Transcript Ch33 - Wells College

Chapter 6: Circular Motion and Other
Applications of Newton’s Laws
Chapter 6 Goals:
• to deepen our understanding of the causes of circular
motion in various scenarios
• to gain skill with applying the FBD method to such
problems
• to learn how to handle the case of velocity-dependent
resistive forces
• note: we will mix things up in dealing with this
chapter, and this lecture will include material from
chapter 4, on relative motion
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Example of curvy motion
A bicycle/rider of mass 100 kg proceeds around a curve
at 25 mi/hr, and experiences a centripetal acceleration of
g/2 while doing so -- any faster and the bicycle skids!!
a) Find the radius of the curve.
b) If static friction is what keeps the bike from skidding,
what is the value of ms? Use FBD method.
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+y
x+
• v = (25 mi/hr)(1/3600 hr/s) (1609 m/mi)
= 11.2 m/s
• ac = v2/r so r = v2/ac = (11.2 m/s)2/(5 m/s2) =
25.1 m
• x: fs = mac  msn = mv2/r  ms = mv2/nr
• y: n – Fg = 0  n = mg
• combining, the m cancels, and ms = v2/gr
• but recall that v2/r = g/2 so ms = .50
fs
n
ac
Fg
Example of curvy motion
A Ferris wheel begins to rotate and ‘tangentially’
accelerates until the speed at the rim is v = 3.20 m/s. Its
radius is 8 m, and it gets ‘up to speed’ in 5.0 seconds. A
child of weight 200 N is riding.
(a) At what instant of time (if any) is ac = at?
(b) What is the child’s apparent weight when at the top?
• v = vi + at t so v = 0 + at
+y
2
 at = v/t = (3.2 m/s)/(5 s)= .65 m/s
n
• at end, ac = v2/r = (3.2) 2(8) = 1.28 m/s2
• ac = (at t)2/r = at  t = (r/at )1/2 = (8/.65) 1/2 = 3.5 s
ac
• y: n – Fg = – mac  n = – mac + mg
• so the weight is ‘reduced’ to 87%: n = 174 N
Fg
• at the bottom the weight is ‘enhanced’ to 113%
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The simple pendulum I
at
+t
+r
• simple means that mass is
the point-like bob mass m
• length of pendulum is L
• operates in gravity g = –jhatg
• FBD is at left: ‘position’ is q
• introduce radial and
tangential coordinates r and t
(t in direction of increasing q)
• at = dv/dt
ar = –v 2/L
• t: –mg sinq = – mat = 
at = g sinq = 0
• r: mg cosq - T = –v 2/L
• T = mg cosq + mv 2/L
{show Active Figure 15.16}
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The simple pendulum II
• a challenging non-linear differential equation for angle
position: since at = L d2q/dt2  d2q/dt2 + (g/L) sinq = 0
• if the amplitude is small, then sinq ≈ q
R
• in radians, q := s/R
h
• but sin q := o/h
s o
q
• s ≈ o; h ≈ R  sin q ≈ s/R = q
• cos q := a/h ≈ 1 –q2/2 tan q := o/h ≈ s/R = q
a, R
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The simple pendulum III
• under this approximation, we get d2q/dt2 + (g/L) q = 0
• this is a well-studied equation for q (t) and it is known
that the motion is simple harmonic oscillations
• for small amplitudes, a pendulum executes SHM in its
variable q, the angle away from the vertical!! Period is
amplitude-independent!! Makes a nice clock!!
• What about the equation for the tension?
• T = mg cosq + mw 2L
• an interesting combination of terms, that allow T to
‘balance’ the bob’s weight [mg cosq] and also to
provide the needed radial (centripetal) force [mw 2L]
• our other equation shows that mg sinq provides the
need tangential (Hooke-like) force to cause the SHM
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Relative Motion as seen from two
inertial frames of reference
• recall that an inertial frame of reference is an arena for
doing physics in which Newton’s first law is correct: no
net force  no acceleration
• [if acceleration is observed then the frame is noninertial (accelerating) and humans invent fictitious
forces to explain the weird motions observed]
• back to section 4.6…
• two observers measuring some kinematics: the motion
of a body at point P
• observer A is not moving, and A measures position
rPA, velocity vPA, and acceleration aPA
• observer B is moving as seen by A (at vBA) and it
measures rPB, vPB, and acceleration aPB
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How do we relate what A and B observe?
• basic kinematical fact:s
• rPA = rPB + rBA
°
rPA
A
P• take one time derivative:
• vPA = vPB + vBA
rBA
• in English: the velocity of a point with
rPB
respect to frame A is the velocity of the
B
point with respect to frame B plus the
velocity of B with respect to A
• we are assuming B is moving at constant velocity vBA
• another time derivative yields aPA = aPB so accelerations
are the same when observed from either inertial frame!!
• N2 is correct in either frame
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An alternative way of writing the same stuff
• vPA = vP/A vPB = vP/B
vBA = vB/A
• we have the Galilean transformation: vP/A = vP/B + vB/A
• note how the algebra works: P/A = (P/B)(B/A)
• addition of vectors multiply subscripts
• thus, vP/B = vP/A – vB/A or P/B = (P/A)/(B/A)
• subtraction of vectors divide subscripts
• the challenge is to identify the frames and the object
Simple 1d example: An empty water bottle rolls toward the back of
a bus at 2.5 m/s. The bus is traveling at 55 mph ≈ 25 m/s. What is
the velocity of the bottle as seen from the ground?
Ground is frame A; bus is frame B; bottle is object P
vP/A = vP/B + v B/A  vP/A = – 2.5 + 25 = 22.5 m/s
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Trickier 2d example: An empty water bottle rolls sideways across
the back of a bus at 5 m/s. The bus is traveling at 55 mph ≈ 25 m/s.
What is the velocity of the bottle as seen from the ground? What is
the speed of the bottle as seen from the ground? In what direction is
the bottle moving as seen from the ground?
Ground is frame A; bus is frame B; bottle is object P
Take x direction to be along bus’s motion as seen from ground
Take y direction to be perpendicular to that: across the road
vP/A = vP/B + v B/A  vP/A = 5 m/s j+ 25 m/s i
Speed is |vP/A | = √(52 + 252) = 25.5 m/s
Direction is Arctan(5/25) = 11.3° away from the x direction
y
y
x
vP/B
As seen from B
x
vP/A
As seen from A
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Trickier 2d example: You are canoeing across a river that is 75 m
wide, and in which the current moves at 2 m/s. You can paddle at
2.5 m/s. How should you paddle to go directly across the river, and
how much time will it take [hint: if you aim straight across, you will
reach the other side in time t = 75 m/2.5 m/s = 30 s but will have
drifted downstream a distance d = 2 m/s(30 s) = 60 m!!!]
Ground is frame A; river is frame B; canoe is object P
Take x direction to be along river’s motion as seen from ground
Take y direction to be perpendicular to that: across the river
vP/A = vP/B+ v B/A  vP/A := vP/A j = vP/B + 2 m/s i
 vP/B= vP/A j – 2 m/s i
But we know vP/B = 2.5 m/s so we can solve:
2
2.5 = vP/A
 2 2  vP/A = 2.52 - 2 2 = 1.5 m/s [3 : 4 : 5 triangle]
 vP/B= 1.5 m/s j – 2 m/s i and vP/A = 1.5 m/s j
Speed is 1.5 m/s with respect to the shore
Canoe is aimed at Arctan(1.5/2) = 36.9° away from UPSTREAM!
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y
y
x
x
vP/B
As seen from B
vP/A
As seen from A
It will require a time t = 75 m/1.5 m/s = 50 s to get across, but at
least you will be where you intended to be!!
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Extremely tricky 2d example: High-altitude winds are blowing at
450 km/hr at an angle of 35° S of E. A plane with airspeed 1100
km/hr desires to fly from New York to Los Angeles, a distance of
5000 km at a bearing of 25°S of W . How much time for the flight?
How should the pilot aim the plane? What will the plane’s speed be
with respect to the ground?
Ground is frame A; air is frame B; plane is object P
Take x direction to be E and y direction to be N
LA's position is rP/A = -5000 cos 25 ˆi - 5000 sin 25 ˆj = -4532ˆi - 2113ˆj

  ˆj
rP/A - 4532 ˆ
 plane's velocityis (where t is the timeof flight) v P/A =
=
i
t
t
wind velocity is v
= 450 cos 35 ˆi - 450 sin 35 ˆj = 368.6ˆi - 258.1ˆj
- 2113
t
B/A
v P/B = vP/B,E ˆi  vP/V, N ˆj
2
2
airspeed is | v P/B |= vP/B,
E  vP/B,N = 1100 km/hr
v P/B = v P/A - v B/A =
 ˆi   ˆj - 368.6ˆi - 258.1ˆj
-4532
t
-2113
t
2
 1100t 2 = - 4532 - 368.6t   - 2113  258.1t 2
Solve for t and substitute back in… quadratic equation!!
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1.210x106 t 2 = 20.54 x106  3.341x106 t  .1359x106 t 2  4.465x106
- 1.091x106 t  .06662x106 t 2 [cancel the106 ]  2.165t 2 - 2.250t - 25.00 = 0
t =
 -2.2502 -4 2.165 -25.00
=
21.007
2.250
 v P/A =
2.25014.885
= 8.51 hr
2.014
 ˆi   ˆj = -533 km/hrˆi - 248 km/hrˆj
-4532
8.51
-2110
8.51
 plane's speed is vP/A =
- 5332  - 2482
v P/B = v P/A - v B/A =
- 4532 ˆ
i  -2110 ˆj - 368.6ˆi - 258.1ˆj
    
8.51

8.51
= -901 km/hrˆi - 10 km/hrˆj
 plane is ' aimed' at angle
Arctan
  = .63  180
-10
-901



= 180.63 = .63 N of W
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= 635 km/hr
Resistive Forces and Fluid Drag I
• a different type of ‘friction’ in which the friction force
depends on how fast the object is moving
• friction we know and love (‘kinetic’) is also dubbed
Coulomb friction
• linear drag obeys R = – b v where b is a constant
• resistance force R is opposed to v and proportional to v
• taking + = ↓ and assuming object
released from rest in gravity  FBD
dv
dv
b
m = -bv  mg  = - v  g
dt
dt
m
• if dv/dt = 0, v is terminal speed vT
• vT = mg/b
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R
a
Fg
+y
Resistive Forces and Fluid Drag II
dv
dv
b
m = -bv  mg  = - v  g
dt
dt
m
• this is a differential equation for v(t)
• we think about what functions have a first derivative
that ‘looks like them’… accounting for the g is issue too
• thought reveals exponential functions of time
T ry v(t ) = Aet  B and stick it into thedifferential equation
b
-b
mg
t
t
 Ae = - ( Ae  B)  g  worksif  =
,B=
= vT
m
m
b

• no information about value of A as yet
• we assumed that the object was released from rest,
so the velocity at t = 0 is 0 [v(0) = 0]
• this is called implementation of initial conditions
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Resistive Forces and Fluid Drag III
v(0) = 0 = Ae ( 0)  B
 0 = A B

mg
 A = -B = = -vT
b
- bt / m
 v(t ) = vT (1 - e
)
dv(t )
a(t ) =
= ge-bt / m
dt
• here is the graph for v(t)
• exponential decay ‘up to a constant’
• the acceleration is simple
exponential decay
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