Chapter 5: Circular Motion
•
•
•
•
•
•
•
•
Uniform circular motion
Radial acceleration
Unbanked turns (banked)
Circular orbits: Kepler’s laws
Non-uniform circular motion
Tangential & Angular acceleration
(apparent weight, artificial gravity)
Hk: CQ 1, 2. Prob: 5, 11, 15, 19, 39, 49.
1
angular measurement
• degrees (arbitrary numbering system, e.g.
some systems use 400)
• radians (ratio of distances)
• e.g. distance traveled by object is product
of angle and radius.
2
Radians
s

r


t


t
rad/s
rad/s
s
s  r
s = arc length
r = radius
3
motion tangent to circle
s  r
s r
v 
 r
t
t
v r
a 
 r
t
t
4
Angular Motion
 avg


t
• radian/second
 avg


t
(radian/second)/second
5
angular conversions
Convert 30° to radians:
30 2rad 
 rad  0.52rad
1 360
6
Convert 15 rpm to radians/s
15rev 2rad 1 min
 1.57 rad / s
min rev 60 s
6
Angular Equations of Motion
Valid for constant- only
  o  t
 
1
2
o   t
  ot 
1
2
t
2
    2
2
2
o
7
Centripetal Acceleration
• Turning is an acceleration toward center of
turn-radius and is called Centripetal
Acceleration
• Centripetal is left/right direction
• a(centripetal) = v2/r
• (v = speed, r = radius of turn)
• Ex. V = 6m/s, r = 4m.
a(centripetal) = 6^2/4 = 9 m/s/s
8
Centripetal Force
FN
f
f
mg
Top View
Back View
Acceleration with Non-Uniform
Circular Motion
• Total acceleration
=
tangential
+ centripetal
• = forward/backward + left/right
• a(total) = r (F/B) + v2/r (L/R)
• Ex. Accelerating out of a turn;
4.0 m/s/s (F) + 3.0 m/s/s (L)
• a(total) = 5.0 m/s/s
Centripetal Force
• required for circular motion
• Fc = mac = mv2/r
•
•
•
•
Example:
1.5kg moves in r = 2m circle v = 8m/s.
ac = v2/r = 64/2 = 32m/s/s
Fc = mac = (1.5kg)(32m/s/s) = 48N
11
Rounding a Corner
• How much horizontal force is required for
a 2000kg car to round a corner, radius =
100m, at a speed of 25m/s?
• Answer: F = mv2/r = (2000)(25)(25)/(100)
= 12,500N
• What percent is this force of the weight of
the car?
• Answer: % = 12,500/19,600 = 64%
12
Mass on Spring 1
• A 1kg mass attached to spring does r =
0.15m circular motion at a speed of 2m/s.
What is the tension in the spring?
• Answer: T = mv2/r = (1)(2)(2)/(.15) = 26.7N
13
Mass on Spring 2
• A 1kg mass attached to spring does r =
0.15m circular motion with a tension in the
spring equal to 9.8N. What is the speed of
the mass?
• Answer: T = mv2/r, v2 = Tr/m
• v = sqrt{(9.8)(0.15)/(1)} = 1.21m/s
14
Kepler’s Laws
15
Kepler’s Laws of Orbits
1. Elliptical orbits
2. Equal areas in equal times (ang. Mom.)
3. Square of year ~ cube of radius
Elliptical Orbits
•
•
•
•
One side slowing, one side speeding
Conservation of Mech. Energy
ellipse shape
simulated orbits
Summary
•
•
•
•
•
s = r
v = r
a(tangential) = r.
a(centripetal) = v2/r
F(grav) = GMm/r2
Kepler’s Laws, Energy, Angular
Momentum
18
Centrifugal Force
• The “apparent” force on an object, due to
a net force, which is opposite in direction
to the net force.
• Ex. A moving car makes a sudden turn to
the left. You feel forced to the right of the
car.
• Similarly, if a car accelerates forward, you
feel pressed backward into the seat.
19
rotational speeds
•
•
•
•
•
rpm = rev/min
frequency “f” = cycles/sec
period
“T” = sec/cycle = 1/f
degrees/sec
rad/sec
 = 2f
20
7-43
•
•
•
•
Merry go round: 24 rev in 3.0min.
W-avg: 0.83 rad/s
V = rw = (4m)(0.83rad/s) = 3.3m/s
V = rw = (5m)(0.83rad/s) = 4.2m/s
Rolling Motion
v = vcm = R
22
Example: Rolling
A wheel with radius 0.25m is rolling at
18m/s. What is its rotational rate?
v  R :   v / R  18m / s  / 0.25m  72 mm / s  72rad / s
72rad 360
72rad / s 
 4127 / s
s 2rad
23
Example
A car wheel angularly accelerates uniformly from
1.5rad/s with rate 3.0rad/s2 for 5.0s. What is the final
angular velocity?
   o  t  1.5rad / s  (3.0rad / s 2)(5.0s)  16.5rad / s
What angle is subtended during this time?
  1 2 o   t
1
2
1.5  16.5(5)  45rad
   o t  1 2 t 2  1.55  1 2 352  45rad
 2   o2  2  1.52  2345  272.25
  272.25  16.5rad / s
24
Ex: Changing Units
1rad 1rev
1rad / s 
 0.1592rev / s
s 2rad
72rad / s 0.1592rev / s
72rad / s 
 11.5rev / s
1rad / s
72rad 1rev 60 s
72rad / s 
 688rpm
s 2rad 1 min
1s
T
 .0869 s
11.5rev
25
Rotational Motion
vt  r
vt
at
at  r
ac
r
vt
vc  0
2
ac
v
v
2
ac 
 r
r
v v 
2
t
2
c
v  0  vt
2
t
26
Convert 50 rpm into rad/s.
• (50rev/min)(6.28rad/rev)(1min/60s)
• 5.23rad/s