Download Circular Motion - Pat-Med Physics AP Exam Regents Exam

Pearson 2007-2008
Uniform Circular Motion
An object moving in a circle with constant speed, v,
experiences a centripetal acceleration with:
*a magnitude that is constant in time and
is equal to
v2
a 
r
*a direction that changes
continuously in time and
always points toward the
center of the circular path
For uniform circular motion, the velocity is tangential
to the circle and perpendicular to the acceleration
Period and Frequency
A circular motion is described in terms of the
period T, which is the time for an object to
complete one revolution.
2r
T
v
The distance traveled in one revolution is
2 r
r
The frequency, f, counts the number of revolutions
per unit time.
1
f
T
Example of Uniform Circular Motion
The moon’s nearly circular orbit about the earth has
a radius of about 384,000 km and a period T of
27.3 days. Determine the acceleration of the Moon
towards the Earth.
2r
2r
T
v
T
v
v
4 r
4 r
a   2 
2
r
T r
T
2
2 2
g 
4
a  2.72  10 m / s 
2   2.78  10 g
9.8m / s 
3
2 
2
Moon…...
*So we find that amoon / g = 0.000278
*Newton noticed that RE2 / R2 = 0.000273
amoon
g
R
RE
*This inspired him to propose that Fgravity  1 / R2
(more on gravity in future lectures)
Uniform Circular Motion
*motion in a circle or circular arc at constant
speed
*the acceleration changes the direction of the
velocity, not the magnitude
*the “center-seeking” or centripetal acceleration
is always orthogonal to the velocity and has
magnitude:
The period of
the motion:
2r
T
v
v2
a
r
Uniform Circular Motion
Newton’s 2nd Law: The net force on a body is equal to the product of the mass of the
body and the acceleration of the body.
*The centripetal acceleration is caused by a
centripetal force that is directed towards the
center of the circle.
F  ma  m
2
v
r
Demo 1D-5
Does the contact force between the wine
glass and red-water remain constant in
uniform circular motion?
Consider the glass directly overhead. Choose the correct statement:
a.
The water doesn’t fall because the centripetal force on the water cancels the
force of gravity.
b.
The water doesn’t fall because there isn’t enough time for it to fall.
c.
The water doesn’t fall because of the horizontal force applied to it by the
glass, plus friction with the glass.
d.
The water is falling, but the glass is falling faster than it would under free
fall.
mac = mv2/r = mg + Ny
When N=0, the centripetal
acceleration is just g.
or
ac = g  N/m
Top
v
N
y
mg
x
 Fy   N  mg  ma
N  m(a  g)
v2
N  m(  g)
r
 Fy   N  mg  ma
Top
N  m(a  g)
v2
N  m(  g)
r
v
N
y
mg
x
 Fy  N  mg  ma
Bottom
mg
N
v
N  m(a  g)
v2
N  m(  g)
r
 Fy   N  mg  ma
Top
N  m(a  g)
v2
N  m(  g)
r
v
N
mg
What speed is needed to lose contact between
wine glass and red-water?
v  rg
2) A person riding a Ferris Wheel moves through positions at (1) the top, (2) the
bottom and (3) midheight. If the wheel rotates at a constant rate, rank
(greatest first) these three positions according to...
(a) the magnitude of persons centripetal acceleration
(a) 2,1,3
(b) 1,2,3
(c) 3,2,1
(1) Top
(d) all tie
(b) The magnitude of the Normal force?
(3) Middle
(2) Bottom
(b) the magnitude of the net centripetal force on the person
1.
1,2,3
2.
3,1,2
3.
3,2,1
4.
all tie
(c) the magnitude of the normal force on the person
1.
all tie
2.
2,3,1
3.
3,2,1
4.
1,2,3
Demo 1D-2 Conical Pendulum
 Fy  0 
q
T cosq  mg
 Fr  ma 
H
2
v
tan q 
T
T sin q  mv 2 / R
gR
v  Rg tan q  R g / H
*as q 90, v increases.
*v is independent of mass.
R
mg
Period  T  2R / v  2 H / g
The period, T, is independent of mass and
depends only on H.
A car of mass, m, is traveling at a constant speed, v, along a
flat, circular road of radius, R. Find the minimum µs required
that will prevent the car from slipping
 Fr  ma  fs  mv
 Fy  N  mg
fs   s N
s 
2
v
gR
2
/R
A mass, m, on a frictionless table is attached to a hanging
mass, M, by a cord through a hole in the table. Find the
speed with which m must move in order for M to stay at
rest.
Tm
2
v
r
T  Mg  0  T  Mg
2
v
Mg  m
r
M
v
gr
m
A car of mass, m, is traveling at a constant speed, v, along a
curve that is now banked and has a radius, R. What bank
angle, q, makes reliance on friction unnecessary?
N
q
 Fy  0 N cosq  mg
2
F

ma

N
sin
q

mv
/R
 r
mg
tan q 
R
2
v
2
v
gR
g tan q
An airplane is flying in a horizontal circle with a speed of 480 km/hr. If the
wings of the plane are tilted 40o to the horizontal, what is the radius of the
circle in which the plane is flying? (Assume that the required force is
provided entirely by an “aerodynamic lift” that is perpendicular to the wing
surface.)
v2
 Fr  0 2L sin 40  M r
 Fy  0 2L cos 40  Mg
v=480 km/hr
L
Mg
L
2 cos 40
Mg
v2
2
sin 40  M
r
2 cos 40
2
v
r
g tan 40
L
W