Transcript Physics 131: Lecture 14 Notes

REVIEW
Phys-151

Lectures 1-11
» Displacement, velocity, acceleration
» Vectors
» Newton’s laws of motion
» Friction, pulleys
Physics 151: Lecture 12, Pg 1
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 ?
12g/mol
-23 g/atom
M carbon 
=
2
x
10
6 1023 atoms/mol
Physics 151: Lecture 12, Pg 2
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
x  x( t )
dx
dt
dv
d 2x
a 

dt
dt 2
v 
v
a
t
t
t
Physics 151: Lecture 12, Pg 3
1-D Motion with constant acceleration
n
 t dt 
1
t n 1  const
n 1

High-school calculus:

Also recall that

Since a is constant, we can integrate this using the above
rule to find:
v   a dt  a  dt  at  v 0
•
a 
dv
dt
Similarly, since v 
dx
we can integrate again to get:
dt
1
x   v dt   ( at  v 0 )dt  at 2  v 0 t  x0
2
Physics 151: Lecture 12, Pg 4
Derivation:
v  v 0  at

Solving for t:
t
v  v0
a
1
x  x0  v 0 t  at 2
2

Plugging in for t:
 v  v0  1  v  v0 
x  x0  v 0 
  a

 a  2  a 
2
v 2  v 0  2a( x  x0 )
2
Physics 151: Lecture 12, Pg 5
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
Alice
(c) vA > vB
v0
Bill
v0
H
vA
vB
Physics 151: Lecture 12, Pg 6
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:
rx = x = r cos q
ry = y = r sin q
y
ry
(x,y)
r
r  x2  y2
q
qarctan( y / x )
rx
x
• In cylindrical coordinates, r is the same as the
magnitude of the vector
Physics 151: Lecture 12, Pg 7
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
y

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
j
k
i
x
z
Physics 151: Lecture 12, Pg 8
Multiplication of vectors / Recap

There are two common ways to multiply vectors
“scalar product”: result is a scalar
A B = |A| |B| cos(q)
q
A B =0
A B =0
“vector product”: result is a vector
|A B| = |A| |B| sin(q)
q
A B =0
A B =0
Physics 151: Lecture 12, Pg 9
Problem 1:
1)
We need to find how high the ball is at a
distance of 113m away from where it starts.
Animation
v
yo
h
q
D
Physics 151: Lecture 12, Pg 10
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
Physics 151: Lecture 12, Pg 11
Problem 1
3) For projectile motion,
 Equations of motion are:
y
vx = v0x
vy = v0y - g t
x = vx t
y = y0 + v0y t - 1/ 2 g t2
v
y0
And, use geometry to find vox and voy
q
v0y
v0x
Find
and
g
v0x = |v| cos q.
v0y = |v| sin q.
x
Physics 151: Lecture 12, Pg 12
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 ?
t=0
TARGET
y
v0
t=0
t = t1
q
PROJECTILE
x
Physics 151: Lecture 12, Pg 13
See text: 4-4
What is Uniform Circular Motion (UCM) ?

Motion in a circle with:
y
Constant Radius R
v
(x,y)
a
 Constant Speed v = |v|
R
x
acceleration ?
a
=0
a
= const.
Physics 151: Lecture 12, Pg 14
Polar Coordinates...



In Cartesian co-ordinates we say velocity dx/dt = v.
 x = vt
In polar coordinates, angular velocity dq/dt = .
 q = t
y
 has units of radians/second.
v
Displacement s = vt.
R
but s = Rq = Rt, so:
qt
v = R
s
x
Physics 151: Lecture 12, Pg 15
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 ?
Physics 151: Lecture 12, Pg 16
Acceleration in UCM:


This is called Centripetal Acceleration.
Now let’s calculate the magnitude:
v
v2
v2
R
R
v1
Similar triangles:
v1
v R

v
R
But R = vt for small t
v vt

So:
v
R
v v 2

t
R
v2
a
R
Physics 151: Lecture 12, Pg 17
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
Physics 151: Lecture 12, Pg 18
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 ?
Physics 151: Lecture 12, Pg 19
See text: Chapter 5
Dynamics

Isaac Newton (1643 - 1727) 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.)
Physics 151: Lecture 12, Pg 20
See text: 5-4
Newton’s Second Law
The acceleration of an object is directly proportional to the
net force acting upon it. The constant of proportionality is
the mass.
 
F  F
NET


 ma
Units
The units of force are kg m/s2 = Newtons (N)
The English unit of force is Pounds (lbs)
Physics 151: Lecture 12, Pg 21
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
Physics 151: Lecture 12, Pg 22
Lecture 7, ACT 1
F= 2N
t1 =1s, v1
to , vo =
0
Cart
x1
t2 =2s, v2
Cart
Cart
Air Track
x2
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
Physics 151: Lecture 12, Pg 23
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
Air Track
Cart
Cart
at rest
A) 8 x as long
B) 4 x as long
C) Same
D) 2 as long
E) can’t determine
Physics 151: Lecture 12, Pg 24
Lecture 7, ACT 1a
What is the distances traveled after Fapp removed ?
Fapp= 2N
Cart
Air Track
Fapp = 0
to , vo1
Ftot = 0 ?otherwise v1=v01, cart keeps moving !
t1 , v1 =
0
Cart
Cart
at rest
x1
Fapp= 2N
Fapp = 0
to , vo2
Ftot = 0 ?
Cart
Cart
Air Track
t2 , v2 =
0
Cart
x2
at rest
B) 4 x as long
Physics 151: Lecture 12, Pg 25
Lecture 8, Act 2

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
A
rope
(C) NO
a= 2.5 m/s2
B
Physics 151: Lecture 12, Pg 26
Lecture 8, Act 2
Solution:

What are the relevant forces ?
mB
mA
mA
rope
T
Fapp = a (mA + mB)
Fapp = 2.5m/s2( 4kg+6kg)
= 25 N
a= 2.5 m/s2
Fapp
aA = a = 2.5 m/s2
T
-T
-T
T = a mA
T = 2.5m/s2 4kg = 10 N
T > 9 N, rope will brake
ANSWER (A)
a = 2.5 m/s2
a = 2.5 m/s2
mB
total mass !
Fapp
Fapp - T = a mB
T = 25N - 2.5m/s2 6kg=10N
T > 9 N, rope will brake
THE SAME ANSWER -> (A)
Physics 151: Lecture 12, Pg 27
See text: Example 5.7
Inclined plane...

Consider x and y components separately:
i: mg sin q = ma
a = g sin q

j: N - mg cos q 0.

m
N = mg cos q
ma
j
mg sin q
N q
mg cos q
mg
i
q
Physics 151: Lecture 12, Pg 28
Free Body Diagram
T2
T1
q2
q1
Eat at Bob’s
mg
Add vectors :
Vertical (y):
mg = T1sinq1 + T2sinq2
Horizontal (x) :
T1cosq1 = T2cosq2
y
T2
T1
q1
q2
x
mg
Physics 151: Lecture 12, Pg 29
Example-1 with pulley


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.
If M1 > M2 find :
Acceleration of M1 ?
Acceleration of M2 ?
Tension on the rope ?
T1
T2
Video
M2
Animation
M1
a
Free-body diagram for each object
Physics 151: Lecture 12, Pg 30
Example-2 with pulley


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.
T4
T1
We use the 5 step method.
F
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
T3
T2
<
T5
M
Free-body diagram for each object
Physics 151: Lecture 12, Pg 31
Pulleys: continued

FBD for all objects
T4
T2
T3
T4
T1
T3
T2
5
F
<
T
T5
T3
F=T1
T5
M
T2
M
Mg
Physics 151: Lecture 12, Pg 32
Pulleys: finally

Step 3: Plan the solution (what are the relevant equations)
F=ma , static (no acceleration: mass is held in place)
T5
T5=Mg
M
T2
T1+T2+T3=T4
T4
Mg
F=T1
T3
T2+T3=T5
T3
F=T1
T
5
T2
Physics 151: Lecture 12, Pg 33
Pulleys: really finally!

Step 4: execute the plan (solve in terms of variables)
We have (from FBD):
F=T1
T5=Mg
T2+T3=T5
T1+T2+T3=T4
Pulleys are massless and frictionless
T1=T3
T4
T2=T3
T2+T3=T5 gives T5=2T2=Mg
T1
T2=Mg/2

T2
F
T1=T2=T3=Mg/2 and T4=3Mg/2
T5=Mg
T3
and F=T1=Mg/2
Step 5: evaluate the answer (here,
dimensions are OK and no numerical values)
<
T5
M
Physics 151: Lecture 12, Pg 34
See text: 6-1
Force of friction acts to oppose motion:

Dynamics:
i:
j:
F  KN = m a
N = mg
F  Kmg = m a
so
j
N
F
ma
K mg
i
mg
Physics 151: Lecture 12, Pg 35
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.
0.20
0.18
0.15
0.13
0.27
Physics 151: Lecture 12, Pg 36
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.
m2
m1
a)
b)
T1
m3
What is the magnitude and direction of acceleration on the
three blocks ?
What is the tension on the two cords ?
Physics 151: Lecture 12, Pg 37
T12
T23
m2
T1
T1
2
T23
m1
T12
T12
k m2g
a
m2g
m1g
T23
a
T23
m2
m1
a
m3
m3
m3g
T23 - m3g = m3a
T12 - m1g = - m1a
-T12 + T23 + k m2g = - m2a
SOLUTION: T12 = = 30.0 N ,
T23 = 24.2 N
,a
= 2.31 m/s2 left for m2
Physics 151: Lecture 12, Pg 38
An example before we considered a race car going
around a curve on a flat track.
N
Ff
mg
What’s differs on a banked curve ?
Physics 151: Lecture 12, Pg 39
Example
Gravity, Normal Forces etc.
Consider a women on a swing:
Animation
1. When is the tension on the rope largest ?
2. 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)
Physics 151: Lecture 12, Pg 40