Physics 321 Hour 28 Extended Object: Angular Momentum and the Inertia Tensor

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Transcript Physics 321 Hour 28 Extended Object: Angular Momentum and the Inertia Tensor 

Physics 321
Hour 28
Extended Object: Angular Momentum and
the Inertia Tensor
Bottom Line
β€’ Angular Momentum and Angular Velocity
𝐿 = π‘šπœ”π‘Ÿ 2 βˆ’ π‘šπ‘Ÿ π‘Ÿ βˆ™ πœ”
β€’ Therefore 𝐿 = πΌπœ” is not always valid.
β€’ We can write 𝐿 = πˆπœ” where 𝐈 is the inertia tensor:
𝐼π‘₯π‘₯
𝐈 = 𝐼π‘₯𝑦
𝐼π‘₯𝑧
𝐼π‘₯𝑦
𝐼𝑦𝑦
𝐼𝑦𝑧
𝐼π‘₯𝑧
𝐽𝑦𝑦 + 𝐽𝑧𝑧
𝐼𝑦𝑧 =
βˆ’π½π‘₯𝑦
𝐼𝑧𝑧
βˆ’π½π‘₯𝑧
𝐽π‘₯𝑦 =
βˆ’π½π‘₯𝑦
𝐽𝑧𝑧 + 𝐽π‘₯π‘₯
βˆ’π½π‘¦π‘§
π‘₯π‘¦πœŒπ‘‘π‘‰
βˆ’π½π‘₯𝑧
βˆ’π½π‘¦π‘§
𝐽π‘₯π‘₯ + 𝐽𝑦𝑦
Center of Mass
β€’ Center of mass
π‘Ÿβ€²
cm
𝑀𝑅 =
π‘Ÿ
𝑅
=
π‘Ÿπ‘‘π‘š =
βˆ†π‘šπ‘– π‘Ÿπ‘–
π‘ŸπœŒπ‘‘π‘‰
β€’ A useful result:
𝑀𝑅 = π‘ŸπœŒπ‘‘π‘‰β€² = π‘…πœŒπ‘‘π‘‰β€²+ π‘Ÿβ€²πœŒπ‘‘π‘‰β€² = 𝑀𝑅 + π‘Ÿβ€²πœŒπ‘‘π‘‰β€²
β†’
π‘Ÿβ€²πœŒπ‘‘π‘‰β€² = 0
More Conclusions
π‘π‘‘π‘œπ‘‘π‘Žπ‘™ = 𝑀𝑅
πΏπ‘‘π‘œπ‘‘π‘Žπ‘™
πΉπ‘‘π‘œπ‘‘π‘Žπ‘™ = 𝑀𝑅
= πΏπ‘π‘š + πΏπ‘Žπ‘π‘œπ‘’π‘‘ π‘π‘š
A Little Math
β€’ Angular Momentum and Angular Velocity
𝐿 = π‘Ÿ × π‘ = π‘šπ‘Ÿ × π‘£ = π‘šπ‘Ÿ × π‘£βŠ₯
where 𝑣βŠ₯ is the component of 𝑣 perpendicular to π‘Ÿ
= π‘šπ‘Ÿ × πœ” × π‘Ÿ
= π‘šπœ”π‘Ÿ 2 βˆ’ π‘šπ‘Ÿ π‘Ÿ βˆ™ πœ”
βˆ†π‘ŸβŠ₯ βˆ†π‘Ÿ
βˆ†πœ—
π‘Ÿ
βˆ†π‘ŸβŠ₯ = π‘Ÿβˆ†πœ—
βˆ†π‘ŸβŠ₯ π‘Ÿβˆ†πœ—
=
βˆ†π‘‘
βˆ†π‘‘
𝑣βŠ₯= π‘Ÿπœ”
𝑣βŠ₯ = πœ” × π‘Ÿ
A Little Math
In the instantaneous rotation of a solid body about an
axis, 𝑣 is always perpendicular to π‘Ÿ so 𝑣βŠ₯ = 𝑣.
A Little Math
β€’ Angular Momentum and Angular Velocity
𝐿 = π‘Ÿ × π‘ = π‘šπ‘Ÿ × πœ” × π‘Ÿ
= π‘šπœ”π‘Ÿ 2 βˆ’ π‘šπ‘Ÿ π‘Ÿ βˆ™ πœ”
β€’ Conclusions:
β€’ 𝐿 is perpendicular to π‘Ÿ, but not necessarily parallel
to πœ”.
β€’ Therefore 𝐿 = πΌπœ” is not always valid.
β€’ If π‘Ÿ is perpendicular to πœ”:
π‘Ÿ×𝑣
πœ”= 2
π‘Ÿ
An Example
y
x
Find vectors about the origin:
πœ” = πœ”π‘§ π‘Ÿ = βˆ’π‘Žπ‘¦
𝑣 = πœ” × π‘Ÿ = π‘Žπœ”π‘₯
𝐿 = π‘Ÿ × π‘šπ‘£ = π‘šπ‘Ž2 πœ”π‘§
𝐿 = πΌπœ” 𝐼 = π‘šπ‘Ž2
Ξ“=π‘Ÿ×𝐹 =0
y
Another Example
x
Find vectors about the origin:
πœ” = πœ” 𝑧 π‘Ÿ = 𝑧𝑧 βˆ’ π‘Ž 𝑦
𝑣 = πœ” × π‘Ÿ = π‘Žπœ”π‘₯
𝐿 = π‘Ÿ × π‘šπ‘£ = π‘šπ‘Ž2 πœ”π‘§ + π‘šπ‘Žπ‘§πœ”π‘¦
𝐿 β‰  πΌπœ”
Ξ“=π‘Ÿ×𝐹 β‰ 0
Cross Products and the Antisymmetric Tensor
𝐢 =𝐴×𝐡
3
𝐢𝑖 =
πœ–π‘–π‘—π‘˜ 𝐴𝑗 π΅π‘˜
𝑗,π‘˜=1
πœ–π‘–π‘—π‘˜
0 𝑖𝑓 π‘Žπ‘›π‘¦ 𝑖𝑛𝑑𝑒π‘₯ 𝑖𝑠 π‘Ÿπ‘’π‘π‘’π‘Žπ‘‘π‘’π‘‘
+1 𝑖𝑓 π‘–π‘—π‘˜ 𝑖𝑠 𝑐𝑦𝑐𝑙𝑖𝑐
=
βˆ’1 𝑖𝑓 π‘–π‘—π‘˜ 𝑖𝑠 π‘Žπ‘›π‘‘π‘–π‘π‘¦π‘π‘™π‘–π‘
Angular Momentum of a Point Mass II
𝐿 = π‘Ÿ × π‘ = π‘šπ‘Ÿ × πœ” × π‘Ÿ
3
𝐿𝑖 =
πœ–π‘–π‘—π‘˜ π‘šπ‘Ÿπ‘— πœ” × π‘Ÿ
π‘˜
𝑗,π‘˜=1
3
=
3
πœ–π‘–π‘—π‘˜ π‘šπ‘Ÿπ‘—
𝑗,π‘˜=1
πœ–π‘˜π‘—β€²π‘˜β€² πœ”π‘—β€² π‘Ÿπ‘˜β€²
𝑗′,π‘˜β€²=1
𝐿𝑖 =
πœ–π‘–π‘—π‘˜ πœ–π‘˜π‘—β€²π‘˜β€² π‘šπ‘Ÿπ‘— π‘Ÿπ‘˜β€² πœ”π‘—β€²
𝑗′
𝑗,π‘˜,π‘˜β€²
πˆπ‘–π‘—β€²
The Inertia Tensor of a Point Mass I
𝐿 = πˆπœ”
πˆπ‘–π‘—β€² =
πœ–π‘–π‘—π‘˜ πœ–π‘˜π‘—β€²π‘˜β€² π‘šπ‘Ÿπ‘— π‘Ÿπ‘˜β€²
𝑗,π‘˜,π‘˜β€²
𝐈12 =
πœ–1π‘—π‘˜ πœ–π‘˜2π‘˜β€² π‘šπ‘Ÿπ‘— π‘Ÿπ‘˜β€² = πœ–123 πœ–321 π‘šπ‘Ÿ2 π‘Ÿ1 = βˆ’π‘šπ‘¦π‘₯
𝑗,π‘˜,π‘˜β€²
𝐈11 =
πœ–1π‘—π‘˜ πœ–π‘˜1π‘˜β€² π‘šπ‘Ÿπ‘— π‘Ÿπ‘˜β€² = πœ–132 πœ–213 π‘šπ‘Ÿ3 π‘Ÿ3
𝑗,π‘˜,π‘˜β€²
πœ–123 πœ–312 π‘šπ‘Ÿ2 π‘Ÿ2 = π‘šπ‘¦ 2 + π‘šπ‘§ 2 = π‘š π‘Ÿ 2 βˆ’ π‘₯ 2
The Inertia Tensor of a Point Mass II
𝐿 = πˆπœ”
π‘š(π‘Ÿ 2 βˆ’ π‘₯ 2 )
𝐈=
βˆ’π‘šπ‘¦π‘₯
βˆ’π‘šπ‘§π‘₯
βˆ’π‘šπ‘₯𝑦
π‘š(π‘Ÿ 2 βˆ’ 𝑦 2 )
βˆ’π‘šπ‘§π‘¦
βˆ’π‘šπ‘₯𝑧
βˆ’π‘šπ‘¦π‘§
π‘š(π‘Ÿ 2 βˆ’ 𝑧 2 )
The Inertia Tensor of an Extended Object
𝐿 = πˆπœ”
𝐽π‘₯𝑦 =
𝐼π‘₯π‘₯
𝐈 = 𝐼π‘₯𝑦
𝐼π‘₯𝑧
𝐼π‘₯𝑦
𝐼𝑦𝑦
𝐼𝑦𝑧
π‘₯π‘¦πœŒπ‘‘π‘‰
𝐼π‘₯𝑧
𝐽𝑦𝑦 + 𝐽𝑧𝑧
𝐼𝑦𝑧 =
βˆ’π½π‘₯𝑦
𝐼𝑧𝑧
βˆ’π½π‘₯𝑧
1
𝑇 = πœ”βˆ™πˆβˆ™πœ”
2
βˆ’π½π‘₯𝑦
𝐽𝑧𝑧 + 𝐽π‘₯π‘₯
βˆ’π½π‘¦π‘§
βˆ’π½π‘₯𝑧
βˆ’π½π‘¦π‘§
𝐽π‘₯π‘₯ + 𝐽𝑦𝑦