Centripetal Acceleration and Centripetal Force

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Centripetal Acceleration and Centripetal Force

• Physics
• Montwood High School
• R. Casao

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Circular Motion

• When an object moves in a circle at constant
  speed, we describe it as undergoing uniform
  circular motion.
• Its speed is constant, but its velocity is not
  because velocity includes direction and the
  objects direction is clearly changing.

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Circular Motion

• A changing velocity means acceleration.
• The pull on the string is always directed
  perpendicular to the velocity.
• The pull accelerates the ball into a circular
  path, even though the ball does not speed up or
  slow down.
• The pull changes only the direction of the
  velocity, not the magnitude.

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Centripetal Acceleration

• The acceleration arising from the change in
  direction of the velocity vector is called the
  centripetal acceleration and is determined
  mathematically by

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Centripetal Acceleration
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Centripetal Acceleration

• Centripetal means center-seeking.
• Centripetal acceleration is always directed
  toward the center of the circle of motion.
• It is this centripetal acceleration that is
  responsible for the change in the direction of
  the velocity the magnitude of the velocity
  remains constant.
• Any change in the tangential acceleration causes
  a change in the speed of the particle as it
  travels around the circle. In uniform circular
  motion, aT 0, so the acceleration is completely
  radial (ar) or centripetal.

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Centripetal Force

• Newtons Second Law explains that an object
  undergoing acceleration is experiencing a net
  force. The net force on an object undergoing
  uniform circular motion is called the centripetal
  force Fc.
• The centripetal force necessary for an object of
  mass m to travel with constant speed v in a
  circle of radius r is given by

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Centripetal Force

• The centripetal force always points toward the
  center of the circle about which the object moves
  with uniform speed.
• If the centripetal force applied to the object is
  removed, the object will move in a straight-line
  tangent to the curved path at the point where the
  centripetal force ceases. When the centripetal
  force ceases, the object has no unbalanced forces
  acting upon it and thus moves in a straight line
  at constant speed.

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Circular Motion

• If the string breaks, the ball flies off in a
  straight line. It is the force of the string
  that causes the acceleration in this example of
  uniform circular motion.

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Centripetal Force

• Centripetal force is the name given to any force
  that is directed at right angles to the path of a
  moving object and that tends to produce circular
  motion.
• Examples
• the gravitational force directed toward the
  center of the Earth holds the Moon in an almost
  circular orbit about the Earth
• an electromagnetic force that is directed toward
  the nucleus holds the electrons that revolve
  about the nucleus of the atom.
• Directions in centripetal force problems
• Positive direction is inwards toward center of
  circle.
• Negative direction is outward away from center of
  circle.

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• Radius r is the distance from the center of the
  mass to the axis of rotation.

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• The centripetal force is not a force and
• does not belong in a free-body diagram.
• The force F in the picture would provide the
  centripetal force needed to maintain the circular
  path.

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Motion On A Flat Curve

• The net force on a car traveling around a curve
  is the centripetal force.
• As a car travels around a curve, the net force on
  the car must be the centripetal force, directed
  toward the center of the circle the curve is a
  portion of.

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Motion On A Flat Curve

• On a flat, level curve, the friction between the
  tires and the road supplies the centripetal
  force.
• If the tires are worn smooth or the road is icy
  or oily, this friction force will not be
  available.
• The car will not be able to move in a circle, it
  will keep going in a straight line and therefore
  go off the road.

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Motion On A Flat Curve

• Accelerations
• ay 0 m/s2
• ax ac
• Equation Fc FF

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Motion On A Banked Curve

• Some curves are banked to compensate for slippery
  conditions.
• In addition to any friction forces that may or
  may not be present, the road exerts a normal
  force perpendicular to its surface.
• The downward force of the cars weight is also
  present.
• These two forces add as vectors to provide a net
  force Fnet that points toward the center of the
  circle this is the centripetal force.
• The centripetal force is directed toward the
  center of the circle, it is not parallel to the
  banked road.

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Motion On A Banked Curve

• The effect of banking is to tilt the normal force
  Fn toward the center of curvature of the road so
  that the inward radial component FN?sin ? can
  supply the required centripetal force.
• Vehicles can make a sharp turn more safely if the
  road is banked. If the vehicle maintains the
  speed for which the curve is designed, no
  frictional force is needed to keep the vehicle on
  the road.

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Motion On A Banked Curve

• The effect of banking is to tilt the normal force
  FN toward the center of curvature of the road so
  that the inward radial component FN?sin ? can
  supply the required centripetal force.

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Motion On A Banked Curve

• There is no acceleration along the y axis, so the
  sum of the forces in the y plane is zero
• The horizontal component of the normal force FN,
  the force the road exerts against the car,
  provides the necessary centripetal force.
  Because the only force in the x plane is the
  centripetal force

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Motion On A Banked Curve

• This equation gives the banking angle that allows
  a car to travel in a curve of radius r with
  constant speed v and require no friction force.
• Goldilocks Just Right!

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Motion On A Banked Curve

• A banked curve is designed for one specific
  speed, called the design speed.
• If the banked curve is icy so that there is no
  friction force at all, then traveling at a speed
  higher than the design speed means the car will
  slide out, up, and over the edge.
• Traveling at a lower speed than the design speed
  means the car will slide in, down, and off the
  bank.
• When a banked curve has to be negotiated at a
  speed above the design speed (or if you are asked
  to find the maximum speed), friction is required.
  The frictional force Ff acts parallel to the
  road surface.

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Motion On A Banked Curve (w/ Friction)Goldilocks
Too Fast!
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Motion On A Banked Curve (w/ Friction)Goldilocks
Too Fast!
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Motion On A Banked Curve (w/ Friction)

• Resolve the normal force FN and friction FF into
  horizontal and vertical components.
• The horizontal components of FN and FF are both
  directed inward toward the center of the curve,
  therefore, these two force components combine to
  determine the centripetal force SFx mac.

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Motion On A Banked Curve (w/ Friction)

• Because there are no unbalanced forces in the
  vertical direction, the upward forces must equal
  the downward forces, therefore
• Substitute into both
  equations.

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Motion On A Banked Curve (w/ Friction)

• Solve both equations for Fn and set these
  equations equal to each other (because FN FN as
  there is only one force acting normal to the
  surface). From this equation, the unknown
  variable can be determined.

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Motion On A Banked Curve (w/ Friction)

• For problems involving a minimum speed for the
  vehicle to travel around the curve without
  skidding, the frictional force is directed up the
  incline to keep the vehicle from sliding down to
  the bottom of the banked curve.
• Resolve the normal force FN and friction FF into
  horizontal and vertical components.

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Motion On A Banked Curve (w/ Friction)Goldilocks
Too Slow!
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Motion On A Banked Curve (w/ Friction)Goldilocks
Too Slow!
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Motion On A Banked Curve (w/ Friction)

• The horizontal component of FN is inward toward
  the center of the curve and is positive FF is
  directed outward away from the center of the
  curve and is negative. These two force
  components combine to determine the centripetal
  force SFx mac.

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• Because there are no unbalanced forces in the
  vertical direction, the upward forces must equal
  the downward forces, therefore
• Substitute into both
  equations.

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• Solve both equations for Fn and set these
  equations equal to each other (because FN FN as
  there is only one force acting normal to the
  surface). From this equation, the unknown
  variable can be determined.

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Vertical Circles

• The force of gravity causes the speed of an
  object in a vertical circular path to vary. The
  object accelerates on the downward portion of its
  circular path and decelerates on the upward
  portion of the circular path.
• At the top and bottom of a vertical circular
  path, the weight and the normal force (or an
  equivalent supporting force, such as tension) are
  the only forces acting on an object. The
  centripetal force is supplied by the resultant of
  the weight and a supporting force (often the
  normal force).

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Vertical Circles

• The forces acting on a person sitting in a roller
  coaster car are shown. The persons weight FW is
  present and so is the normal force FN that the
  seat exerts on him (this is your apparent
  weight).

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Vertical Circles

• The normal force FN, the force you feel on the
  seat of your pants, can be positive, negative, or
  zero.
• A negative value for FN means the passenger has
  to be strapped in, with the straps exerting an
  upward force. Such a situation would be
  dangerous, and roller coaster designers avoid
  this.
• If FN 0 N, the person seems to be weightless as
  well as upside down.

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Vertical Circles

• The forces on an airplane pilot at the bottom of
  a dive can be quite large.
• Gravity pulls downward and the seat exerts its
  usual normal force FN, this time upward.

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Vertical Circles

• At the bottom of the dive, the normal force can
  only be positive, must be greater than the
  weight, and can become very large. A roller
  coaster at the bottom of the loop provides the
  same forces.
• The acceleration can be expressed as
• This acceleration can be expressed in terms of
  gs, where gs are determined by dividing the
  centripetal acceleration by gravity. One g is
  9.8 m/s2. The number of gs represents the
  relative pull of gravity on the body that the
  person experiences.

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Vertical Circles

• Experiencing a significant number of gs makes
  the work of the heart more difficult.
  Accelerations of eight to ten gs make it
  difficult for the circulatory system to get
  enough blood to the brain and may result in
  blackouts. Pressure suits that squeeze on the
  legs push blood back into the rest of the body,
  including the brain, and help prevent blackouts.

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Vertical Circles

• For the Ferris wheel, the only difference occurs
  at the top where the seat is facing upward.
• Top
• This equation is also for a car passing over the
  top of a curve.
• Bottom

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Tension

• For an object attached to a string and moving in
  a vertical circle, the centripetal force is at a
  minimum at the top of its vertical path and at a
  maximum at the bottom of its vertical path.

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Tension

• Top the centripetal force on the object equals
  the tension of the string plus the weight of the
  ball, both acting toward the center of the
  vertical circle. Mathematically
• Bottom the centripetal force on the object is
  equal to the difference between the tension of
  the string and the weight of the object. The
  tension is exerted inward toward the center of
  the vertical circle, while the weight is directed
  away from the center of the vertical circle.
  Mathematically

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Critical Velocity (vmin)

• Critical velocity velocity below which an
  object moving in a vertical circle will not
  describe a circular path.
• Critical velocity depends on the acceleration due
  to gravity and the radius of the vertical circle,
  not on the mass of the object.

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Tension at an Angle Horizontal Circles

• Vertical component of the tension is equal to the
  weight.
• Horizontal component of the weight is equal to
  the centripetal force.
• the radius is the distance from the center of the
  mass to the dot at the center of the horizontal
  circle.

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Conical Pendulum

• For conical pendulums, centripetal force is
  provided by a component of the tension.
• ? is the angle between the vertical and the cord.

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Conical Pendulum
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Fn Difference?

• In every previous inclined plane problem such as
  a skier on a hill or a block on a ramp, the
  normal force is given by Fn mgcos ?.
• In every banked curve problem the normal force is
  given by Fnmg/cos ?.
• Both equations are correct!
• The difference is due to the direction of the
  acceleration.
• the direction of the acceleration in the inclined
  plane problem is down toward the bottom of the
  incline. There is no component of the
  acceleration in the normal direction.

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• the direction of the acceleration in the banked
  curve is horizontally toward the center of the
  circle. This produces a component of
  acceleration in the normal direction.
• In the inclined plane problems, I taught you to
  rotate the x- and y- axes so that the x-axis is
  parallel to the surface and the y-axis is
  perpendicular to the surface. We did not do this
  with the banked turn problems.
• This results in the acceleration being zero on
  the y axis for both cases.
• In the inclined plane problems, the acceleration
  is zero along the normal perpendicular to the
  incline thus the normal force equals a
  component of gravity.)

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• In the frictionless banked turn, the acceleration
  is zero vertically thus the force of gravity
  equals a component of the normal force.
• In the inclined plane problems, one component of
  gravity causes the acceleration (the other
  component cancels out the normal force).
• In the banked turn, one component of the normal
  force causes the acceleration (the other
  component cancels out the force of gravity).

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For Uniform Circular Motion

• n number of revolutions (rotations)
• r radius
• t time
• Period time for one revolution (rotation)

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Web Sites

• Amusement Park Physics
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• CoasterQuest.com