Physics 151: Lecture 12, Pg 1
REVIEWREVIEWPhys-151Phys-151
Lectures 1-11
» Displacement, velocity, acceleration» Vectors» Newton’s laws of motion» Friction, pulleys
Physics 151: Lecture 12, Pg 2
Atomic DensityAtomic 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.
M carbon 12g /mol
6 1023 atoms /mol
See text : 1-3
What is the mass of a single carbon atom ?
= 2 x 10-23 g/atom
Physics 151: Lecture 12, Pg 3
Displacement, Velocity, AccelerationDisplacement, 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!
v
t
x
t
a
t
adv
dt
d x
dt
2
2
vdx
dt
x x t ( )
Physics 151: Lecture 12, Pg 4
High-school calculus:
Also recall that
1-D Motion with 1-D Motion with constantconstant acceleration acceleration
constt1n
1dtt 1nn
adv
dt
0vatdtadtav
Since a is constant, we can integrate this using the above rule to find:
vdx
dt
002
0 xtvat21
dt)vat(dtvx
• Similarly, since we can integrate again to get:
Physics 151: Lecture 12, Pg 5
Derivation:Derivation:
Plugging in for t:
atvv 0 200 at
21
tvxx
Solving for t:
avv
t 0
200
00 avv
a21
avv
vxx
)xx(a2vv 02
02
Physics 151: Lecture 12, Pg 6
Lecture 2, Lecture 2, ACT 4ACT 4
AliceAlice and and BillBill are standing at the top of a cliff of are standing at the top of a cliff of heightheight HH. Both throw a ball with initial speed. Both throw a ball with initial speed vv00, ,
Alice straightAlice straight downdown and Bill straightand Bill straight upup. The . The speed of the balls when they hit the ground arespeed of the balls when they hit the ground are vvAA andand vvBB respectively.respectively.
vv00
vv00
BillBillAliceAlice
HH
vvAA vvBB
Which of the following is true:Which of the following is true:(a) (a) vvAA < < vvBB
(b) (b) vvAA = = vvBB
(c) (c) vvAA > > vvBB
Physics 151: Lecture 12, Pg 7
Converting Coordinate SystemsConverting Coordinate Systems In circular coordinates the vector R = (r,)
In Cartesian the vector R = (rx,ry) = (x,y)
We can convert between the two as follows:
See text: 3-1
• In cylindrical coordinates, r is the same as the magnitude of the vector
rx = x = r cos
ry = y = r sin
arctan( y / x )
22 yxr
y
x
(x,y)
rrry
rx
Physics 151: Lecture 12, Pg 8
Unit Vectors:Unit Vectors:
A Unit Vector Unit Vector is a vector having length 1 and no units.
It is used to specify a direction. Unit vector uu points in the direction of UU.
Often denoted with a “hat”: uu = û
UU
û û
See text: 3-4
x
y
z
ii
jj
kk
Useful examples are the cartesian unit vectors [ i, j, ki, j, k ]
point in the direction of the x, y and z axes.
R = rxi + ryj + rzk
Physics 151: Lecture 12, Pg 9
Multiplication of vectors / RecapMultiplication of vectors / Recap
There are two common ways to multiply vectors“scalar product”: result is a scalar
A B = |A| |B| cos()
|A B| = |A| |B| sin()
“vector product”: result is a vector
A B = 0A B = 0
A B = 0 A B = 0
Physics 151: Lecture 12, Pg 10
Problem 1:Problem 1:
1) We need to find how high the ball is at a distance of 113m away from where it starts.
vv
h
D
yo
Animation
Physics 151: Lecture 12, Pg 11
Problem 1Problem 1 Variables
vo = 36.5 m/s
yo = 1 m
h = 3 m
= 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 12
Problem 1Problem 1
3) For projectile motion, Equations of motion are:
vx = v0x vy = v0y - g t
x = vx t y = y0 + v0y t - 1/ 2 g t2
And, use geometry to find vox and voy
y
x
g
vv
v0x
v0yy0
Find v0x = |v| cos .
and v0y = |v| sin .
Physics 151: Lecture 12, Pg 13
Problem 3 (correlated motion of 2 objects in 3-D)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.
t = t1
y
x
vv00
t = 0
t = 0TARGET
PROJECTILE
Would the projectile now miss or hit the target ?
Physics 151: Lecture 12, Pg 14
What is Uniform Circular Motion (UCM) ?What is Uniform Circular Motion (UCM) ?
Motion in a circle with:
Constant Radius R
Constant Speed v = |vv|
acceleration ?
R
vv
x
y
(x,y)
See text: 4-4
= 0aa
aa
= const.aa
Physics 151: Lecture 12, Pg 15
Polar Coordinates...Polar Coordinates...
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 = Rt, so:RR
vv
x
y
st
v = R
Physics 151: Lecture 12, Pg 16
Lecture 5, Lecture 5, ACT 2ACT 2Uniform Circular MotionUniform 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 17
Acceleration in UCM:Acceleration in UCM:
This is called This is called Centripetal Acceleration. Now let’s calculate the magnitude:
vv1vv2
vv
vv2
vv1RR
RR
vv
RR
Similar triangles:
But R = vt for small t
So: vv
v tR
vt
vR
2
avR
2
Physics 151: Lecture 12, Pg 18
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.
Lecture 6,Lecture 6, ACT 2ACT 2Uniform Circular MotionUniform Circular Motion
Answer:• 7,580 m/s • 5,800 s
Physics 151: Lecture 12, Pg 19
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 ?
Lecture 6,Lecture 6, ACT 3ACT 3Uniform Circular MotionUniform Circular Motion
Physics 151: Lecture 12, Pg 20
DynamicsDynamics
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, FFNET = FF = maa
Law 3: Forces occur in pairs: FFA ,B = - FFB ,A
(For every action there is an equal and opposite reaction.)
See text: Chapter 5
Physics 151: Lecture 12, Pg 21
Newton’s Second LawNewton’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.
See text: 5-4
amFFNET
UnitsThe units of force are kg m/s2 = Newtons (N)The English unit of force is Pounds (lbs)
Physics 151: Lecture 12, Pg 22
Lecture 7,Lecture 7, ACT 1ACT 1Newton’s Second LawNewton’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).
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
Air Track
CartF= 2N
Physics 151: Lecture 12, Pg 23
Lecture 7Lecture 7, , ACT 1ACT 1
B) 4 x as long
Air Track
CartF= 2N
t1 =1s, v1 t2 =2s, v2to , vo = 0
Cart Cart
x1 x2
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 7Lecture 7, , ACT 1aACT 1a
Air Track
Cart CartCartFapp
at rest
What is the distances traveled What is the distances traveled afterafter F Fappapp removed in the two cases: removed in the two cases:
(i) after applying F(i) after applying Fappapp for 1 s for 1 s vs. vs.
(ii) after aplying F(ii) after aplying Fappapp for 2 s ? for 2 s ?
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 25
Lecture 7Lecture 7, , ACT 1aACT 1aWhat is the distances traveled after FWhat is the distances traveled after Fappapp removed ? removed ?
Air Track
Cart
to , vo1
to , vo2
Cart
Air Track
Cart Cart
t1 , v1 = 0
t2 , v2 = 0
Cart
Cart
Fapp= 2N
x1
x2
Fapp = 0
Fapp= 2NFapp = 0
Ftot = 0 ?
Ftot = 0 ?
at rest
at rest
otherwise v1=v01, cart keeps moving !
B) 4 x as long
Physics 151: Lecture 12, Pg 26
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
Lecture 8, Lecture 8, Act 2Act 2
A Ba= 2.5 m/s2rope
Physics 151: Lecture 12, Pg 27
What are the relevant forces ?
Lecture 8, Lecture 8, Act 2Act 2 Solution:Solution:
mAmB
Fapp
a= 2.5 m/s2
mA
rope
mB
T
-T T
-T
Fapp = a (mA + mB)Fapp = 2.5m/s2( 4kg+6kg) = 25 N
total mass !
aA = a = 2.5 m/s2
T = a mA T = 2.5m/s2 4kg = 10 NT > 9 N, rope will brakeANSWER (A)
Fapp - T = a mB T = 25N - 2.5m/s2 6kg=10NT > 9 N, rope will brake
a = 2.5 m/s2
Fapp
a = 2.5 m/s2
THE SAME ANSWER -> (A)
Physics 151: Lecture 12, Pg 28
Inclined plane...Inclined plane...
Consider x and y components separately: ii: mg sin = ma a = g sin
jj: N - mg cos . N = mg cos
mgg
NN
mg sin
mg cos
maa
ii
jj
See text: Example 5.7
m
Physics 151: Lecture 12, Pg 29
Free Body DiagramFree Body Diagram
Eat at Bob’s
T1
mg
T2
Add vectors :
x
yT1
T2
mg
Vertical (y):mg = T1sin1 + T2sin2
Horizontal (x) :T1cos1 = T2cos2
Physics 151: Lecture 12, Pg 30
Example-1 with pulleyExample-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 ?
Free-body diagram for each object
M1
T2T1
M2
aAnimation
Video
Physics 151: Lecture 12, Pg 31
Example-2 with pulleyExample-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.
M
T5
T4
T3T2
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
Physics 151: Lecture 12, Pg 32
Pulleys: continuedPulleys: continued FBD for all objects
M
T5
T4
T3T2
T1
F
T4
F=T1
T2
T3
T2 T3
T5
M
T5
Mg
Physics 151: Lecture 12, Pg 33
Pulleys: finallyPulleys: finally
Step 3: Plan the solution (what are the relevant equations)F=ma , static (no acceleration: mass is held in place)
M
T5
Mg
T5=Mg
T2 T3
T5
T2+T3=T5
T4
F=T1
T2
T3
F=T1
T1+T2+T3=T4
Physics 151: Lecture 12, Pg 34
Pulleys: really finally!Pulleys: really finally! Step 4: execute the plan (solve in terms of variables)
We have (from FBD):
T5=MgF=T1 T2+T3=T5 T1+T2+T3=T4
M
T5
T4
T3T2
T1
F
T2=T3T1=T3
T2=Mg/2
T2+T3=T5 gives T5=2T2=Mg
F=T1=Mg/2
T1=T2=T3=Mg/2 and T4=3Mg/2
T5=Mg and
Pulleys are massless and frictionless
Step 5: evaluate the answer (here, dimensions are OK and no numerical values)
Physics 151: Lecture 12, Pg 35
Force of friction acts to oppose motion:Force of friction acts to oppose motion:
Dynamics:
i : F KN = m a
j : N = mg
so F Kmg = m a
maaFF
mgg
NN
ii
j j
K mg
See text: 6-1
Physics 151: Lecture 12, Pg 36
Lecture 9, Lecture 9, ACT 4ACT 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. 0.20B. 0.18C. 0.15D. 0.13E. 0.27
Physics 151: Lecture 12, Pg 37
ExampleExampleProblem 5.40 from the bookProblem 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.
a) What is the magnitude and direction of acceleration on the three blocks ?
b) What is the tension on the two cords ?
m1
T1m2
m3
Physics 151: Lecture 12, Pg 38
m1
T1m2
m3
m1m2 m3
m2g
T23T12
m1g
T12
m3g
T23
T12
T12 T23
T23
T12 - m1g = - m1a T23 - m3g = m3a
k m2ga
a
a
-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 39
An example before we considered a race car going around a curve on a flat track.
N
mg
Ff
What’s differs on a banked curve ?
Physics 151: Lecture 12, Pg 40
ExampleExampleGravity, Normal Forces etc.Gravity, Normal Forces etc.
Consider a women on a swing:
1. When is the tension on the rope largest ? 2. Is it : A) greater than
B) the same asC) less than
the force due to gravity acting on the woman(neglect the weight of the swing)
Animation
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