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# Chapter13 rigid body rotation

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• I = 0.120 kg m2I = 0.0600 kg m2
• a = 3.92m/s2
• Work = 392 JPower = 98 W
• ### Chapter13 rigid body rotation

1. 1. 11B: Rigid Body Rotation<br />
2. 2. Inertia of Rotation<br />Consider Newton’s first law for the inertia of rotation to be patterned after the law for translation.<br />Moment of Inertia or Rotational Inertia is the rotational analog of mass.<br />Units of I: kg-m2 ; g-cm2 ; slug-ft2<br />
3. 3. Common Rotational Inertias<br />
4. 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.<br />I = 0.3 kg m2<br />I = 0.1 kg m2<br />hoop<br />R<br />R<br />Disk<br />
5. 5. Inertia of Rotation<br />ΣF = 20 N<br />a = 4m/s2<br />F = 20 N<br />R = 0.5 m<br />a = 2 rad/s2<br />Consider Newton’s second law for the inertia of rotation to be patterned after the law for translation.<br />Linear Inertia, m<br />Rotational Inertia, I<br />Force does for translation while torque does for rotation<br />
6. 6. Newton’s 2nd Law for Rotation<br />V<br />f<br />t<br />w<br />wo = 50 rad/s t = 40 N m<br />R<br />4 kg<br />Important Analogies<br />For many problems involving rotation, there is an analogy to be drawn from linear motion.<br />I<br />A resultant torque t produces angular acceleration a of disk with rotational inertia I.<br />A resultant force F produces negative acceleration a for a mass m.<br />
7. 7. F<br />wo = 50 rad/s<br />R = 0.20 m<br />F = 40 N<br />w<br />R<br />4 kg<br />Newton’s 2nd Law for Rotation<br />How many revolutions are required to stop?<br />
8. 8. Seatwork: ½ sheet of pad paper (crosswise)<br />Time allotted: 40 minutes <br />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<br /> <br />Fig. 1<br />
9. 9. Work and Power for Rotation<br />∆𝑠<br /> <br />∆𝜃<br /> <br />F<br />F<br />∆𝑠=𝑅∆𝜃<br /> <br />𝑊=𝐹∆𝑠<br /> <br />since<br /> <br />𝑅<br /> <br />𝐹𝑅=𝜏<br /> <br />𝑊=𝐹𝑅∆𝜃<br /> <br />since<br />∆𝑠=𝑅∆𝜃<br /> <br />𝑊𝑟𝑜𝑡=𝜏∆𝜃<br /> <br />𝑃𝑟𝑜𝑡=𝑊𝑟𝑜𝑡∆𝑡<br /> <br />𝑃𝑟𝑜𝑡=𝜏∆𝜃∆𝑡<br /> <br />∆𝜃∆𝑡=𝜔<br /> <br />since<br />
10. 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.<br />s<br />q<br />F<br />6 kg<br />2 kg<br />F=W<br />s = 20 m<br />
11. 11. Rotational Kinetic Energy<br />v = wR<br />m<br />m4<br />w<br />m3<br />m1<br />m2<br />axis<br />Object rotating at constant w.<br />Consider tiny mass m:<br />𝐾=12𝑚𝑣2<br /> <br />𝑣=𝜔𝑅<br /> <br />since<br />𝐾=12𝑚𝜔𝑅2<br /> <br />𝐾=12𝑚𝑅2𝜔2<br /> <br />To find the total kinetic energy:<br />𝑚𝑅2=𝐼<br /> <br />since<br />𝑲𝒓𝒐𝒕=𝟏𝟐𝑰𝝎𝟐<br /> <br />(½w2 same for all m )<br />
12. 12. Recall for linear motion that the work done is equal to the change in linear kinetic energy:<br />Using angular analogies, we find the rotational work is equal to the change in rotational kinetic energy:<br />The Work-Energy Theorem<br />
13. 13. F<br />wo = 60 rad/s<br />R = 0.30 m<br />F = 40 N<br />w<br />R<br />4 kg<br />Example:Applying the Work-Energy Theorem:<br />What work is needed to stop wheel rotating in the figure below?<br />