Physics 2211: Lecture 11 Todays Agenda

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Physics 2211 Lecture 11Todays Agenda

• More friction
• Motion in a circle

2
Friction...

• Force of friction acts to oppose motion
• Parallel to surface.
• Perpendicular to Normal force.

j
N
F
i
ma
fF
W
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Model for Sliding Friction

• The direction of the frictional force vector is
  perpendicular to the normal force vector N.
• The magnitude of the frictional force vector fF
  is proportional to the magnitude of the normal
  force N .
• fF ?K N ( ?K??W in the previous
  example)
• The heavier something is, the greater the
  friction will be...makes sense!
• The constant ?K is called the coefficient of
  kinetic friction.

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Model...

• Dynamics
• i F ? ?KN ma
• j N mg
• so F ???Kmg ma

j
N
F
i
ma
?K mg
W W mg
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Inclined Plane with Friction

• Draw free-body diagram

ma
?KN
j
N
?
mg
?
i
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Inclined plane...

• Consider i and j components of FNET ma

?KN
ma
j
N
?
a / g sin ?????Kcos ?
?
mg
mg cos ??
i
mg sin ??
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Static Friction...

• So far we have considered friction acting when
  something moves.
• We also know that it acts in un-moving static
  systems
• In these cases, the force provided by friction
  will depend on the forces applied on the system.

j
N
F
i
fF
W W mg
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Static Friction...

• Just like in the sliding case except a 0.
• i F ??fF 0
• j N mg

While the block is static fF ??F
j
N
F
i
fF
W W mg
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Static Friction...

• The maximum possible force that the friction
  between two objects can provide is fMAX ?SN,
  where ?s is the coefficient of static friction.
• So fF ? ?S N.
• As one increases F, fF gets bigger until fF ?SN
  and the object starts to move.

j
N
F
i
fF
W W mg
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Static Friction...

• ?S is discovered by increasing F until the block
  starts to slide
• i FMAX ???SN 0
• j N mg
• ?S ??FMAX / mg

j
N
FMAX
i
?Smg
W W mg
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Static Friction

• We can also consider ?S on an inclined plane.
• In this case, the force provided by friction will
  depend on the angle ? of the plane.

?
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Static Friction...

• The force provided by friction, fF , depends on ?.

fF
ma 0 (block is not moving)
mg sin ????ff ???
N
?
(Newtons 2nd Law along x-axis)
mg
?
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Static Friction...

• We can find ?s by increasing the ramp angle until
  the block slides

mg sin ????ff????
In this case
?ff????SN ? ??Smg cos ?M
?SN
mg sin ?M????Smg cos ?M????
N
mg
??M
?S???tan ?M?
?
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Additional comments on Friction

• Since fF ?N , the force of friction does not
  depend on the area of the surfaces in contact.
• By definition, it must be true that ?S gt ?K
  for any system (think about it...).

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Review Centripetal Acceleration

• UCM results in acceleration
• Magnitude a v2 / R ?? R
• Direction - r (toward center of circle)

v ? R
Useful stuff f rotations / sec T 1 / f ?
2? / T 2? f rad/sec
a
R
?
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Problem Motion in a Circle

• A boy ties a rock of mass m to the end of a
  string and twirls it in the horizontal plane. The
  distance from his hand to the rock is R. The
  speed of the rock is constant and equal to v.
• What is the tension T in the string?

v
T
R
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Motion in a Circle...

• Draw a Free Body Diagram (pick y direction to
  be down)
• We will use FNET ma (surprise)
• First find FNET in y direction
• FNET T

y
T
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Motion in a Circle...

• RECALL CENTRIPETAL ACCELERATION a v2 / R
• SO THE CENTRIPETAL FORCE IS
• ma mv2 / R
• FNET ma T
• T mv2 / R

v
y
T
R
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Problem Rotating puck weight.

• A mass m1 slides in a circular path with constant
  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.
• What is the tension (T) in the string?
• What is the speed (v) of the sliding mass?

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Problem Rotating puck weight...
T

• Draw FBD of hanging mass
• Since R is constant, a 0.
• so T m2g

m2
m2g
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Problem Rotating puck weight...
T m2g
N
T m2g

• Draw FBD of sliding mass

m1
Use F T m1a where a v2 / R
m1g
m2g m1v2 / R
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Problem Motion in a Vertical Circle

• A boy ties a rock of mass m to the end of a
  string and twirls it in the vertical plane. The
  distance from his hand to the rock is R. The
  speed of the rock at the top of its trajectory is
  v.
• What is the tension T in the string at the top of
  the rocks trajectory?

v
T
R
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Motion in a Vertical Circle...

• Draw a Free Body Diagram (pick y-direction to
  be down)
• We will use FNET ma (surprise again!)
• First find FNET in y direction
• FNET mg T

y
mg
T
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Motion in a Vertical Circle...

• FNET mg T
• Acceleration in y direction
• ma mv2 / R
• mg T mv2 / R
• T mv2 / R - mg

v
y
mg
T
F ma
R
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Motion in a Circle...

• What is the minimum speed of the mass at the top
  of the trajectory such that the string does not
  go limp?
• i.e. find v such that T 0.
• mv2 / R mg T
• v2 / R g
• Notice that this doesnot depend on m.

v
mg
T 0
R
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Lecture 14, Act 1Motion in a Circle

• A skier of mass m goes over a mogul having a
  radius of curvature R. How fast can she go
  without leaving the ground?

(a) (b)
(c)
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Lecture 14, Act 1Solution

• mv2 / R mg - N
• For N 0

v
N
mg
R
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Problem Accelerometer

• A weight of mass m is hung from the ceiling of a
  car with a massless string. The car travels on a
  horizontal road, and has an acceleration a in the
  x direction. The string makes an angle ? with
  respect to the vertical (y) axis. Solve for ? in
  terms of a and g.

a
?
i
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Accelerometer...

• Draw a free body diagram for the mass
• What are all of the forces acting?

i
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Accelerometer...

• Using components (recommended)
• i FX TX T sin ? ma
• j FY TY - mg
• T cos ??- mg 0

TX
?
TY
T
?
m
ma
mg
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Accelerometer...

• Using components
• i T sin ? ma
• j T cos ??- mg 0
• Eliminate T

TX
TY
T
?
m
ma
T sin ??? ma
T cos ??? mg
mg
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Accelerometer...

• Alternative solution using vectors (elegant but
  not as systematic)
• Find the total vector force FNET

T (string tension)
T
?
mg
?
m
FTOT
mg (gravitational force)
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Accelerometer...

• Alternative solution using vectors (elegant but
  not as systematic)
• Find the total vector force FNET
• Recall that FNET ma
• So

T (string tension)
?
T
mg
?
m
ma
mg (gravitational force)
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Accelerometer...

• Lets put in some numbers
• Say the car goes from 0 to 60 mph in 10 seconds
• 60 mph 60 x 0.45 m/s 27 m/s.
• Acceleration a ?v/?t 2.7 m/s2.
• So a/g 2.7 / 9.8 0.28 .
• ? arctan (a/g) 15.6 deg

a
?
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Recap of Todays lecture

• Circular Motion
• Centripetal Force
• Motion in a vertical circle