Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

- Physics 11 by M.T.H Corporation 363 views
- Physics Chapter 8- Rotational of a ... by Muhammad Solehin 8890 views
- Retibus socialibus et legibus momen... by Daniel Gayo Avello 4396 views
- Lecture Ch 08 by rtrujill 3820 views
- Newton's Second Law for Rotation by Worawarong Rakreu... 443 views
- Motion,moment of inertia,torque by Shomitro Kumar 193 views

3,658 views

Published on

No Downloads

Total views

3,658

On SlideShare

0

From Embeds

0

Number of Embeds

196

Shares

0

Downloads

196

Comments

0

Likes

1

No embeds

No notes for slide

- 1. 11B: Rigid Body Rotation<br />
- 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. Common Rotational Inertias<br />
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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 />

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment