Chapter13 rigid body rotation
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1) Moment of inertia is the rotational analog of mass and describes an object's resistance to changes in its rotation. It depends on the object's mass and how it is distributed. 2) Newton's second law of rotation states that the torque applied to an object produces angular acceleration proportional to the torque and inversely proportional to the object's moment of inertia. 3) The work-energy theorem states that the work done on an object equals its change in kinetic energy. This applies to both linear and rotational motion, with rotational kinetic energy depending on the moment of inertia and angular velocity.
Chapter13 rigid body rotation
- 1. 11B: Rigid Body Rotation
- 2. Inertia of RotationConsider 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 m2I = 0.1 kg m2hoopRRDisk
- 5. Inertia of RotationΣF = 20 Na = 4m/s2F = 20 NR = 0.5 ma = 2 rad/s2Consider Newton’s second law for the inertia of rotation to be patterned after the law for translation.Linear Inertia, mRotational Inertia, IForce does for translation while torque does for rotation
- 6. Newton’s 2nd Law for RotationVftwwo = 50 rad/s t = 40 N mR4 kgImportant AnalogiesFor many problems involving rotation, there is an analogy to be drawn from linear motion.IA 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. Fwo = 50 rad/sR = 0.20 mF = 40 NwR4 kgNewton’s 2nd Law for RotationHow many revolutions are required to stop?
- 8. Seatwork: ½ sheet of pad paper (crosswise)Time allotted: 40 minutes A disk of mass 𝑀=6.00 kg and radius 𝑅=0.500 m is used to draw water from a well (Fig. 1). A bucket of mass 𝑚=2.00 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: 𝐼𝑑𝑖𝑠𝑘=12𝑀𝑅2 Fig. 1
- 9. Work and Power for Rotation∆𝑠 ∆𝜃 FF∆𝑠=𝑅∆𝜃 𝑊=𝐹∆𝑠 since 𝑅 𝐹𝑅=𝜏 𝑊=𝐹𝑅∆𝜃 since∆𝑠=𝑅∆𝜃 𝑊𝑟𝑜𝑡=𝜏∆𝜃 𝑃𝑟𝑜𝑡=𝑊𝑟𝑜𝑡∆𝑡 𝑃𝑟𝑜𝑡=𝜏∆𝜃∆𝑡 ∆𝜃∆𝑡=𝜔 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.sqF6 kg2 kgF=Ws = 20 m
- 11. Rotational Kinetic Energyv = wRmm4wm3m1m2axisObject rotating at constant w.Consider tiny mass m:𝐾=12𝑚𝑣2 𝑣=𝜔𝑅 since𝐾=12𝑚𝜔𝑅2 𝐾=12𝑚𝑅2𝜔2 To find the total kinetic energy:𝑚𝑅2=𝐼 since𝑲𝒓𝒐𝒕=𝟏𝟐𝑰𝝎𝟐 (½w2 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. Fwo = 60 rad/sR = 0.30 mF = 40 NwR4 kgExample:Applying the Work-Energy Theorem:What work is needed to stop wheel rotating in the figure below?