Download Lecture 13 - UD Physics

Phys 207
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Labs as usual this week
No quizzes in DSC this week; review Exam 1
Exam 1 (max = 69) : average = 43 (62%), median 42
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Points
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Today’s Agenda
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Friction
ÍInclined plane
ÍDrag Forces
» Terminal velocity
More application of Newton’s Laws to Circular motion
ÍRotating puck and weight
ÍConical pendulum
ÍCar on level, curved road
ÍCar on banked, curved ramp
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Page 1
Model for Surface Friction
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The direction of the frictional force vector fF is perpendicular to the
normal force vector N, in the direction opposing relative motion of
the two surfaces.
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Kinetic (sliding): The magnitude of the frictional force vector is
proportional to the magnitude of the normal force N.
fF = µ KN
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Static: The frictional force balances the net applied forces such that
the object doesn’t move. The maximum possible static frictional
force is proportional to N.
fF ≤ µ SN
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Static Friction…
(with one surface stationary)
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Just like in the sliding case except a = 0.
i:
F − fF = 0
j:
N = mg
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While the block is static: fF = F
N
F
j
i
fF
mg
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Page 2
Static Friction...
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µS is discovered by increasing F until the block starts to
slide:
i:
FMAX − µSN = 0
j:
N = mg
µS = FMAX / mg
j
N
FMAX
µSmg
i
mg
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Problem: Box on Truck
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A box with mass m sits in the back of a truck. The coefficient
of static friction between the box and the truck is µS.
ÍWhat is the maximum acceleration a that the truck can
have without the box slipping?
m
µS
a
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Page 3
Problem: Box on Truck
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Draw Free Body Diagram for box:
ÍConsider case where fF is max...
(i.e. if the acceleration were any
larger, the box would slip).
N
j
i
fF = µSN
mg
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Problem: Box on Truck
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Use FNET = ma for both i and j components
Íi
µSN = maMAX
Íj
N = mg
aMAX = µS g
N
j
aMAX
i
fF = µSN
mg
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Page 4
Inclined Plane with Friction:
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Draw free-body diagram:
ma
µKN
j
N θ
mg
θ
i
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Inclined plane...
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Consider i and j components of
µKN
ma
j
FNET = ma :
i
mg sin θ − µKN = ma
j
N = mg cos θ
mg sin θ − µKmg cos θ = ma
N θ
mg
i
θ
a = g(sin θ − µKcos θ)
mg cos θ
mg sin θ
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Page 5
Static Friction:
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We can also consider µS on an inclined plane.
θ
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In this case, the force provided by friction will depend on the
angle θ of the plane.
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Static Friction...
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We can find µs by increasing the ramp angle until the block
slides:
mg sin θ − ff = 0
In this case:
ff = µSN = µSmg cos θM
µSN
mg sin θM − µSmg cos θM = 0
j
N
θM mg
θ
µS = tan θM
i
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Page 6
Lecture 13, Act 1
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A block of mass m, when placed on a rough inclined plane
(µ > 0) and given a brief push, keeps moving down the plane
with constant speed.
ÍIf a similar block (same µ) of mass 2m were placed on the
same incline and given a brief push, it would:
(a) stop
(b) accelerate
(c) move with constant speed
m
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Lecture 13, Act 1
Solution
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Draw FBD and find the total force in the x-direction
FNET,X = mg sin θ − µKmg cos θ
µKN
= ma = 0 (first case)
Doubling the mass will simply
double both terms…net force
will still be zero!
j
N θ
Speed will still be constant!
mg
θ
i
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Page 7
Friction in Fluids: Drag Forces
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When an object moves through a viscous medium, like air or
water, the medium exerts a “drag” or “retarding” force that
opposes the motion of the object relative to the medium.
FDRAG
j
v
Fg = mg
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Drag Forces:
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This drag force is typically proportional to the speed v of
the object raised to some power. This will result in a
maximum (terminal) speed.
FD = bvn
j
feels like n=2
v
Fg = mg
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Page 8
Terminal Speed:
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Suppose FD = bv2. Sally jumps out of a plane and after
falling for a while her downward speed is a constant v.
ÍWhat is FD after she reaches this terminal speed?
ÍWhat is the terminal speed v?
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FTOT = FD - mg = ma = 0.
ÍFD = mg
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Since FD = bv2
Íbv2 = mg
FD = bv2
j
v
Fg = mg
v=
mg
b
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Drag Forces
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Increase with cross-sectional area, A
larger at higher fluid (air) densities
approx. linear in velocity at low speed
approx. quadratic in velocity at high speed
r
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Fdrag = DρAv 2
2
1
∑ F = mg − 2 DρAv 2
⎛ DρA ⎞ 2
a = g −⎜
⎟v
⎝ 2m ⎠
vT =
D = drag coefficient
same A, but Dcircle > Dsphere
Fdrag
j
mg
2mg
DρA
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Page 9
Approach to Terminal Velocity
FD = bv1
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let n = 1 for simplicity
dv
mg − bv = ma = m
dt
dv
b
=g− v
dt
m
dv
dv
= dt ; ∫
= ∫ dt
b
b
g− v
g− v
m
m
mg
1 − e − bt m )
v=
(
b
j
v
Fg = mg
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Problem: Rotating puck & weight.
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A mass m1 slides in a circular path with speed v on a
horizontal frictionless table. It is held at a radius R by a
string threaded through a frictionless hole at the center of
the table. At the other end of the string hangs a second
mass m2.
ÍWhat is the tension (T) in the string?
ÍWhat is the speed (v) of the sliding mass?
v
m1
R
m2
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Page 10
Problem: Rotating puck & weight...
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Draw FBD of hanging mass:
ÍSince R is constant, a = 0.
so
T
m2
T = m2g
m2g
v
m1
R
T
m2
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T = m2g
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Problem: Rotating puck & weight...
N
Draw FBD of sliding mass:
T = m2g
m1
Use F = T = m1a
m1g
where a = v2 / R
v = gR
m2g = m1v2 / R
m2
m1
v
m1
R
T
m2
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Page 11
Conical Pendulum
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Page 13