Download Slide 1

An exposure to
Newtonian mechanics: part II
• Solving 17 problems
Motivation
Newton’s concept of the Universe was
one of crystalline beauty.
The future is predictable.
The past can be reconstructed.
The present can be completely
deconstructed.
Today, we will explore some examples.
1
Our tools
Key definitions
v= Δx/Δt
a= Δv/Δt
vtotal= v1 + v2
Newton’s basics
F = ma F1 = -F2
F12 = Gm1m2/R122
Expressions of energy
K.E.=½mv2 P.E.=mgh
P.E.=-GMm/R
Circular force
Fc = mvc2/R
Kinematics
x= =½a Δt2
R=(v2/g)sin2θ
2
Velocities
1) A juvenile delinquent skateboarding at
10 km/hr, throws a bottle forwards at
20 km/hr relative to him. How fast is the bottle travelling with
respect to the Halloween pumpkin that it hits?
Easy! Just use the velocity addition formula.
vtotal = v1 + v2
vtotal = 10 km/hr + 20 km/hr = 30 km/hr
2) What if you were travelling at 2/3 the speed of light, and fired a
probe at 2/3 the speed of light. How fast does the star P-Umpkin see
the probe coming towards it?
vtotal = v1 + v2
vtotal = (2/3)c + (2/3)c = (4/3)c = 4×108 m/s. NOTE THE USE OF “c”
3
Kinematics
3) A car drives 150 km in 4 hours. What is its
speed?
v= Δx/Δt = 150 km / 4 hr = 37.5 km/hr
4) A car accelerates from a stop to 100 km/hr in 6 seconds.
What is its average acceleration?
First, convert the change in velocity to m/sec…
100 km/hr × (1000 m/km) × (1hr/3600 s) = 27.8 m/s
a= Δv/Δt = 27.8 m/s / 6 s = 4.6 m/s/s = 4.6 m/s2
4
Kinematics
5) Thelma and Louise
plummet off the edge of
the Grand Canyon. How
long until they pancake?
Δx= ½at2 → t2=2Δx/g = (2 × 1800 m)/(9.8 m/s2) = 367 s2
t=19 s
How fast are they going at impact?
v=at = gt = (9.8 m/s2) × 19 s
v=186 m/s = 420 miles/hr
5
Kinematics
6) At the strike of midnight, a new-year’s reveler shoots his
.357 Remington, aimed upwards at a 45° angle. How far away
does the bullet land?
vmuzzle = 46 m/s
R=(v2/g)sin2θ = [(46 m/s)2/9.8 m/s] × (sin90°)
R=216 m
At what time does the bullet go through your living room
window?
(Magic: vhorizontal = vmuzzle × cosθ° = 32.5 m/s)
t=Δx/v = (216 m)/(32.5 m/s) = 6.6 s
6
T=12:00:07
Newton’s Laws
7) The space shuttle’s main engines and 2 SRBs provided a total
thrust of about 31,000,000 N (force). The shuttle (unloaded) had a
mass of about 2×106 kg (2,000,000 kg).
If an empty shuttle, with two SRBs, began travelling in space, what
would be its acceleration?
F=ma → a = F/m
a = 31,000,000 N / 2,000,000 kg = 15.5 m/s2
7
Newton’s Laws
8) Your mass is 80 kg. How hard is the pull of gravity, from an
object as massive as the Earth, from a distance of the Earth’s radius?
F = GMEm/RE2
= (6.67×10-11 N m2/kg2) × (5.97×1024 kg) × (80 kg)
(6,378,000 m)2
= 783 N.
Since 4.45 N = 1 pound,
F=176 pounds
8
Kepler’s Third Law (1627)
Planets orbiting the Sun follow the law that
P2=Da3
Where P is the orbital period and a is the distance
of the object from the Sun.
9) Let us look at this for the circular orbit case.
Is it that the orbits are simply larger, and take longer to traverse?
v = Δx/Δt,
so Δt = Δx/v
Δx = C = 2πa,
Δt = P =2πa/v, so
P2 = 4π2a2/v2 = (4π2/v2)a2 = Da2
Close, but not quite Kepler’s Third Law.
9
Newton and Kepler’s Third Law
P2=Da3
Where P is the orbital period and a is the distance
of the object from the Sun.
10) Now let us try balancing the gravitational force of
attraction with the inward centripetal force needed
to maintain a circular orbit.
F12 = Gm1m2/R2
and
Fc = m1vc2/R, so
Gm1m2/R2 = m1vc2/R
Gm/a = v2 (I substituted “a” for “R”, as in Kepler’s Third Law)
Since P =2πa/v, therefore v =2πa/P, and so
Gm/a = 4π2a2/P2, so
P2 = (4π2/Gm)a3
Kepler’s Third Law!
10
Energy and free fall
11) How fast would something be moving
if it accelerated at 1g, for a distance equal
to the Earth’s diameter?
Easy! Just balance P.E. and K.E.
mgh = ½mv2
gD = ½v2
v2 = 2gD, so
v = (2gD)½
v= (2gD)½ = (2×9.8 m/s2×6.4×106 m)½
v = 11,200 m/s = 25,000 mph!
i.e, Mach 73 (the speed of sound is about 343 m/s).
The Space Shuttle orbited at about 8000 m/s.
11
Energy and free fall
12) How far would our object in 1g have
to fall, to reach the speed of light?
How does this distance compare to
interplanetary and interstellar space?
Easy! Just balance K.E. and P.E.
mgh = ½mv2
gD =½c2
D = c2/2g
D = (3×108 m/s)2 /(2 × 9.8 m/s2 ) = 4.6 ×1015 m
Since 1 AU = 1.5 ×1011 m, and 1 LY = 9.46 ×1015 m…
D= 3.1
×104 AU
= 0.49 LY
12
Length measure
13) You’re sitting at a railroad
crossing, and the train—passing
at 50 km/hour—takes five minutes to go by. How long is the train?
Easy! Just use the rate equation.
(“The length of the train” is the same as just asking how far does the
front of the train travel, in the time it took for the rear of the train to
arrive.)
v = Δx/Δt
Δx = vΔt
Δx = (50 km/hr × 1 hr/60 min) × 5 min = 4.2 km = 4200 m
13
More gravity
14) Professional basketball players can typically
jump upwards about 1.10 m. How fast are they
jumping?
Easy! Just balance K.E. and P.E.
mgh = ½mv2
gh = ½v2
v2 = 2gh, so
v= (2gh)½ = (2 × 9.8 m/s2·1.1 m)½
v = 4.6 m/s = 10.4 mph
15) How fast is an elevator falling if it plummets three stories?
(Estimate 1 story as 3.7 m; 12 ft.)
v= (2gh)½ = (2 × 9.8 m/s2 × 11.1 m)½
v = 14.7 m/s = 32.9 mph
14
Weight loss
16) How does your weight compare on a
planet with a different mass and radius?
Make a ratio of the law of gravity.
FE = GMEm/RE2
FP = GMPm/RP2, so
(FP/ FE) = (GMPm/RP2)/(GMEm/RE2)
(FP/ FE) = (MP/RP2)/(ME/RE2) = (MP/ME) × (RE/RP)2
Mars, M= 0.107 ME, R= 0.53 RE,
FMars = (0.107 ME/ME) × (RE/0.53 RE)2 = 0.38 FE
15
Monkey in a tree
17) Consider a monkey in a tree. It knows you intend to shoot it
with your gun, which is pointed directly at it. The monkey drops
from the tree the moment you fire the gun. What happens?
16
Not all physics is easy
Rotations complicate things!
Consider a spinning book!
17