Chapter 6 Circular Motion and Other Applications of Newton

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Chapter 6Circular Motion and Other Applications
of Newtons Laws
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Review of Circle Terms
Circumference     2pr     pd Circle Area  
      pr²    
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6.1 Newtons 2nd Law Applied to Uniform Circular
Motion

• Centripetal acceleration (UCM)
• ac ? v and radially inward always!!
• Newtons 1st Law
• There must be a force acting
• A force, Fr , is directed toward the center of
  the circle
• This force is associated with an acceleration, ac

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Newtons 2nd Law UCM

• Applying Newtons 2nd Law along the radial
  direction gives
• F ma mac ?
• (6.1)
• (magnitude)
• Direction The total force must be radially
  inward always!

Fr is TENSION in the String
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Newtons 2nd Law UCM

• A force causing a centripetal acceleration acts
  toward the center of the circle
• It causes a change in the direction of the
  velocity vector
• If the force vanishes, the object would move in a
  straight-line path tangent to the circle

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Centripetal Force

• The Force causing ac is sometimes called
  Centripetal Force (Center directed force).
• So far we know forces in nature
• Friction, Gravity, Normal, Tension.
• Should we add Centripetal Force to this List?
  NO!!!!!
• Force causing ac is NOT a new kind of force!
• It is a New Role for force!!!
• It is simply one or more of the forces we know
  acting in the role of a force that causes a
  circular motion.
• Book will not use the term Centripetal Force!!!

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Centripetal Force as New Role

• Earth-Sun Motion
• Centripetal Force Gravity
• Object sitting on a rotating turntable
• Centripetal Force Friction
• Rock-String (horizontal plane)
• Centripetal Force Tension
• Wall-Person (rotating circular room)
• Centripetal Force Normal

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Centripetal Force as New Role

• Ferris Wheel (lowest point)
• Centripetal Force Normal Gravity
• Ferris Wheel (highest point)
• Centripetal Force Normal Gravity
• Rock-String (vertical plane)
• Centripetal Force Tension Gravity

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Centrifugal Force

• Centrifugal Force (Outward) is
• Another Misconception
• Force on the ball is NEVER outward Centrifugal
  Force
• Force is ALWAYS inward

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Centrifugal Force
(a)
(b)

• If Centrifugal Force existed, the ball would
• Fly Off as in (a) when released.
• Ball Flies off as in figure (b). Similar to
  sparks flying in straight line from the edge of
  rotating grinding wheel.

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Example 6.1 Conical Pendulum

• Find an expression for v?
• The object is in equilibrium in the vertical
  direction and undergoes UCM in the horizontal
  direction
• Components of T
• Tx Tsin? Ty Tcos?
• From Example 6.2
• Tx maC mv2/r
• Ty mg ?

Ty
?
Tx
v
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Example 6.1 Conical Pendulum, final

• Tx Tsin? mv2/r (1)
• Ty Tcos? mg (2)
• Dividing (1) by (2)
• Sin?/cos? tan? v2/rg
• Since r Lsin? ?
• tan? v2/Lsin?g
• Solving for v
• v is independent of m

v
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Circular Motion Animation

• http//www.mhhe.com/physsci/physical/giambattista/
  circular/circular.html

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Motion in a Horizontal Circle
v
T

• The speed at which the object moves depends on
  the mass of the object and the tension in the
  cord
• The centripetal force the tension.
• F ma ? T max maC mv2/r ?

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Force on revolving Ball (Horizontal)
v
T

• Find T ? If m 0.150-kg, r 0.600 m, f 2
  rev/s (T 0.500s)
• Assumption Circular path is ? in horizontal
  plane, so
• ? ? 0 ? cos(?) ? 1
• Newtons 2nd Law F ma ? T max maC mv2/r
• But v 2prf 7.54 m/s ?
• T (0.150kg)(7.54m/s)2/0.600m) ?
• T 14.2N (Tension)

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Horizontal (Flat) Curve

• Force of static friction (ƒs,max) supplies the
  centripetal force
• Radial (x) SFx mac
• Car stays in the curve if and only if ƒs,max
  mac ?
• ƒs,max mac mv2/r
• Vertical (y) n mg
• Since ƒs,max ?sn ?smg ?
• ?smg mv2/r (cancel m)
• Solving for v the maximum speed at which the car
  can negotiate the curve is
• Again v is independent of m

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Example 6.3 Highway Curves(Example 6.4 Text
Book)
A 1000 kg car rounds a curve on a flat road of
radius 50.0m at a speed of 50 km/hr (14m/s).
Find the Coefficient of Static Friction (?s) if
the car follows the curve and make the turn
successfully.
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Example 6.3 Highway Curves, 2

• Radial (x) SFx mac
• Car stays in the curve if and only if ƒs,max
  mac ?
• ƒs,max mac mv2/r
• Vertical (y) n mg
• Since ƒs,max ?sn ?smg ?
• ?smg mv2/r (cancel m)
• Solving for ?s

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Example 6.4 Banked Curve (Example 6.5 Text Book )

• These are designed with friction equaling zero
• A car is traveling with speed v around a curve of
  radius r, determine a formula for the angle (?)
  at which a road should be banked so that no
  friction is required.
• Solution
• There is a component of the normal force (nsin?)
  that supplies the centripetal force
• Car is moving along a horizontal circle so ac is
  horizontal too.

?
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Example 6.4 Banked Curve, final

• Components of n
• nx nsin? ny ncos?
• x direction
• SFx max ? nx maC mv2/r ?
• nsin? mv2/r (1)
• y direction
• SFy 0 ? ncos? mg (2)
• Dividing (1) by (2)
• Sin?/cos? tan? v2/rg
• (Similar to The Conical Pendulum) ?

nx nsin?
ny ncos?
?
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Example 6.5 Force on Revolving Ball. Top and
Bottom of Circle (Vertical)

• Similar to Example 6.6 Change T by n
• Given r, v, m
• Use SF mac mv2/r
• Find tensions Ttop Tbot
• The tension at the bottom is a maximum
• Tbot mg mv2/r ?
• Tbot mv2/r mg

ac
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Example 6.5 Force on Revolving Ball. Top and
Bottom of Circle, final

• The tension at the top is a minimum
• Ttop mg mv2/r ?
• Ttop mv2/r mg
• If Ttop 0, then v2/r g ?

ac
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Loop-the-Loop (Example 6.6 Text Book)

• This is an example of a vertical circle
• At the bottom of the loop (b), the upward force
  experienced by the object is greater than its
  weight

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Loop-the-Loop, final

• At the top of the circle (c), the force exerted
  on the object is less than its weight

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Example 6.6 Quizzes 6.1 6.2

• 6.1
• Centripetal Acceleration a
• is always toward the
• Center of the Circular Path!!
• 6.2
• Normal Force is always
• Perpendicular to the surface
• that applies the force.
• Because the car maintains
• its orientation at all points of
• the ride, the Normal Force
• Is always upward!!!

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6.2 Non-Uniform Circular Motion

• The acceleration and force have tangential
  components
• Fr produces the centripetal acceleration
• Ft produces the tangential acceleration
• SF SFr SFt

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Vertical Circle with Non-Uniform Speed

• The gravitational force exerts a tangential force
  on the object
• Look at the components of Fg
• The tension at any point can be found
• T mgcos? mv2/R ?
• T mv2/R mgcos? ?