Physics 1710 Chapter 11 Rolling Motion and Angular Momentum

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Quiz
• What torque must be applied to a discus (assume
  a short cylindrical shape) that has a diameter
  of 20 cm and a mass of 3.0 kg, if it leaves the
  hand of the discus thrower spinning at a rate of
  180 rpm if he takes 0.5 sec to throw the disc?

? 2??(180)/60 18.7 rad/s I ½ MR2 ½ (3.0)
(0.10)2 0.015 kg m
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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Quiz
• What torque must be applied to a discus (assume
  a short cylindrical shape) that has a diameter
  of 20 cm and a mass of 3.0 kg, if it leaves the
  hand of the discus thrower spinning at a rate of
  180 rpm if he takes 0.5 sec to throw the disc?

? 18.7 rad/s I ½ MR2 0.015 kg m ? 18.7
rad/s/(0.5 sec) 37.4 rad/s/s T I? (0.015)
(37.4) N?m 0.56 N?m F T/R 0.56/0.1 5.6 N
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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• 1' Lesson
• The total Kinetic energy of a rotating system is
  the sum of the rotational energy about the Center
  of Mass and the translational KE of the CM.
• T r x F
• Angular momentum L is the vector product of the
  moment arm and the linear momentum.
• The net externally applied torque is equal to
  the time rate of change in the angular momentum.

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• 1' Lesson (contd)
• Angular momentum about an axis z is equal to
  the product of the moment of inertia of the body
  about that axis and the angular velocity about z.
• In the absence of torques, the angular momentum
  is conserved.

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Rolling Motion
• The compound motion of a rolling object is a
  translation plus a rotation.
• vCM R? aCM R?
• Cylinder rolling down a ramp
• F Mg sin ? T RF I ?
• aCM R 2 (Mg sin ?)/I x ½ aCM t 2
• Which will win? A large disk or a small one?
• A heavy one or a light one?

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Total Energy of Motion
• K ½ Ip ? 2 ½ ICM ? 2 ½ MR 2 ? 2
• By parallel axis theorem.
• Thus, the total kinetic energy is the energy of
  motion of the center of mass plus the energy of
  rotation about the CM

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Vector Product
• C A x B
• Cx Ay Bz Az By
• Cyclically permute (xyz), (yzx), (zxy)
• C vCx2 Cy2 Cy2
• AB sin ?
• Directed by RH Rule.

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Vector Product
• A x B - B x A
• A x ( B C ) A x B A x C
• d/dt ( A x B ) d A /dt x B A x d B/dt
• i x i j x j k x k 0
• i x j - j x i k
• j x k - k x j i
• k x i - i x k j

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Angular Momentum
• L r x p
• The angular momentum is the vector product of
  the moment arm and the linear momentum.
• ? T d L/dt
• The net torque is equal to the time rate of
  change in the angular momentum.

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Proof
• ? T r x ?F r x d p/dt
• And
• d L/dt d( r x p) /dt
• d r/dt x p r x d p/dt.
• But p m d r/dt , therefore d r/dt x p 0
• d L/dt r x d p/dt
• And thus
• ? T d L/dt.

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Rotating Platform Demonstration

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Analysis
• Why does an ice skater increase her angular
  velocity without the benefit of a torque?
• L r x p
• r x ( m v)
• r x ( m r x ?)
• Li mi ri 2 ?
• Lz (?i mi ri 2 ) ?
• Lz I ? ? Lz / I
• Therefore, a decrease in I ( by reducing r) will
  result in an increase in ?.

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Torque and Angular Momentum (redux)
• L I ?
• The quantity I is a tensor .
• Changing I changes the magnitude and alignment
  of L and ?.

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• If I is constant then for rotation about an axis
  z
• ? Tz d Lz /dt Iz ?
• In the absence of net external torque the angular
  momentum is conserved.

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Spinning Bicycle Wheel
• Demonstration

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Gyroscope Demonstration

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Fundamental Angular Momentum
• Fundamental unit of angular momentum ?
• ? 1.054 x 10 -34 kg?m/s2
• ICM? ?
• ? ? / ICM
• 1.054 x 10 -34 kg?m/s2 / (1.95 x 10 -46 kg?m)
• 5.41 x 10 11 rad/s

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Summary
• The total Kinetic energy of a rotating system is
  the sum of the rotational energy about the Center
  of Mass and the translational KE of the CM.
• K ½ ICM ? 2 ½ MR 2 ? 2
• T r x F

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Summary Question
• Which will win a race down an incline?
• A ball rolling without slipping?
• A ball sliding with no friction?
• Why?

K ½ ICM ? 2 ½ MR 2 ? 2
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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Summary
• Angular momentum L is the vector product of the
  moment arm and the linear momentum.
• L r x p
• The net externally applied torque is equal to
  the time rate of change in the angular momentum.
• ? Tz d Lz /dt Iz ?

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum

• Summary
• Angular momentum about an axis z is equal to
  the product of the moment of inertia of the body
  about that axis and the angular velocity about z.
• L I ?
• Lz Iz ?
• In the absence of torques, the angular momentum
  is conserved.

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Physics 1710Chapter 11 Rolling Motion and
Angular Momentum