Physics 221, February 2
Key Concepts:
•Motion with constant acceleration
•Drag
•Hooke’s law
•Harmonic motion
Electronic Devices
Please separate your professional from your social life
Do not use social media during class time.
Physics
Nobody can read it to you. You can only learn it by doing it.
There are not that many things to remember.
There are a few concepts that you must understand.
You start to understand the concepts by applying them.
• Observations and experiment
• Questions
• Problems
It is usually not the answer itself that helps you understand the
concept, but you questioning the answer.
“What does that answer tell me?” “What does it mean?”
Forces
On macroscopic scale, the most common forces
(interactions) we experience on a daily basis are gravity and
contact forces.
Gravity near the surface of Earth:
• proportional to mass
• constant as a function of position
• pointing straight downward
Fg = mg
constant force  constant acceleration
Motion with constant acceleration in one dimension:
Kinematic equations
a = constant.
vf = vi + a ∆t,
xf - xi = vi∆t + (1/2)a∆t2,
vf2 = vi2 + 2a(xf – xi).
This includes a = 0.
A gymnast jumps upward with an initial speed of 10 m/s. She is
in the air for a total time of
1.
2.
3.
4.
5.
1s
2s
4s
0.5s
0.25 s
0%
1
0%
2
0%
0%
3
4
0%
5
30
Solution:
At her highest point her velocity is zero.
v = v0 – gt, let g = 10 m/s2.
0 = 10 m/s – gt, therefore t = 1 s is the time it takes her to
reach her highest point.
It takes her the same amount of time to come back down, so
she is in the air for 2 s.
or
Her final position is equal to her initial position.
xf – xi = v0t - (1/2)gt2
0 = (10 m/s) t - (1/2)gt2
gt = 20 m/s, t = ~2s, she is in the air for ~ 2 s.
A particle is moving with velocity v = 60 m/s at t = 0.
Between t = 0 and t = 15 s the velocity decreases uniformly to
zero. What was the acceleration during this 15 s time interval?
1.
2.
3.
4.
5.
4 m/s2
8.6 m/s2
0.53 m/s2
-0.27 m/s2
-4 m/s2
0%
1
0%
2
0%
0%
3
4
0%
5
30
Solution:
Since the velocity decreases uniformly, the
acceleration is constant. We therefore have
a = (vf – vi) /(tf – ti)
= (0 m/s – 60 m/s)/(15 s)
= - 4m/s2
Which position versus time graph does NOT represent motion with constant
acceleration?
1.
2.
4.
5.
3.
0%
1
0%
2
0%
3
0%
4
30
0%
5
Extra Credit: Look at the plots of position and velocity of a falling ball as a
function of time, similar to those obtained in lab 3.
The same data points
are represented to the
right in a plot of
velocity versus
1.
2.
3.
acceleration.
position.
square root of time.
30
0%
1
0%
2
0%
3
A rock is thrown downward with an initial
velocity of 10 m/s from a height of 30 m
above the ground. How long does it take the
rock to fall the last 10 m before it hits the
ground?
How do you solve this problem?
Solution:
• Find time for the rock to fall 30 m.
• Find time for the rock to fall 20 m.
• Subtract t(30 m) – t(20 m) to get time to fall the last 10 m.
Let g = 10 m/s2.
30 m = (10 m/s)*t + ½ (10 m/s2) t2
5t2 + 10t – 30 = 0, t2 + 2t - 6 = 0.
t(30 s) = [-1 + sqrt(7)] s = 1.65 s
20 m = (10 m/s)*t + ½ (10 m/s2) t2
5t2 + 10t – 20 = 0, t2 + 2t – 4 = 0.
t(20 s) = [-1 + sqrt(5)] s = 1.24 s
t(30 m) – t(20 m) = 0.41 s
Contact forces:
pushes, pulls, friction, drag
Friction and drag are forces that always points in a direction
opposite to the direction of the velocity of the object. They
reduce the speed of the object.
The magnitude of drag force increases with
the speed. An object acted on by a constant
applied force and drag will reach
terminal velocity.
The net force on an object can be zero, because the vector sum of the
forces acting on different parts of the object is zero.
The material then is under tension or compression.
It expands or contracts.
No material is perfectly hard. As long as the external
pushes or pulls are not too strong, the contraction
or expansion is proportional to the magnitude of the
pushing or pulling forces.
The material pushes back with force
Fext = kx
F = -kx
This is Hooke’s law.
Please understand that Hooke's law is NOT a law of nature.
It is a rule of thumb that is often holds over a limited range of expansion and
contraction.
Harmonic motion
If the only force acting on an object with mass m is a Hooke’s law force,
F = -kx
then the motion of the object is simple harmonic motion.
With x being the displacement from equilibrium we have
x(t) = Acos(ωt + φ)
v(t) = -ωAsin(ωt + φ),
a(t) = -ω2Acos(ωt + φ) = -ω2x.
ω = sqrt(k/m) = 2πf = 2π/T.
In terms of k and m:
A = amplitude
ω = angular frequency
f = frequency
T = period
φ = phase constant
Extra Credit :
You have always wondered how much one of your friends weighs and devise a scheme
to measure his weight secretly. You have him sit in a tubular steel chair. This popular
style of chair consists of a single steel tube that is bent into a frame and that supports
a seat bottom and a back. The empty chair weighs 10 pounds and is 30 inches tall.
The frame acts as a spring and bends downward slightly when the chair is occupied.
When you sit properly in the chair yourself, it bends downward 1 inch. When your
friend sits properly in the chair, it bends downward 2 inches. From that observation,
you know that your friend weighs about
1.
2.
3.
4.
5.
200 pounds.
150 pounds.
four times as much as you.
twice as much as you.
one half as much as you.
0%
1
0%
2
0%
3
0%
4
0%
5
30
Extra credit:
Demo:
Measure the frequency or period of a 0.02 kg mass oscillating on
a spring.
What is the spring constant of this spring?
1.
2.
3.
4.
5.
~0.65 N/m
~0.95 N/m
~3 N/m
~6.5 N/m
~80 N/m
0%
1
0%
2
0%
3
0%
4
0%
5
30
Position, velocity and
acceleration of an oscillating
mass on a spring are shown
as a function of time.
What is the period of the
oscillations?
1.
2.
3.
4.
~1.5 s
~2.5 s
~4 s
~0.4 s
0%
1
0%
2
0%
3
0%
4
30
Extra Credit: Position,
velocity and acceleration of
an oscillating mass on a
spring are shown as a
function of time.
If the mass is 0.5 kg, what is
the spring constant in N/m?
1.
2.
3.
4.
~6
~3
~1
~0.5
0%
1
0%
2
0%
3
0%
4
30