Transcript Rotational Motion and the Law of Gravity

Rotational Motion and the Law
of Gravity
-Study of linear motion: Δxdisplacement, v-velocity,
a-acceleration
-Study of rotational motion:
Δθ-angular
displacement,
ω-angular velocity, αangular acceleration
- Angular quantities in
Physics must be express
in radians
• Radian θ=s/r –tha arc
length s along a circle
divided by the radius r
• Ex: 360o= 2π radians
180o= π radians
45o=2πrad/360o=(π/4)rad
• If OP-reference line,
t=0; P-on reference line,
t=Δt; P-new position
In this time, the line OP has
moved the angle θ with
respect of the reference
line.
θ (SI rad)- angular position
(linear motion was –x)
An object angular
displacement:
Δθ= θf- θi SI: radians
The average angular speed
ωav of a rotating rigid object
during the time Δt is defined
as the angular displacement
Δω divided by Δt:
ωav= (θf-θi) /(tf -ti)= Δθ/Δt
SI: rad/s
The instantaneous speed ω
of a rotating rigid object is
defined as the limit of
average speed:
ω =lim(Δt→0)Δθ/Δt
An object average angular acceleration αav, during an
interval Δt is: αav= (ωf- ωi)/ (tf-ti) =Δω/Δt
SI: rad/s2
The instntaneous angular acceleration α
α = lim(Δt→0) Δω/Δ t
SI: rad/s2
• When a rigid objects rotates about a fixed axis,
as does the bicycle wheel, every portion of the
object has a same angular speed and the
same angular acceleration
Rotational motion under constant angular
acceleration
• Relations between angular and linear
quantities:
Δθ=Δs/r
Δθ/Δt=1/r(Δs/Δt)
ω= v/r
The tangential speed of a point on a
rotating object= the distance oft hat
point from the axis of rotation multiply
by an angular speed
vt= ωr
Δvt=rΔω
Δvt /Δt=r Δω/Δt
at= r α
Tha tangential acceleration of a point on
a rotating object= distance from the
axis of rotation multiplied with α
• Centripetal acceleration
Fig shows a car moving in a
circular path with constant
linear speed
Even through the car moves
at a constant speed, it still
has an acceleration
aav= (vf-vi) /(tf-ti)= Δv /Δt
Δv = (vf-vi) –change of velocity
When Δt is very small, Δs and
Δθ are very small, vf and viare
almost parallel, Δv is
perpendicular, and points
toward the center of the circle
• In fig, the triangle with
sides Δs and r are
similar to the one
formed by the vectors:
Δv/v = Δs/r
Δv =v/r Δs
aav = Δv/ Δt
aav = v/r Δv /Δt
If Δt becomes very small,
Δs /Δt approches
instantaneous value of
tangential speed v
For circular motion at a constant speed, the
acceleration vector always points toward the
center of the circle, the acceleration is called a
centripetal (center-seeking) acceleration
aav approaches ac, the instantaneous
centripetal acceleration:
ac =v2/r
but vt=rω
ac =r2ω2/r =rω2 at =rω
The tangential and centripetal components of
acceleration are perpendicular to each other,
total acceleration a=√(at2+ac2)
• Angular quantities are
vectors
• The direction of angular
velocity ω can be found
with right-hand rule
(wrap the axis of rotation
with your right hand so
that fingers wrap in the
direction of rotation, your
tumbpoints the diretion of
ω
• Forces causing
Centripetal Acceleration
An object can have a
centripetal acceleration
only if some external
force acts on it.
-for a ball whirling in a
circle at the end of a
sting, that force is
tension in string
-for a car moving on a flat
circular track, the force is
friction between the car
and the track
- A satellite in circular orbit around the Earth has a
centripetal acceleration due to the gravitational
force between the satellite and Earth.
- “ centripetal force” –mean the force in action
acts toward the center
Fc= m ac= mv2/r
A net force causing a centripetal acceleration, acts
toward the center of a circular path and effects a
change in the direction of the velocity vector, if
the force vanish, the object will leave the circular
path, and move a straight line tangent to the
circle.
• Pb solving strategy:
1.Draw a free-body diagram, labeling the
forces act on it
2. Choose a coordinate system (one axis
perpendicular to the circular path
followed by the object (radial direction)
one axis tangent to the circular path (the
tangential, or angular direction) The
normal direction, perpendicular to the
plane of motion, also needed
3. Find the net force Fc toward the center
Fc =ΣFr, where ΣFr is the sum of the radial
components of the forces (this cause the
centripetal acceleration)
4. Use Newton’s 2nd Law, for the radial,
tangential, and normal directions:
ΣFr=m ac, ΣFt= m at, ΣFr=m an
Remember, ac=vt2/r
5. Solve for the unknown quantities
• Newtons Law of Universal
Gravitation:
If two particles with
masses m1 and m2 are
separated by a distance
r, then a gravitational
force acts along a line
joining them with the
magnitude :
F=G( m1m2/ r2)
G=6.673x10-11kg-1m3s-2 is
constant of universal
gravitation
F- always an attractive force
• Gauss’s Law: the
gravitational force
exerted by a uniform
sphere on a particle
outside the sphere is the
same as the force
exerted in the entire
mass of the sphere was
concentrated at its
center.
• Near the surface of Earth,
F= mg, g varies
considerably with altitude
(article page 207)
• Gravitational PE:
PE= mgh –if object near
Earth’s surface
PE= -G MEm/r; SI: Joules(J)
PE2-PE1
= -GMEm/(RE+r)-(-GMEm/RE)
=- GMEm[ 1/(RE+r)-1/RE]
= GMEmh/[RE(RE+h)]
(1/[RE(RE+h)≈1/RE2)
≈GMEmh/RE2
(g = GME/RE2)
≈mgh
• Escape speed- if an object is projected
upward from Earth’s surface with a large
enough speed, it can soar off into space
and never return
• Escape speed- can be found apllying
conservation of energy:
KEi +PEi= 1/2mv2- GMEm/RE =0 (vf=0)
1/2mvesc2- GMEm/RE =0
vesc=√ 2GME/RE
•
Keplers’s Law (article:
history/214)
1.All planets move in elliptical
orbits with the Sun at one
of the focal points.
2.A line drawn from the Sun
to any planet sweeps out
equal areas in equal
intervals
3. The square of the orbital
period of any planet is
proportional with the cube
of the average distance
from the planet to the Sun
• 1.All planets move in elliptical orbits
with the Sun at one of the focal points.
Newton’s Law of gravitaion: any object
bound by a force that varies as 1/r2, will
move in elliptical orbit
Sun is in one of the foci. The distance from
the sun to the planet continuosly changes
• 2.
In an interval Δt. The planet
moves from A to B, more
slowly because it is far away
from the Sun
In the same interval Δt from C
to D the planet moves faster
because it is closer to the
sun
( conservation of angular
momentum)
• 3.
If a planet of mass MP moving around Sun
(MS), because the orbit is circular, the
planet moves with a ct. speed v
Newton’s Law:
MP ac= MPv2/r =(MP/r)(GMS/r)= MPGMS/r2
The speed of the planet = circumference of
the orbit/ time required for one revolutions
v= 2πr / T (T- period of the planet)
GMS /r2= v2/r = (2πr / T)2/r
T2 =(4π2/ GMS )r3 = KS r3
T2 = KS r3
Ks = ct. = 4π2/ GMS = 2.97x10-19 s2/m2
The mass of the Sun :
MS =4πr / G KS
(see table page 216)