Rolling, Torque, and Angular Momentum

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Rolling, Torque, and Angular Momentum

• AP Physics C
• Montwood High School
• R. Casao

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Rolling

• When a wheel moves along a straight track, the
  center of the wheel moves forward in pure
  translation.
• A point on the rim of the wheel traces out a
  complex path called a cycloid.

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• For a wheel passing at constant speed while
  rolling smoothly (no sliding)
• the center of mass O of the wheel moves forward
  at constant speed vcom.
• the point P where the wheel makes contact with
  the surface also moves forward at speed vcom so
  that it remains directly below the center of mass
  O.
• During time interval t, both
• O and P move forward by
• a distance s. Observers
• see the wheel rotate thru an
• angle q about the center of
• the wheel.

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• The point on the wheel P that was touching the
  surface at t 0 s moves through an arc length s.
• Equations s R? vcom ds/dt ? d?/dt
• Differentiating the arc length equation wrt time
  ds/dt Rd?/dt to give us vcom R?.

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• The rolling motion of a wheel is a combination of
  purely translational and purely rotational
  motions.
• Every point on the wheel rotates about the center
  with angular speed ?.
• Every point on the outside edge of the wheel has
  linear speed vcom.
• Purely translational motion every point on the
  wheel moves forward with speed vcom.
• Combination of rotational motion and
  translational motion produces the rolling motion
  of the wheel.

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• The portion of the wheel at the bottom (point P)
  is stationary and the top of the wheel is moving
  at speed 2vcom. The wheel is moving fastest
  near the top than near the bottom because the
  spokes are more blurred at the top than at the
  bottom.

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Kinetic Energy of Rolling

• Consider the rolling motion about an axis thru
  point P K 0.5Ip?²
• Ip is the rotational inertia of wheel about axis
  thru point P and ? is the angular speed of the
  wheel.
• Apply the parallel-axis theorem to determine Ip
• Ip Icom Mh² for the wheel, h R
• Substituting
• K 0.5Icom?² 0.5MR²?²
• Because vcom R?,
• K 0.5Icom?² 0.5Mvcom²

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• A rolling object has two types of kinetic energy
  a rotational kinetic energy (0.5Icom?²) due to
  its rotation about its center of mass and a
  translational kinetic energy (0.5Mvcom²) due to
  translation of its center of mass.
• Forces of Rolling
• If a wheel rolls at constant speed it has no
  tendency to slide at the point of contact P and
  no frictional force acts there acom 0 at point
  P.
• If a net force acts on the rolling wheel to speed
  it up or slow it down, the net force causes an
  acceleration of the center of mass along the
  direction of travel.

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• The wheel also rotates faster or slower, so an
  angular acceleration ? occurs about the center of
  mass.
• These accelerations tend to make the wheel slide
  at point P and a frictional force must act on the
  wheel at P to oppose the tendency to slide.
• If the wheel does NOT slide, the force is a
  static frictional force fs and the motion is
  smooth rolling motion.
• Smooth rolling motion acom R?
• If the wheel does slide when the net force acts
  on it, the frictional force that acts at P is a
  kinetic frictional force and the motion is not
• smooth rolling motion.

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Rolling Down a Ramp

• For a uniform body of mass M and radius R rolling
  smoothly down a ramp at angle ? along an x axis.
• The normal force N acts at
• point P but has been shifted
• up to the center of mass.
• Static frictional force fs acts
• at point P and is directed
• up the incline. If the body
• were to slide down the
• incline, the frictional force
• would oppose that motion.

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• Motion along the x-axis (up as positive)
• fs Mgsin ? Macom
• Angular motion Torque I?
• N and Fgsin ? pass thru the pivot at the center
  of mass and do not produce a torque.
• fs applied perpendicularly a distance R at point
  P and produces a torque.
• Torque F?r? fsR fsR Icom?
• -acom R? acom is negative because it is
  directed in the negative direction on the x-axis.
  Unlike the linear motion problems we have done
  in the past where we took the direction of motion
  as positive, with the rotation involved, we take
  the positive direction as the positive direction
  of the rotation (ccw positive cw negative).
  

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• Solve for ? and substitute ? -acom/R
• fsR Icom -acom/R

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Yo-Yo

• As yo-yo rolls down its string from height h,
  potential energy mgh is converted to both
  rotational (0.5Icom?²) and translational
  kinetic energy (0.5Mvcom²). As it climbs back
  up, kinetic energy is converted to potential
  energy.
• The string is looped around the axle and when the
  yo-yo hits the bottom of the string, an upward
  force on the axle from the string stops the
  descent.
• The yo-yo spins with the axle inside the loop
  with only rotational kinetic energy.

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• The yo-yo spins until you jerk on the string to
  cause the string to catch on the axle and allow
  the yo-yo to climb back up.
• The rotational kinetic energy of the yo-yo at the
  bottom of the string can be increased by throwing
  the yo-yo downward so that it starts down the
  string with initial speeds vcom and ? instead of
  rolling down from rest.
• The yo-yo rolls on the axle of radius Ro.
• The yo-yo is slowed by the tension force T from
  the string on the axle.

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• Acceleration equation
• the yo-yo has the same downward acceleration
  when it is climbing back up the string because
  the forces acting on it are still those shown in
  the figure.
• Torque Revisited
• We previously defined torque for a rigid body
  that rotated about a fixed axis in a circle.
• To define the torque for an individual particle
  that moves along any path relative to a fixed
  point (rather than a fixed axis), the path
  doesnt have to be a circle and the torque has to
  be written as a vector.

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• The figure shows a particle at point A in the xy
  plane with a single force F acting on the
  particle.
• The particles position relative to the origin O
  is given by the position vector r.
• Torque equation

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• To find the direction of the torque, slide the
  force vector without changing its direction until
  its tail is at the origin O so that it is tail to
  tail with the position vector.
• Use the right hand rule to rotate the position
  vector r into the force vector F. The thumb
  points in the direction of the torque.

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• The magnitude of the torque is given by
• ? is the angle between the position vector r and
  the force vector F.
• Angular Momentum
• The figure shows a particle of mass
• m with linear momentum p mv
• as it passes thru point a in the xy
• plane. The angular momentum l of
• the particle with respect to the
• origin O is

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• r is the position vector of the particle wrt O.
• As the particle moves wrt O in the direction of
  its momentum p, the position vector r rotates
  around O.
• To have angular momentum, the
• particle itself does not have to
• rotate around O.
• Unit for angular momentum
• kgm²/s Js
• The direction of the angular
• momentum vector is found by
• sliding the vector p until its tail
• is at the origin O.

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• Use the right-hand rule, rotating the vector r
  into vector p. The thumb points in the direction
  of the angular momentum vector.
• The magnitude of the angular momentum vector is
  l rmvsin ?, where ? is the angle between r
  and p when the two vectors are tail to tail.
• Angular momentum only has meaning wrt a specific
  origin.
• The direction of the angular momentum vector is
  always perpendicular to the plane formed by the
  position vector and the linear momentum vector.

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Newtons Second Law in Angular Form

• Fnet dp/dt expresses the close relation between
  force and linear momentum for a single particle.
• There is also a close relationship between torque
  and angular momentum Tnet dl/dt
• The vector sums of all the torques acting on a
  particle is equal to the time rate of change in
  the angular momentum of that particle.
• The torques and angular momentum are both defined
  wrt the same origin.

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Angular Momentum of a System of Particles

• For a system of particles with respect to an
  origin, the total angular momentum L of the
  system is the vector sum of the angular momenta l
  of the individual particles
• Over time, the angular momenta of individual
  particles may change due to interactions within
  the system between individual particles or
  because of outside influences on the system.

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• Since Tnet dl/dt, then
• The rate of change of the systems angular
  momentum L is equal to the vector sum of the
  torques on the individual particles.
• These torques include internal torques due to
  forces between particles and external torques due
  to forces on the particles from bodies outside of
  the system.
• The forces between particles always occur in 3rd
  law pairs so their torques cancel each other, so
  the only torques that can change the total
  angular momentum L of a system are external
• torques.

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• 2nd law for rotation of a system of particles
  the net external torque Tnet acting on a system
  of particles is equal to the time rate of change
  of the systems total angular momentum L.
• The torque and the systems angular momentum must
  be measured from the same origin.
• Angular Momentum of a Rigid Body Rotating About a
  Fixed Axis
• For a system of particles that form a rigid body
  that rotates about a fixed axis rotating with
  constant angular speed ? L I?

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Conservation of Angular Momentum

• If the net external torque acting on a system is
  zero, the angular momentum L of the system
  remains constant, no matter what changes take
  place within the system Li Lf Iiwi Ifwf
• Because momentum is a vector quantity , the
  conservation of momentum has to be considered in
  all three dimensions (x, y, z). Depending on the
  torques acting on a system, the angular momentum
  of the system might be conserved in only 1 or 2
  directions but not in all directions.

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