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11B: Rigid Body Rotation
Inertia of Rotation Consider Newton’s first law for the inertia of rotation to be patterned after the law for translation. Moment of Inertia or Rotational Inertia is the rotational analog of mass. Units of I:   kg-m2     ;      g-cm2     ; slug-ft2
Common Rotational Inertias
Example 2: A circular hoop and a disk each have a mass of 3 kg and a radius of 30 cm. Compare their rotational inertias. I = 0.3 kg m2 I = 0.1 kg m2 hoop R R Disk
Inertia of Rotation Ξ£F = 20 N a = 4m/s2 F = 20 N R = 0.5 m a = 2 rad/s2 Consider Newton’s second law for the inertia of rotation to be patterned after the law for translation. Linear Inertia, m Rotational Inertia, I Force does for translation while torque does for rotation
Newton’s 2nd Law for Rotation V f t w wo = 50 rad/s t = 40 N m R 4 kg Important Analogies For many problems involving rotation, there is an analogy to be drawn from linear motion. I A resultant torque t produces angular acceleration a  of disk with rotational inertia I. A resultant force F produces negative acceleration a for   a mass m.
F wo = 50 rad/s R = 0.20 m F = 40 N w R 4 kg Newton’s 2nd Law for Rotation How many revolutions are required to stop?
Seatwork:   Β½ sheet of pad paper (crosswise) Time allotted: 40 minutes  A disk of mass 𝑀=6.00Β kgΒ and radius 𝑅=0.500Β mΒ is used to draw water from a well (Fig. 1). A bucket of mass π‘š=2.00Β kg is attached to a cord that is wrapped around the disk.a.  What is the tension T in the cord and acceleration a of the bucket?b.  If the bucket starts from rest at the top of the     well and falls for 3.00 s before hitting the water,    how far does it fall? c.  About how many revolutions are made     by the disk in the situation stated in (b)?Note:    πΌπ‘‘π‘–π‘ π‘˜=12𝑀𝑅2 Β  Fig. 1
Work and Power for Rotation βˆ†π‘  Β  βˆ†πœƒ Β  F F βˆ†π‘ =π‘…βˆ†πœƒ Β  π‘Š=πΉβˆ†π‘  Β  since Β  𝑅 Β  𝐹𝑅=𝜏 Β  π‘Š=πΉπ‘…βˆ†πœƒ Β  since βˆ†π‘ =π‘…βˆ†πœƒ Β  π‘Šπ‘Ÿπ‘œπ‘‘=πœβˆ†πœƒ Β  π‘ƒπ‘Ÿπ‘œπ‘‘=π‘Šπ‘Ÿπ‘œπ‘‘βˆ†π‘‘ Β  π‘ƒπ‘Ÿπ‘œπ‘‘=πœβˆ†πœƒβˆ†π‘‘ Β  βˆ†πœƒβˆ†π‘‘=πœ” Β  since
Example:The rotating disk has a radius of 40 cm and a mass of  6 kg. Find the work and power needed in lifting the 2-kg mass 20 m above the ground in 4 s in uniform motion. s q F 6 kg 2 kg F=W s = 20 m
Rotational Kinetic Energy v = wR m m4 w m3 m1 m2 axis Object rotating at constant w. Consider tiny mass m: 𝐾=12π‘šπ‘£2 Β  𝑣=πœ”π‘… Β  since 𝐾=12π‘šπœ”π‘…2 Β  𝐾=12π‘šπ‘…2πœ”2 Β  To find the total kinetic energy: π‘šπ‘…2=𝐼 Β  since 𝑲𝒓𝒐𝒕=πŸπŸπ‘°πŽπŸ Β  (Β½w2 same for all m )
Recall for linear motion that the work done is equal to the change in linear kinetic energy: Using angular analogies, we find the rotational work is equal to the change in rotational kinetic energy: The Work-Energy Theorem
F wo = 60 rad/s R = 0.30 m F = 40 N w R 4 kg Example:Applying the Work-Energy Theorem: What work is needed to stop wheel rotating in the figure below?

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Chapter13 rigid body rotation

  • 1. 11B: Rigid Body Rotation
  • 2. Inertia of Rotation Consider Newton’s first law for the inertia of rotation to be patterned after the law for translation. Moment of Inertia or Rotational Inertia is the rotational analog of mass. Units of I: kg-m2 ; g-cm2 ; slug-ft2
  • 4. Example 2: A circular hoop and a disk each have a mass of 3 kg and a radius of 30 cm. Compare their rotational inertias. I = 0.3 kg m2 I = 0.1 kg m2 hoop R R Disk
  • 5. Inertia of Rotation Ξ£F = 20 N a = 4m/s2 F = 20 N R = 0.5 m a = 2 rad/s2 Consider Newton’s second law for the inertia of rotation to be patterned after the law for translation. Linear Inertia, m Rotational Inertia, I Force does for translation while torque does for rotation
  • 6. Newton’s 2nd Law for Rotation V f t w wo = 50 rad/s t = 40 N m R 4 kg Important Analogies For many problems involving rotation, there is an analogy to be drawn from linear motion. I A resultant torque t produces angular acceleration a of disk with rotational inertia I. A resultant force F produces negative acceleration a for a mass m.
  • 7. F wo = 50 rad/s R = 0.20 m F = 40 N w R 4 kg Newton’s 2nd Law for Rotation How many revolutions are required to stop?
  • 8. Seatwork: Β½ sheet of pad paper (crosswise) Time allotted: 40 minutes A disk of mass 𝑀=6.00Β kgΒ and radius 𝑅=0.500Β mΒ is used to draw water from a well (Fig. 1). A bucket of mass π‘š=2.00Β kg is attached to a cord that is wrapped around the disk.a. What is the tension T in the cord and acceleration a of the bucket?b. If the bucket starts from rest at the top of the well and falls for 3.00 s before hitting the water, how far does it fall? c. About how many revolutions are made by the disk in the situation stated in (b)?Note: πΌπ‘‘π‘–π‘ π‘˜=12𝑀𝑅2 Β  Fig. 1
  • 9. Work and Power for Rotation βˆ†π‘  Β  βˆ†πœƒ Β  F F βˆ†π‘ =π‘…βˆ†πœƒ Β  π‘Š=πΉβˆ†π‘  Β  since Β  𝑅 Β  𝐹𝑅=𝜏 Β  π‘Š=πΉπ‘…βˆ†πœƒ Β  since βˆ†π‘ =π‘…βˆ†πœƒ Β  π‘Šπ‘Ÿπ‘œπ‘‘=πœβˆ†πœƒ Β  π‘ƒπ‘Ÿπ‘œπ‘‘=π‘Šπ‘Ÿπ‘œπ‘‘βˆ†π‘‘ Β  π‘ƒπ‘Ÿπ‘œπ‘‘=πœβˆ†πœƒβˆ†π‘‘ Β  βˆ†πœƒβˆ†π‘‘=πœ” Β  since
  • 10. Example:The rotating disk has a radius of 40 cm and a mass of 6 kg. Find the work and power needed in lifting the 2-kg mass 20 m above the ground in 4 s in uniform motion. s q F 6 kg 2 kg F=W s = 20 m
  • 11. Rotational Kinetic Energy v = wR m m4 w m3 m1 m2 axis Object rotating at constant w. Consider tiny mass m: 𝐾=12π‘šπ‘£2 Β  𝑣=πœ”π‘… Β  since 𝐾=12π‘šπœ”π‘…2 Β  𝐾=12π‘šπ‘…2πœ”2 Β  To find the total kinetic energy: π‘šπ‘…2=𝐼 Β  since 𝑲𝒓𝒐𝒕=πŸπŸπ‘°πŽπŸ Β  (Β½w2 same for all m )
  • 12. Recall for linear motion that the work done is equal to the change in linear kinetic energy: Using angular analogies, we find the rotational work is equal to the change in rotational kinetic energy: The Work-Energy Theorem
  • 13. F wo = 60 rad/s R = 0.30 m F = 40 N w R 4 kg Example:Applying the Work-Energy Theorem: What work is needed to stop wheel rotating in the figure below?

Editor's Notes

  1. I = 0.120 kg m2I = 0.0600 kg m2
  2. a = 100 rad/s2 = 12.5 rad or 1.99 rev
  3. a = 3.92m/s2
  4. Work = 392 JPower = 98 W