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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
2
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.
3
Radians
r
s
s = arc length
r = radius
rs rad/s
t
s
rad/s t
4
motion tangent to circle
rs
r
t
r
t
sv
r
t
r
t
va
5
Angular Motion
• radian/second (radian/second)/second
tavg
tavg
6
angular conversions
Convert 30° to radians:
radradrad
52.06360
2
1
30
Convert 15 rpm to radians/s
sradsrev
radrev/57.1
60
min12
min
15
7
Angular Equations of Motion
to
22
1 tto
to 21
222o
Valid for constant- only
8
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
Top View Back View
f f
NF
mg
Centripetal Force
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
11
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
12
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%
13
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
14
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
15
Kepler’s Laws
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
18
Summary
• s = r v = r
• a(tangential) = r.
• a(centripetal) = v2/r
• F(grav) = GMm/r2
• Kepler’s Laws, Energy, Angular Momentum
19
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.
20
rotational speeds
• rpm = rev/min
• frequency “f” = cycles/sec
• period “T” = sec/cycle = 1/f
• degrees/sec
• rad/sec = 2f
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
22
Rolling Motion
v = vcm = R
23
Example: Rolling
A wheel with radius 0.25m is rolling at 18m/s. What is its rotational rate?
RvRv /: sradsmsm mm /72/7225.0//18
srads
radsrad /4127
2
36072/72
24
Example
to
A car wheel angularly accelerates uniformly from 1.5rad/s with rate 3.0rad/s2 for 5.0s. What is the final angular velocity?
sradssradsrad /5.16)0.5)(2/0.3(/5.1
What angle is subtended during this time?
to 21 rad45)5(5.165.12
1
22
1 tto rad455355.1 22
1
222o 25.27245325.1 2
srad /5.1625.272
25
Ex: Changing Units
srevrad
rev
s
radsrad /1592.0
2
11/1
srevsrad
srevsradsrad /5.11
/1
/1592.0/72/72
rpms
rad
rev
s
radsrad 688
min1
60
2
172/72
srev
sT 0869.
5.11
1
26
Rotational Motion
rvt
ttct vvvvv 0222
rat
22
rr
vac
vt at
ac
ac
vt
r
0cv
Convert 50 rpm into rad/s.
• (50rev/min)(6.28rad/rev)(1min/60s)
• 5.23rad/s
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