Radius R
Constant Speed v = |v|
acceleration ? R v x y
(x,y) a
= 0 a
= const. a

15. Polar Coordinates... v = R
In Cartesian co-ordinates we say velocity dx/dt = v.
x = vt
In polar coordinates, angular velocity d/dt = .
 = t
 has units of radians/second.
Displacement s = vt.
but s = R = Rt, so: y v R s
t
v = R x

16. Lecture 5, ACT 2 Uniform Circular Motion
A fighter pilot flying in a circular turn will pass out if the centripetal acceleration he experiences is more than about 9 times the acceleration of gravity g. If his F18 is moving with a speed of 300 m/s, what is the approximate diameter of the tightest turn this pilot can make and survive to tell about it ?

17. Acceleration in UCM: This is called Centripetal Acceleration.
Now let’s calculate the magnitude:
Similar triangles: v1 v2 v
But R = vt for small t v2 v1 R
So: R

18. Lecture 6, ACT 2 Uniform Circular Motion
A satellite is in a circular orbit 600 km above the Earth’s surface. The acceleration of gravity is 8.21 m/s2 at this altitude. The radius of the Earth is 6400 km.
Determine the speed of the satellite, and the time to complete one orbit around the Earth.
Answer:
7,580 m/s
5,800 s

19. Lecture 6, ACT 3 Uniform Circular Motion
A stunt pilot performs a circular dive of radius 800 m. At the bottom of the dive (point B in the figure) the pilot has a speed of 200 m/s which at that instant is increasing at a rate of 20 m/s2.
What acceleration does the pilot have at point B ?

20. Dynamics See text: Chapter 5
Isaac Newton ( ) proposed three “laws” of motion:
Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame.
Law 2: For any object, FNET = F = ma
Law 3: Forces occur in pairs: FA ,B = - FB ,A
(For every action there is an equal and opposite reaction.)

21. Newton’s Second Law See text: 5-4
The acceleration of an object is directly proportional to the net force acting upon it. The constant of proportionality is the mass.
Units
The units of force are kg m/s2 = Newtons (N)
The English unit of force is Pounds (lbs)

22. Lecture 7, ACT 1 Newton’s Second Law
I push with a force of 2 Newtons on a cart that is initially at rest on an air table with no air. I push for a second. Because there is no air, the cart stops after I finish pushing. It has traveled a certain distance (before removing the force).
F= 2N
Cart
Air Track
For a second shot, I push just as hard but keep pushing for 2 seconds. The distance the cart moves the second time versus the first is (before removing the force) :
A) 8 x as long B) 4 x as long C) Same
D) 2 as long E) can’t determine

23. A) 8 x as long B) 4 x as long C) Same D) 2 as long E) can’t determine
Lecture 7, ACT 1
to , vo = 0
t1 =1s, v1
t2 =2s, v2
F= 2N
Cart
Cart
Cart
Air Track
Dx1
Dx2
A) 8 x as long B) 4 x as long C) Same
D) 2 as long E) can’t determine
B) 4 x as long

24. Lecture 7, ACT 1a
What is the distances traveled after Fapp removed in the two cases: (i) after applying Fapp for 1 s vs. (ii) after aplying Fapp for 2 s ?
Fapp
Cart
Cart
Cart
Air Track
at rest
A) 8 x as long B) 4 x as long C) Same
D) 2 as long E) can’t determine

25. Lecture 7, ACT 1a What is the distances traveled after Fapp removed ?
Ftot = 0 ?
otherwise v1=v01, cart keeps moving !
Fapp = 0
to , vo1
t1 , v1 = 0
Fapp= 2N
Cart
Cart
Cart
Air Track
at rest
Dx1
Fapp = 0
Ftot = 0 ?
Fapp= 2N
to , vo2
t2 , v2 = 0
Cart
Cart
Cart
Air Track
at rest
Dx2
B) 4 x as long

26. A B Lecture 8, Act 2 rope a= 2.5 m/s2
You are going to pull two blocks (mA=4 kg and mB=6 kg) at constant acceleration (a= 2.5 m/s2) on a horizontal frictionless floor, as shown below. The rope connecting the two blocks can stand tension of only 9.0 N. Would the rope break ?
(A) YES (B) CAN’T TELL (C) NO
rope
a= 2.5 m/s2 A B

27. Lecture 8, Act 2 Solution: total mass ! Fapp = a (mA + mB) Fapp
What are the relevant forces ?
Fapp = a (mA + mB)
Fapp = 2.5m/s2( 4kg+6kg)
= 25 N
Fapp
a= 2.5 m/s2 mA mB
T = a mA
T = 2.5m/s2 4kg = 10 N
T > 9 N, rope will brake T mA
aA = a = 2.5 m/s2
ANSWER (A) -T T
rope
a = 2.5 m/s2
Fapp - T = a mB
T = 25N - 2.5m/s2 6kg=10N
T > 9 N, rope will brake mB
a = 2.5 m/s2 -T
Fapp
THE SAME ANSWER -> (A)

28. Inclined plane... See text: Example 5.7
Consider x and y components separately:
i: mg sin  = ma a = g sin 
j: N - mg cos  = N = mg cos  ma i j m
mg sin 
mg cos  N mg  

29. Free Body Diagram T2 T1 Eat at Bob’s mg Add vectors : x y T2q1 q2
Eat at Bob’s mg
Add vectors : x y T2 q2
Vertical (y):
mg = T1sinq1 + T2sinq2
Horizontal (x) :
T1cosq1 = T2cosq2 T1 q1 mg

30. Example-1 with pulley Video Animation
Two masses M1 and M2 are connected by a rope over the pulley as shown.
Assume the pulley is massless and frictionless.
Assume the rope massless. M1 T2 T1 M2 a
If M1 > M2 find :
Acceleration of M1 ?
Acceleration of M2 ?
Tension on the rope ?
Video
Animation
Free-body diagram for each object

31. < Example-2 with pulley F
A mass M is held in place by a force F. Find the tension in each segment of the rope and the magnitude of F.
Assume the pulleys massless and frictionless.
Assume the rope massless.
< M T5 T4 T3 T2 T1 F
We use the 5 step method.
Draw a picture: what are we looking for ?
What physics idea are applicable ? Draw a diagram and list known and unknown variables.
Newton’s 2nd law : F=ma
Free-body diagram for each object

32. < Pulleys: continued F FBD for all objects T4 F=T1 T2 T3 M T5 T4 T3Mg

33. Pulleys: finally
Step 3: Plan the solution (what are the relevant equations)
F=ma , static (no acceleration: mass is held in place) M T5 Mg T4
F=T1 T2 T3
T5=Mg
T1+T2+T3=T4
F=T1 T2 T3 T5
T2+T3=T5

34. Pulleys: really finally!
Step 4: execute the plan (solve in terms of variables)
We have (from FBD):
T5=Mg
F=T1
T2+T3=T5
T1+T2+T3=T4
Pulleys are massless and frictionless
< M T5 T4 T3 T2 T1 F
T2=T3
T1=T3
T2+T3=T5 gives T5=2T2=Mg
T2=Mg/2
F=T1=Mg/2
T1=T2=T3=Mg/2 and T4=3Mg/2
T5=Mg
and
Step 5: evaluate the answer (here, dimensions are OK and no numerical values)

35. Force of friction acts to oppose motion:
See text: 6-1
Force of friction acts to oppose motion:
Dynamics:
i : F  KN = m a
j : N = mg
so F Kmg = m a j N F i ma
K mg mg

36. Lecture 9, ACT 4
In a game of shuffleboard (played on a horizontal surface), a puck is given an initial speed of 6.0 m/s. It slides a distance of 9.0 m before coming to rest. What is the coefficient of kinetic friction between the puck and the surface ? A B C D E

37. Example Problem 5.40 from the book
Three blocks are connected on the table as shown. The table has a coefficient of kinetic friction of 0.350, the masses are m1 = 4.00 kg, m2 = 1.00kg and m3 = 2.00kg. m1 T1 m2 m3
What is the magnitude and direction of acceleration on the three blocks ?
What is the tension on the two cords ?

38. T12
T23 m1 T1 m2 m3
T12
T23
m1g
T12
m3g
T23
T12
T23 a m2 m1 m3 a
mk m2g a
m2g
T12 - m1g = - m1a
T23 - m3g = m3a
-T12 + T23 + mk m2g = - m2a
SOLUTION: T12 = = 30.0 N , T23 = 24.2 N , a = 2.31 m/s2 left for m2

39. An example before we considered a race car going around a curve on a flat track.Ff mg
What’s differs on a banked curve ?

40. Example Gravity, Normal Forces etc.
Consider a women on a swing:
Animation
When is the tension on the rope largest ?
Is it : A) greater than
B) the same as
C) less than
the force due to gravity acting on the woman
(neglect the weight of the swing)