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, N A = 6.02 x 10 23 .
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 ?
12
g
/
mol
6 10 23
atoms
/
mol
= 2 x 10 -23 g/atom 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
v a
dx dt dv dt
dt
2
v t t a t
Physics 151: Lecture 12, Pg 3
•
1-D Motion with
constant
acceleration
High-school calculus: Also recall that
a
t n dt
n 1
1 t n
1
const dv dt
Since
a
is constant , we can integrate this using the above rule to find:
v
a dt
a
dt
at
v 0
dx dt x
v dt
( at
v 0 ) dt
1 2 at 2
v 0 t
x 0
Physics 151: Lecture 12, Pg 4
v
v 0
at
Solving for t:
t
v
v 0 a
Derivation:
x
x 0
v 0 t
1 2 at 2
Plugging in for t:
x
x 0
v 0 a v 0
1 2 a
a v 0
2 v 2
v 0 2
2 a ( x
x 0 )
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 Alice straight down and Bill straight up . The
v 0
, speed of the balls when they hit the ground are
v A
and
v B
respectively.
Which of the following is true: (a)
v A
(b)
v A
(c)
v A
<
v B
=
v B
>
v B
Alice
v 0 v 0
Bill
H v A v B
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
= (r x ,r y ) = (x,y) We can convert between the two as follows:
r x r y = x = r cos = y = r sin
q q
y
r
r y
r
x
2
y
2 q arctan( y / x ) r q x
(x,y)
• In cylindrical coordinates, r is the same as the magnitude of the vector
x
Physics 151: Lecture 12, Pg 7
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
See text: 3-4
û U
Useful examples are the cartesian unit vectors [
i, j, k
] point in the direction of the
x
,
y
and
z
axes.
R = r x
i
+ r y
j
+ r z
k
z
k
y
j i
x
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 “ vector product ”: result is a vector
A B
= 0
A B
= 0 |
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
y o v
q
D h
Physics 151: Lecture 12, Pg 10
Problem 1
Variables
v o
= 36.5 m/s
y
o
= 1 m
h = 3 m
q o
= 30 º
D
= 113 m
a = (0,a y )
a y
t = unknown, = -g
Y f – 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 v x = v 0x x = v x t v y = v y = y 0 0y - g t + v 0y t 1 / 2 g t 2 y 0 g
v
q
v 0x v 0y
And, use geometry to find v ox and v oy Find and
v 0x v 0y = |v| cos
q .
= |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 v 0
q
t = 0
PROJECTILE
t = t 1 x
Physics 151: Lecture 12, Pg 13
See text: 4-4
What is Uniform Circular Motion (UCM) ?
Motion in a circle with: Constant Radius
R
Constant Speed
v = |v| y
a
R
v
(x,y) x
acceleration ?
a a
= 0 = 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
d
q
/dt =
.
q =
t
has units of
radians/second
.
y
v
Displacement
s = vt
.
but
s = R
q
= R
t,
so:
R
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 2 v
2 v v
1
Similar triangles:
v v
R R
But
R = v
t
for small
t
R
R v 1
So:
v
v R
v t
v 2 R a
v 2 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/s 2 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/s 2 . 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,
F
NET =
F = ma
Law 3
: Forces occur in pairs:
F
A ,B = - F B ,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
m a
Units The units of force are kg m/s 2 = 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
t o , v o 0
Cart
=
x 1
t 1 =1s, v 1 t 2 =2s, v 2
Cart
Air Track x 2
Cart
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 removed in the two cases: F app (i) after applying F app vs. (ii) after aplying F app for 1 s for 2 s ?
Cart
F app 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 F app removed ?
Cart
F app = 2N Air Track
F app = 0
t o , v o1
Cart F tot = 0 ?
otherwise v 1 =v 01 , cart keeps moving !
t 1 , v 1 0
Cart
=
at rest
x 1 F app = 2N
Cart
Air Track B) 4 x as long
F app = 0
t o , v o2
Cart F tot = 0 ?
x 2
t 2 , v 2 0
Cart at rest
=
Physics 151: Lecture 12, Pg 25
Lecture 8,
Act 2
You are going to pull two blocks (m A =4 kg and m B =6 kg) at constant acceleration (a= 2.5 m/s 2 ) 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
A
rope
B
a= 2.5 m/s 2 Physics 151: Lecture 12, Pg 26
Lecture 8,
Act 2
Solution:
m A
total mass !
What are the relevant forces ?
m A m B
F app a= 2.5 m/s 2 F app F app = a (m A + m B ) = 2.5m/s 2 ( 4kg+6kg) = 25 N rope -T T a A = a = 2.5 m/s 2 T a = 2.5 m/s 2 T = a m A T = 2.5m/s 2 4kg = 10 N T > 9 N, rope will brake ANSWER (A)
m B
-T a = 2.5 m/s 2 F app F app - T = a m B T = 25N - 2.5m/s 2 6kg=10N T > 9 N, rope will brake THE SAME ANSWER -> (A) Physics 151: Lecture 12, Pg 27
Inclined plane...
Consider
x
i
: mg sin
q and
y = ma
components separately:
a = g sin
q
j
: N mg cos
q 0 .
N = mg cos
q See text: Example 5.7
N
ma
m
mg sin
q q
mg mg cos
q q
j i
Physics 151: Lecture 12, Pg 28
Free Body Diagram T
2
T
1 q 1
Eat at Bob’s
m
g
Add vectors : Vertical (y): mg = T 1 sin q 1 + T 2 sin q 2 Horizontal (x) : T 1 cos q 1 = T 2 cos q 2 q 2
T
1 q 1 y q 2 m
g T
2 x Physics 151: Lecture 12, Pg 29
Example-1 with pulley
Two masses M 1 and M 2 are connected by a rope over the pulley as shown. Assume the pulley is massless and frictionless.
Assume the rope massless.
If M 1 > M 2 find : Acceleration of M 1 ?
Acceleration of M 2 ?
Tension on the rope ?
T 1 Video Animation M 2 M 1 a T 2 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 rope and the magnitude of
F
.
F
. Find the tension in each segment of the Assume the pulleys massless and frictionless.
Assume the rope massless.
T 4 T 1 T 2 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 2 nd law :
F
= m
a F
< M T 5 T 3 Free-body diagram for each object Physics 151: Lecture 12, Pg 31
Pulleys: continued
FBD for all objects T 4 T 1 T 2
F
< M T 5 T 3
T
2
T
3 M
T
5 M
g T
5
T
4
F=T
1
T
2
T
3 Physics 151: Lecture 12, Pg 32
Pulleys: finally
Step 3: Plan the solution (what are the relevant equations)
F
= m
a
, static (no acceleration: mass is held in place)
T
5 M T 5 =Mg M
g T
4 T 1 +T 2 +T 3 =T 4 F=T 1
T
2
T
3 T 2 +T 3 =T 5
F=T
1
T
3
T
5
T
2 Physics 151: Lecture 12, Pg 33
Pulleys: really finally!
Step 4: execute the plan (solve in terms of variables) We have (from FBD): F=T 1 T 5 =Mg T 2 +T 3 =T 5 Pulleys are massless and frictionless T 1 +T 2 +T 3 =T 4 T 4 T 1 =T 3 T 2 =T 3 T 2 +T 3 =T 5 gives T 5 =2T 2 =Mg T 2 =Mg/2 T 1 T 2 T 3 T 1 =T 2 =T 3 =Mg/2 and T 4 =3Mg/2 T 5 =Mg and F=T 1 =Mg/2 Step 5: evaluate the answer (here, dimensions are OK and no numerical values)
F
< M T 5 Physics 151: Lecture 12, Pg 34
See text: 6-1
Force of friction acts to oppose motion:
Dynamics:
i : j : F
K
N
=
m a N = mg so F
K
mg
=
m a
N j F
ma
K
mg mg
i
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. 0.20
0.18
0.15
D. 0.13
E. 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 m 1 = 4.00 kg , m 2 = 1.00kg
and m 3 = 2.00kg
. m 2 T 1 m 1 m 3 a) b) 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
T 1 2 m 1 T 12 m 2 T 23 T 1 m 3 T 23 a T 12 m 1 m 1 g T 12 m 1 g = - m 1 a T 12 a T 23 m 2 m 2 g k m 2 g a T 23 m 3 m 3 g T 23 m 3 g = m 3 a -T 12 + T 23 + k m 2 g = - m 2 a SOLUTION: T 12 = = 30.0 N , T 23 = 24.2 N , a = 2.31 m/s 2 left for m 2 Physics 151: Lecture 12, Pg 38
An example before we considered a race car going around a curve on a flat track.
N F
f What’s differs on a banked curve ?
m
g
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