Chapters 10, 11

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Chapters 10, 11 Rotation and angular momentum
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• Rotation of a rigid body
• We consider rotational motion of a rigid body
  about a fixed axis
• Rigid body rotates with all its parts locked
  together and without any change in its shape
• Fixed axis it does not move during the rotation
• This axis is called axis of rotation
• Reference line is introduced

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• Angular position
• Reference line is fixed in the body, is
  perpendicular to the rotation axis, intersects
  the rotation axis, and rotates with the body
• Angular position the angle (in radians or
  degrees) of the reference line relative to a
  fixed direction (zero angular position)

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• Angular displacement
• Angular displacement the change in angular
  position.
• Angular displacement is considered positive in
  the CCW direction and holds for the rigid body as
  a whole and every part within that body

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• Angular velocity
• Average angular velocity
• Instantaneous angular velocity the rate of
  change in angular position

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• Angular acceleration
• Average angular acceleration
• Instantaneous angular acceleration the rate of
  change in angular velocity

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• Rotation with constant angular acceleration
• Similarly to the case of 1D motion with a
  constant acceleration we can derive a set of
  formulas

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Chapter 10 Problem 6
A rotating wheel requires 3.00 s to rotate
through 37.0 revolutions. Its angular speed at
the end of the 3.00-s interval is 98.0 rad/s.
What is the constant angular acceleration of the
wheel?
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• Relating the linear and angular variables
  position
• For a point on a reference line at a distance r
  from the rotation axis
• ? is measured in radians

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• Relating the linear and angular variables speed
• ? is measured in rad/s
• Period (recall Ch. 4)

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• Relating the linear and angular variables
  acceleration
• a is measured in rad/s2
• Centripetal acceleration (Ch. 4)

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• Rotational kinetic energy
• We consider a system of particles participating
  in rotational motion
• Kinetic energy of this system is
• Then

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• Moment of inertia
• From the previous slide
• Defining moment of inertia (rotational inertia)
  as
• We obtain for rotational kinetic energy

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• Moment of inertia rigid body
• For a rigid body with volume V and density ?(V)
  we generalize the definition of a rotational
  inertia
• This integral can be calculated for different
  shapes and density distributions
• For a constant density and the rotation axis
  going through the center of mass the rotational
  inertia for 9 common body shapes is given in
  Table 10-2 (next slide)

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Moment of inertia rigid body
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• Moment of inertia rigid body
• The rotational inertia of a rigid body depends
  on the position and orientation of the axis of
  rotation relative to the body

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• Parallel-axis theorem
• Rotational inertia of a rigid body with the
  rotation axis, which is perpendicular to the xy
  plane and going through point P
• Let us choose a reference frame, in which the
  center of mass coincides with the origin

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Parallel-axis theorem
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Parallel-axis theorem
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Parallel-axis theorem
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Chapter 10 Problem 22
Rigid rods of negligible mass lying along the y
axis connect three particles. The system rotates
about the x axis with an angular speed of 2.00
rad/s. Find (a) the moment of inertia about the x
axis and the total rotational kinetic energy and
(b) the tangential speed of each particle and the
total kinetic energy. (c) Compare the answers for
kinetic energy in parts (a) and (b).
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• Torque
• We apply a force at point P to a rigid body that
  is free to rotate about an axis passing through O
• Only the tangential component Ft F sin f of
  the force will be able to cause rotation

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• Torque
• The ability to rotate will also depend on how
  far from the rotation axis the force is applied
• Torque (turning action of a force)
• SI unit Nm (dont confuse with J)

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• Torque
• Torque
• Moment arm r- r sinf
• Torque can be redefined as
• force times moment arm
• t F r-

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• Newtons Second Law for rotation
• Consider a particle rotating under the influence
  of a force
• For tangential components
• Similar derivation for rigid body

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Newtons Second Law for rotation
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Chapter 10 Problem 39
An electric motor turns a flywheel through a
drive belt that joins a pulley on the motor and a
pulley that is rigidly attached to the flywheel.
The flywheel is a solid disk with a mass of 80.0
kg and a diameter of 1.25 m. It turns on a
frictionless axle. Its pulley has much smaller
mass and a radius of 0.230 m. The tension in the
upper (taut) segment of the belt is 135 N, and
the flywheel has a clockwise angular acceleration
of 1.67 rad/s2. Find the tension in the lower
(slack) segment of the belt.
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• Rotational work
• Work
• Power
• Work kinetic energy theorem

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Corresponding relations for translational and
rotational motion
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• Smooth rolling
• Smooth rolling object is rolling without
  slipping or bouncing on the surface
• Center of mass is moving at speed vCM
• Point P (point of momentary contact between two
  surfaces) is moving at speed vCM
• s ?R
• ds/dt d(?R)/dt R d?/dt
• vCM ds/dt ?R

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• Rolling translation and rotation combined
• Rotation all points on the wheel move with the
  same angular speed ?
• Translation all point on the wheel move with
  the same linear speed vCM

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Rolling translation and rotation combined
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Chapter 10 Problem 53
A cylinder of mass 10.0 kg rolls without slipping
on a horizontal surface. At a certain instant its
center of mass has a speed of 10.0 m/s. Determine
(a) the translational kinetic energy of its
center of mass, (b) the rotational kinetic energy
about its center of mass, and (c) its total
energy.
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• Rolling pure rotation
• Rolling can be viewed as a pure rotation around
  the axis P moving with the linear speed vcom
• The speed of the top of the rolling wheel will
  be
• vtop (?)(2R)
• 2(?R) 2vCM

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• Friction and rolling
• Smooth rolling is an idealized mathematical
  description of a complicated process
• In a uniform smooth rolling, P is at rest, so
  theres no tendency to slide and hence no
  friction force
• In case of an accelerated smooth rolling
• aCM a R
• fs opposes tendency to slide

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Rolling down a ramp Fnet,x M aCM,x fs M g
sin ? M aCM,x R fs ICM a a aCM,x /
R fs ICM aCM,x / R2
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Rolling down a ramp
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• Vector product of two vectors
• The result of the vector (cross) multiplication
  of two vectors is a vector
• The magnitude of this vector is
• Angle f is the smaller of the two angles between
  and

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• Vector product of two vectors
• Vector is perpendicular to the plane that
  contains vectors and and its direction is
  determined by the right-hand rule
• Because of the right-hand rule, the order of
  multiplication is important (commutative law does
  not apply)
• For unit vectors

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Vector product in unit vector notation
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• Torque revisited
• Using vector product, we can redefine torque
  (vector) as

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• Angular momentum
• Angular momentum of a particle of mass m and
  velocity with respect to the origin O is
  defined as
• SI unit kgm2/s

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Chapter 11 Problem 15
A particle of mass m moves in a circle of radius
R at a constant speed. The motion begins at point
Q at time t 0. Determine the angular momentum
of the particle about point P as a function of
time.
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Newtons Second Law in angular form
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Angular momentum of a system of particles
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• Angular momentum of a rigid body
• A rigid body (a collection of elementary masses
  ?mi) rotates about a fixed axis with constant
  angular speed ?
• ?mi is described by

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Angular momentum of a rigid body
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• Conservation of angular momentum
• From the Newtons Second Law
• If the net torque acting on a system is zero,
  then
• If no net external torque acts on a system of
  particles, the total angular momentum of the
  system is conserved (constant)
• This rule applies independently to all components

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Conservation of angular momentum
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Conservation of angular momentum
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More corresponding relations for translational
and rotational motion
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Chapter 11 Problem 50
A projectile of mass m moves to the right with a
speed v. The projectile strikes and sticks to the
end of a stationary rod of mass M, length d,
pivoted about a frictionless axle through its
center. (a) Find the angular speed of the system
right after the collision. (b) Determine the
fractional loss in mechanical energy due to the
collision.
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Answers to the even-numbered problems Chapter 10
Problem 4 - 226 rad/s2
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• Answers to the even-numbered problems
• Chapter 10
• Problem 16
• 54.3 rev
• (b) 12.1 rev/s

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Answers to the even-numbered problems Chapter 10
Problem 26 11mL2/12
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Answers to the even-numbered problems Chapter 10
Problem 32 168 Nm clockwise
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• Answers to the even-numbered problems
• Chapter 10
• Problem 34
• 1.03 s
• 10.3 rev

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Answers to the even-numbered problems Chapter 10
Problem 48 276 J
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• Answers to the even-numbered problems
• Chapter 11
• Problem 4
• 168
• 11.9 principal value
• Only the first is unambiguous.

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Answers to the even-numbered problems Chapter 11
Problem 12 (- 22.0 kgm2/s)ˆk