Angular Kinetic Energy and Momentum AP Physics

Work-Energy Theorem for Rotation End Slide 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

Work-Energy Theorem for Rotation End Slide 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

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

Carousel 𝟐𝟔𝟒𝟎=𝟒𝟔.𝟕𝟓 𝝎 𝒇 𝟐 𝑾=∆ 𝑲𝑬 𝒓 𝝎 𝒇 𝟐 =𝟓𝟔.𝟒𝟕 End Slide 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

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

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

Conserved Energy ∆ 𝑷𝑬 𝒈 +∆ 𝑲𝑬 𝒕 +∆ 𝑲𝑬 𝒓 =𝟎 A hollow sphere of mass 1.02 kg and radius 0.142 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

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 0.142 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

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

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

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