1. Angular Kinetic Energy and Momentum
AP Physics

2. Work-Energy Theorem for Rotation
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Work-Energy Theorem for Rotation
𝑾𝒐𝒓𝒌=𝑵𝒆𝒕 𝑭𝒐𝒓𝒄𝒆∗𝑫𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕
𝑭𝒐𝒓 𝒂 𝒓𝒐𝒕𝒂𝒕𝒊𝒏𝒈 𝒐𝒃𝒋𝒆𝒄𝒕,
𝒊𝒕 𝒘𝒊𝒍𝒍 𝒉𝒂𝒗𝒆 𝒂 𝒓𝒂𝒅𝒊𝒖𝒔
𝒇𝒓𝒐𝒎 𝒂𝒏 𝒂𝒙𝒊𝒔 𝒐𝒇 𝒓𝒐𝒕𝒂𝒕𝒊𝒐𝒏.
=𝚺𝑭∗ 𝒓 𝒓 ∗∆𝒙
𝑾=𝚺𝑭∗∆𝒙
𝚺𝑭∗𝒓=𝚺𝝉
𝑾=𝚺𝝉∗∆𝜽
∆𝒙 𝒓 =∆𝜽

3. Work-Energy Theorem for Rotation
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Work-Energy Theorem for Rotation
𝑨𝒍𝒔𝒐…
𝚺𝝉=𝑰𝜶
𝑾=𝚺𝝉∗∆𝜽
=𝑰𝜶∆𝜽
𝝎 𝒇 𝟐 = 𝝎 𝒐 𝟐 +𝟐𝜶∆𝜽
𝑾=𝑰 𝟏 𝟐 𝝎 𝒇 𝟐 − 𝟏 𝟐 𝝎 𝒐 𝟐
𝜶∆𝜽= 𝟏 𝟐 𝝎 𝒇 𝟐 − 𝟏 𝟐 𝝎 𝒐 𝟐
𝑾= 𝟏 𝟐 𝑰 𝝎 𝒇 𝟐 − 𝟏 𝟐 𝑰 𝝎 𝒐 𝟐
𝑲𝑬 𝒓 = 𝟏 𝟐 𝑰 𝝎 𝟐
𝑾= 𝑲𝑬 𝒓𝒇 − 𝑲𝑬 𝒓𝒐
𝑾=∆ 𝑲𝑬 𝒓

4. Work-Energy Theorem for Rotation
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Work-Energy Theorem for Rotation
𝑾=𝚺𝝉∗∆𝜽
= 𝟏 𝟐 𝑰 𝝎 𝒇 𝟐 − 𝝎 𝒐 𝟐
𝑾=∆ 𝑲𝑬 𝒓

5. Carousel 𝟐𝟔𝟒𝟎=𝟒𝟔.𝟕𝟓 𝝎 𝒇 𝟐 𝑾=∆ 𝑲𝑬 𝒓 𝝎 𝒇 𝟐 =𝟓𝟔.𝟒𝟕
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Carousel
A rope at the edge of a small carousel (𝑰=𝟗𝟑.𝟓 𝒌𝒈∙ 𝒎 𝟐 ) is pulling with an average force 𝟓𝟓𝟎 𝑵. Assuming the carousel started from rest, what is the angular velocity after 𝟒.𝟖 𝒎 has been pulled from the carousel?
𝟐𝟔𝟒𝟎=𝟒𝟔.𝟕𝟓 𝝎 𝒇 𝟐
𝑾=∆ 𝑲𝑬 𝒓
𝝎 𝒇 𝟐 =𝟓𝟔.𝟒𝟕
𝑭∗∆𝒙= 𝟏 𝟐 𝑰 𝝎 𝒇 𝟐 − 𝟏 𝟐 𝑰 𝝎 𝒐 𝟐
𝝎 𝒇 =𝟕.𝟓𝟏 𝒓𝒂𝒅 𝒔𝒆𝒄
𝟓𝟓𝟎∗𝟒.𝟖= 𝟏 𝟐 ∗𝟗𝟑.𝟓∗ 𝝎 𝒇 𝟐 −𝟎

6. End Slide
Grindstone
A grindstone (a solid disk) has a mass of 18.1 kg, a radius of 20.3 cm, and is rotating with an angular velocity of 𝟔.𝟑 𝒓𝒂𝒅 𝒔𝒆𝒄 . A person puts his hand on the stone to slow it to 𝟎.𝟕 𝒓𝒂𝒅 𝒔𝒆𝒄 .
What is the change in rotational kinetic energy of the stone?
∆ 𝑲𝑬 𝒓 = 𝟏 𝟐 𝑰 𝝎 𝒇 𝟐 − 𝝎 𝒐 𝟐
∆ 𝑲𝑬 𝒓 = 𝟏 𝟐 𝟎.𝟑𝟕𝟑 −𝟑𝟗.𝟐
∆ 𝑲𝑬 𝒓 = 𝟏 𝟐 𝟏 𝟐 𝒎 𝒓 𝟐 𝝎 𝒇 𝟐 − 𝝎 𝒐 𝟐
∆ 𝑲𝑬 𝒓 =−𝟕.𝟑𝟏 𝑱
∆ 𝑲𝑬 𝒓 = 𝟏 𝟐 𝟏 𝟐 ∗𝟏𝟖.𝟏∗ 𝟎.𝟐𝟎𝟑 𝟐 𝟎.𝟕 𝟐 − 𝟔.𝟑 𝟐

7. End Slide
Grindstone
A grindstone (a solid disk) has a mass of 18.1 kg, a radius of 20.3 cm, and is rotating with an angular velocity of 𝟔.𝟑 𝒓𝒂𝒅 𝒔𝒆𝒄 . A person puts his hand on the stone to slow it to 𝟎.𝟕 𝒓𝒂𝒅 𝒔𝒆𝒄 .
What is the force of friction applied if the stone made 2 rev during this change?
∗ 𝟐𝝅 𝒓𝒂𝒅 𝟏 𝒓𝒆𝒗
𝑾=𝚺𝝉∗∆𝜽=∆ 𝑲𝑬 𝒓
∆𝜽=𝟐 𝒓𝒆𝒗
=𝟏𝟐.𝟓𝟕 𝒓𝒂𝒅
𝑭 𝒇 ∗𝒓∗∆𝜽=∆ 𝑲𝑬 𝒓
𝑭 𝒇 = ∆ 𝑲𝑬 𝒓 𝒓∗∆𝜽
= −𝟕.𝟑𝟏 𝟎.𝟐𝟎𝟑∗𝟏𝟐.𝟓𝟕
𝑭 𝒇 =−𝟐.𝟖𝟕 𝑵

8. Conserved Energy ∆ 𝑷𝑬 𝒈 +∆ 𝑲𝑬 𝒕 +∆ 𝑲𝑬 𝒓 =𝟎
A hollow sphere of mass 1.02 kg and radius m starts from rest at the top of an incline. The height at the beginning of the track is 0.82 m above the end. How fast will the hoop be going at the end?
∆ 𝑷𝑬 𝒈 +∆ 𝑲𝑬 𝒕 +∆ 𝑲𝑬 𝒓 =𝟎
𝒎𝒈∆𝒉+ 𝟏 𝟐 𝒎 𝒗 𝒇 𝟐 − 𝒗 𝒐 𝟐 + 𝟏 𝟐 𝑰 𝝎 𝒇 𝟐 − 𝝎 𝒐 𝟐 =𝟎
𝒎𝒈∆𝒉+ 𝟏 𝟐 𝒎 𝒗 𝒇 𝟐 + 𝟏 𝟐 𝟐 𝟑 𝒎 𝒓 𝟐 𝒗 𝒇 𝒓 𝟐 =𝟎
𝒎𝒈∆𝒉+ 𝟏 𝟐 𝒎 𝒗 𝒇 𝟐 + 𝟏 𝟐 ∗ 𝟐 𝟑 𝒎 𝒓 𝟐 ∗ 𝒗 𝒇 𝟐 𝒓 𝟐 =𝟎

9. Conserved Energy 𝒎𝒈∆𝒉+ 𝟏 𝟐 𝒎 𝒗 𝒇 𝟐 + 𝟏 𝟑 𝒎 𝒗 𝒇 𝟐 =𝟎 𝟏 𝟐 + 𝟏 𝟑 = 𝟓 𝟔
End Slide
Conserved Energy
A hollow sphere of mass 1.02 kg and radius m starts from rest at the top of an incline. The height at the beginning of the track is 0.82 m above the end. How fast will the hoop be going at the end?
𝒎𝒈∆𝒉+ 𝟏 𝟐 𝒎 𝒗 𝒇 𝟐 + 𝟏 𝟑 𝒎 𝒗 𝒇 𝟐 =𝟎
𝟏 𝟐 + 𝟏 𝟑 = 𝟓 𝟔
𝒎𝒈∆𝒉+ 𝟓 𝟔 𝒎 𝒗 𝒇 𝟐 =𝟎
𝒗 𝒇 𝟐 = 𝟔 𝟓 ∗𝟖.𝟎𝟒
𝟗.𝟖∗−𝟎.𝟖𝟐+ 𝟓 𝟔 𝒗 𝒇 𝟐 =𝟎
𝒗 𝒇 = 𝟗.𝟔𝟒
−𝟖.𝟎𝟒+ 𝟓 𝟔 𝒗 𝒇 𝟐 =𝟎
𝒗 𝒇 =𝟑.𝟏𝟏 𝒎 𝒔

10. End Slide
Momentum for Rotation
Like other linear measurements (∆𝒙, 𝒗, 𝐚), momentum (p) of an object moving in a circle has an angular momentum (L).
Angular Momentum for a Particle: 𝑳=𝒑∗𝒓=𝒎𝒗𝒓
Angular Momentum for other types of masses have different distributions. Thus,
Also…
=𝒎∗𝒗∗𝒓∗ 𝒓 𝒓
=𝒎 𝒓 𝟐 ∗ 𝒗 𝒓
𝑳=𝒎∗𝒗∗𝒓
⟹ 𝑳=𝑰𝝎
𝝎= 𝒗 𝒓
𝚺𝑭∗𝒕=∆𝒑
𝚺𝝉∗𝒕=∆𝑳=𝑰 𝝎 𝒇 − 𝝎 𝒐
𝑰=𝒎 𝒓 𝟐
𝚺𝑭∗𝒕∗𝒓=∆𝒑∗𝒓

11. What is the angular momentum of Earth?
𝒎 𝑬 =𝟓.𝟗𝟖× 𝟏𝟎 𝟐𝟒 𝒌𝒈
𝒓 𝑬 =𝟔.𝟑𝟖× 𝟏𝟎 𝟔 𝒎
= 𝟏 𝒓𝒆𝒗 𝟐𝟑.𝟗𝟑 𝒉𝒓𝒔
∗ 𝟏 𝒉𝒓 𝟑𝟔𝟎𝟎 𝒔𝒆𝒄
∗ 𝟐𝝅 𝒓𝒂𝒅 𝟏 𝒓𝒆𝒗
=𝟕.𝟐𝟗× 𝟏𝟎 −𝟓 𝒓𝒂𝒅 𝒔𝒆𝒄
𝝎 𝑬
𝑰 𝑺𝒑𝒉𝒆𝒓𝒆 = 𝟐 𝟓 𝒎 𝑬 𝒓 𝑬 𝟐
= 𝟐 𝟓 𝟓.𝟗𝟖× 𝟏𝟎 𝟐𝟒 𝟔.𝟑𝟖× 𝟏𝟎 𝟔 𝟐
𝑰 𝑬 =𝟗.𝟕𝟒× 𝟏𝟎 𝟑𝟕 𝒌𝒈∙ 𝒎 𝟐

12. What is the angular momentum of Earth?
𝑳 𝑬 = 𝑰 𝑬 ∗ 𝝎 𝑬
𝑰 𝑬 =𝟗.𝟕𝟒× 𝟏𝟎 𝟑𝟕 𝒌𝒈∙ 𝒎 𝟐
𝑳 𝑬 =𝟗.𝟕𝟒× 𝟏𝟎 𝟑𝟕 ∗𝟕.𝟐𝟗× 𝟏𝟎 −𝟓
=𝟕.𝟐𝟗× 𝟏𝟎 −𝟓 𝒓𝒂𝒅 𝒔𝒆𝒄
𝝎 𝑬
𝑳 𝑬 =𝟕.𝟏𝟎× 𝟏𝟎 𝟑𝟑 𝒌𝒈∙ 𝒎 𝟐 𝒔