Transcript Notes PPT

Dynamics Unit
Vocabulary:
Dynamics: The study of Forces and their interactions with matter.
Force: (F) A push or a pull. (A vector quantity, has magnitude and
direction.)
Inertial Mass: (m) The property of matter that makes it resist being
accelerated.
Gravitational Mass: (m) The property of matter that makes it attract
other matter at a distance.
Weight: (w) The force of gravity on an object
W = mg
Weight = mass * acceleration of gravity
1 Newton = (kg) *
(m/s2)
Dynamics Unit
Newton’s Laws of Motion
Newton’s First Law- INERTIA
An object at rest will remain at rest and an object in motion will
remain in motion unless acted on by an unbalanced force.
Newton’s Second Law – FNET = ma
Newton’s Third Law – Action-Reaction
For every applied force there is an equal and opposite reaction
force.
Bell Warmer
A sprinter accelerates uniformly from rest for 4 seconds. If the
sprinter covers a distance of 40 m, find the acceleration of the
sprinter and the final velocity of the sprinter.
a) How much force would you have to apply to a 50 kg mass on a
surface with a coefficient of static friction of 0.8 to make the mass
move?
b) If the mass above has a coefficient of kinetic friction of 0.3, at
what rate would the mass accelerate after it starts sliding (assume
the same force continues to be applied.)
d=?
Height of window = 1 m
A bald alien visiting Earth sees a flower pot fall past their
window and notices that it took 0.25 seconds to travel
from the top of the window to the bottom.
Find d (the distance between the bottom of the top
window and the top of the bottom window.)
100 N
50 kg
d=?
A constant force of 100 N acts on a 50 kg mass for 10
seconds. If the mass started from rest,
a)How far will it move during the ten seconds?
b)How fast will it be moving at the end of 10 seconds?
Table of Coefficients of Friction for plastic.
Surface
Marble
Oak
Glass
ms
0.4
0.6
0.4
mk
0.2
0.3
0.2
A 50 gram block of plastic rests on an oak table.
a) What minimum horizontal force must be applied to
make the block start moving?
b) How much force must be applied to keep the block
moving at a constant speed?
Dynamics Unit
Vocabulary:
Centripetal Acceleration: Acceleration of an object with a velocity
that is changing direction. (Always directed inward toward the
center of the circular motion, perpendicular to the velocity.)
Equilibrium: An object is at equilibrium when its acceleration is
zero. (That includes when the velocity is constant.)
Free Body Diagram: A diagram of one body in a system showing all
the forces that act on that body.
Dynamics Unit
Formulas
W = mg
FNET = ma ( SFforward – SFbackward = ma }
Ffriction = m Fnormal (Remember the difference between Static and
kinetic friction)
acentripetal = v2/R (always directed perpendicular to the velocity,
toward the center of the circle.)
A 100 kg person stands on an electronic scale in an
elevator. If the scale reads the force exerted on the scale
by the person standing on the scale, what is the reading
on the scale when the elevator is accelerating upward at
3 m/s2.
a=3 m/s2
Scale
1. Draw Free Body Diagram.
2. Resolve any oblique forces into x and y
components
3. Use Newton’s 2nd Law to consider the
forces and acceleration in the y direction.
nd
4. Use Newton’s 2 Law to consider the
forces and acceleration in the x direction
Vo = ?
Ho = 4 m
Dx = 5 m
F = 500 N
20o
100 kg
You are pushing a 100 kg wheel barrow at a constant
speed by applying a force of 500N to the handle which
makes an angle of 20 degrees with respect to
horizontal. Find the coefficient of kinetic friction.
Inclined Planes
FN
Ffric
mg cosq =
w
w
q
w
= mg sinq
q
Inclined Planes
FN
w = mg sinq
w = mg cosq
Ffric
q
m
30o
A mass slides down a frictionless incline plane that
is angled 30 degrees above horizontal. Find the
acceleration of the block.
M = 2 kg
30o
A mass slides down an inclined plane that is angled
30 degrees above horizontal with an acceleration of
2 m/s2 What is the coefficient of kinetic friction?
T=?
50o
Some weirdo tied a 10 N block to a wall at the top of an
inclined plane (see figure above). If the coefficient of
static friction between the block and the plane is 0.2,
then what is the tension in the cord?
20o
Mwagon = 25 kg
30o
What minimum force must this person apply to
the handle of this wagon on this inclined plane to
hold it still if the coefficient of static friction is
0.05?
Acceleration = ?
20o
You are riding an airplane. As it takes off, you take out
your yo-yo and hold it and measure that it makes an
angle of 20 degrees with vertical. What is the plane’s
acceleration?
50o
20 cm long piece
of string
How fast is the mass on the end of this string
moving?
17 m
There is enough
information in these
images for you to find the
speed of the people riding
this ride. (So do it!)
42o
Uhhrghmph!
15 kg
30 kg
You apply a horizontal 200 N force to a 15 kg box that is in
turn pushing on a 30 kg box. If the coefficient of friction
between cardboard and the floor is 0.2, find the
acceleration of the boxes and the force that the 15 kg box
exerts on the 30 kg box.
1 kg
T=?
a=?
2 kg
Look at this! Your nemesis
tied a 1 kg block to a 2 kg
block and ran the string
connecting the blocks over
a massless frictionless
pulley! If the coefficient of
friction between the table
and the 1 kg block is 0.6,
what is the acceleration of
the blocks AND what is the
tension in the string?
m=0.4
a=?
4 kg
50o
Someone stuck an
Atwood Machine on an
inclined plane! Find the
acceleration of the
masses and the tension
in the string!
ms= 0.4
F=?
1 kg
a = 4 m/s2
What minimum horizontal force must be applied
to this 1 kg block with a coefficient of static
friction of 0.4, to keep it from sliding down the
wall?
Team A
Team B
At the tug of war National Championships, Team B
beat Team A in the finals, pulling them 10m in
under 20 seconds. If Team B exerted 10,000 N, we
know that: a) Team A exerted more than 10,000N,
b) Team A exerted less than 10,000 N, C) Team A
exerted 10,000N d) Need more information.
Impulse and Momentum
F = ma
F = m (Dv/Dt)
F Dt = m Dv = mvf – mvo (Impulse-Momentum Theorem)
J = Impulse = F Dt (units: N-s or Kg-m/s)
Momentum = p = mv (units: kg-m/s)
ptotal before = ptotal after (Momentum Conservation)
Area Under F-T = Impulse (= Dp)
M1
v1B
v2B
M2
Before A Collision
Ptotal before = m1v1B + m2v2B
During A Collision
M1 M2
J12 = F12 t12 & J21 = F21 t21
But…
t12 = t21 (Common Sense)
F12 = -F21 (If N’s 3rd Law is true)
So J21 = F21 t21 = -F12 t12 = - J12
And J = mDv so in every imaginable collision,
m1Dv1 = -m2Dv2
Ptotal before = Ptotal after
1500 kg truck
traveling 20 m/s
to the East
900 kg car
traveling 25 m/s
to the North
The car and truck shown above collide inelastically. Find their
velocity just after the collision.
800 kg car traveling to the East
with unknown Vbefore
Vf = 24.2 m/s
q1 = 30o
750 kg car
traveling 25 m/s
to the North
Before
After
The two cars shown above collide inelastically.
How fast was the 800 kg car moving just before the collision?
Impulse and Momentum in 2 Dimensions
V1B = 10 m/s
m1 = 1 kg
V2B = 0
m2 = 2 kg
V1A = 6 m/s
q1 = 30o
V2A =?
Step 1 – Resolve velocities into x and y components.
Step 2 – ptotal x before = ptotal x after
Step 3 – ptotal y before = ptotal y after
V1B = 10 m/s
V2B = 0
V1A = 6 m/s
q1 = 30o
V1AY = 6 sin30
V1AX = 6 cos30
m1 = 1 kg
m2 = 2 kg
V2AX =?
V2AY =?
ptotal x before = ptotal x after
m1 V1BX + m2 V2BX = m1 V1AX + m2 V2AX
(1)(10) + (2)(0) = (1)(5.2) + (2)(V2AX)
V2AX = 2.4 m/s
V1B = 10 m/s
V2B = 0
V1A = 6 m/s
q1 = 30o
V1AY = 6 sin30
V1AX = 6 cos30
m1 = 1 kg
m2 = 2 kg
V2AX = 2.4 m/s
V2AY =?
ptotal y before = ptotal y after
m1 V1BY + m2 V2BY = m1 V1AY + m2 V2AY
(1)(0) + (2)(0) = (1)(3) + (2)(V2AY)
V2AY = -1.5 m/s
V2B = 0
V1B = 10 m/s
V1A = 6 m/s
q1 = 30o
V1AY = 6 sin30
V1AX = 6 cos30
m1 = 1 kg
m2 = 2 kg
V2AX = 2.4 m/s
q2 = ?
V2AY = 1.5 m/s
V2A = (2.42 + 1.52)1/2 = 2.8 m/s
V2AX = 2.4 m/s
V2AY = -1.5 m/s
q2 = tan-1 (1.5/2.4)
= 32o
Impulse and Momentum in 2 Dimensions
V1B = 15 m/s
m1 = 2 kg
V2B = 0
m2 = 1 kg
V1A = 8 m/s
q1 = 20o
V2A =?
Step 1 – Resolve velocities into x and y components.
Step 2 – ptotal x before = ptotal x after
Step 3 – ptotal y before = ptotal y after
Force (N)
Time (Sec)
Impulse = Area Under F-t curve
Unit 4 – Work and Energy
Vocabulary
Energy – the ability to do work.
Work = Force Displacement = |F||d|cosq (Scalar)
Units: N m = Joule {calories are also units of energy}
Power = Work/time
(Scalar)
Units: Joule/sec = Watt
Conservative Force–Force that depends only on position.
(Gravity is an example.)
Non-Conservative Force-force that does not depend on only
position. (Friction is an example.)
Unit 4 – Work and Energy
Vocabulary (Cont’d)
Kinetic Energy = KE = (½)mv2 (energy associated with motion)
Gravitational Potential Energy = GPE = mgh
Hooke’s Law = Fs = kx (force vs displacement from equilibrium
length for an “ideal” spring)
Work = Area Under a Force vs Displacement graph.
Spring Potential Energy = Us = ½ kx2
Energy Conservation:
Total Engery = KE + GPE + Us
Work = DTE
Problem :
A 2 kg bouncy ball is dropped from a height of 10
meters, hits the floor and returns to its original
height. What was the change in momentum of the
ball upon impact with the floor? What was the
impulse provided by the floor?
http://www.physicsclassroom.com/calcpad/energy/problems
Mtot= 200 kg
va= 10 m/s
Use Energy Conservation to find the velocity of the
frictionless roller coaster car at each point, A, B, C, D.
A
And find the maximum
compression of the spring at the
bottom of the ride when the car
is momentarily at rest.
ha= 50m
B
hb= 30m
C
D
x=?
hc= 5 m
kspring = 1500 N/m
Vo= 10 m/s
X = maximum compression of spring = 15 cm
M=2kg
d= 5 m
Find k for the spring above if
the floor is frictionless?
If the same spring, same mass, with the same initial velocity
were used but the spring only compressed 10 cm at maximum
compression, how much work was done by friction?
Unit 4 – Work and Energy
Vocabulary (Cont’d)
***A collision is Perfectly Elastic (or Elastic) IF the
KETotal before = KETotal after
To solve an Elastic Collision problem, use Momentum
Conservation to get:
Equation 1:
m1 V1B + m2 V2B = m1 V1A + m2 V2A
THEN use this new formula:
Equation 2:
V2B - V1B = V1A - V2A
5m
40o
T=? When the person is at the
lowest point?
50 kg
vbot=? When the person is at
the lowest point