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Calculation of Surface Velocity, Angular Momentum, and White Dwarf Angular Frequency, Exercises of Classical Mechanics

University of Debrecen Classical Mechanics

Solutions to three physics problems related to surface velocity, angular momentum, and white dwarf angular frequency. The solutions involve calculations using given data and physical concepts such as angular momentum theorem, conservation of angular momentum, and homogeneous spheres.

Typology: Exercises

2019/2020

Uploaded on 05/11/2020

nhuthaoly 🇭🇺

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29.1. What is the surface velocity of the line segment connecting the Earth and the Moon?

Solution:

Let us first calculate the area of the triangle created by 𝑟 and 𝑑𝑟:

∆𝐴=1

2|𝑟×∆𝑟| → 𝐴󰇗=𝑑𝐴

𝑑𝑡=lim

∆𝑡→01

2|𝑟×∆𝑟

∆𝑡|=1

2|𝑟×𝑣|,

or if we define the surface velocity as a vector-type quantity:

𝐴󰇗=1

2𝑟×𝑣.

𝐴󰇘=1

2𝑣×𝑣+1

2𝑟×𝑎=0, that is 𝐴󰇗=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.

Therefore we can get the surface velocity using the minimal value of 𝑟 and at the same

time the maximal value of 𝑣. These two vectors are perpendicular to each other, therefore

the magnitude of their vector (cross) product is 𝑟𝑚𝑖𝑛∙𝑣𝑚𝑎𝑥,

𝐴󰇗=𝑟𝑚𝑖𝑛∙𝑣𝑚𝑎𝑥

2.

𝑟𝑚𝑖𝑛=363 104 𝑘𝑚, 𝑣𝑚𝑎𝑥 =1.082 𝑘𝑚

𝑠 → 𝐴󰇗=196 439 𝑘𝑚2

𝑠≈1.96∙105 𝑘𝑚2

𝑠

29.2. What is the angular momentum of the reader respect to the center of the Earth? (The

radius of Earth is 6373 km.)

Solution:

𝑟

∆𝑟

𝑥

𝑧

𝑦

𝜔

𝛼

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29.1. What is the surface velocity of the line segment connecting the Earth and the Moon?

Solution:

Let us first calculate the area of the triangle created by 𝑟⃗ and 𝑑𝑟⃗:

|𝑟⃗ × ∆𝑟⃗| → 𝐴̇ = 𝑑𝐴

𝑑𝑡 = lim^ ∆𝑡→

2 |𝑟⃗ ×

|𝑟⃗ × 𝑣⃗|,

or if we define the surface velocity as a vector-type quantity:

𝐴⃗̇ =

2 𝑟⃗ × 𝑣⃗.

𝐴⃗̈ = 12 𝑣⃗ × 𝑣⃗ + 12 𝑟⃗ × 𝑎⃗ = 0, that is 𝐴⃗̇ = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡. Therefore we can get the surface velocity using the minimal value of 𝑟⃗ and at the same time the maximal value of 𝑣⃗. These two vectors are perpendicular to each other, therefore the magnitude of their vector (cross) product is 𝑟𝑚𝑖𝑛 ∙ 𝑣𝑚𝑎𝑥,

𝐴̇ =

𝑠 →^ 𝐴̇ = 196 439

𝑘𝑚^2

2 𝑠

29.2. What is the angular momentum of the reader respect to the center of the Earth? (The radius of Earth is 6373 km.)

Solution:

𝑅 = 6373𝑘𝑚 = 6.73 ∙ 10^6 𝑚 𝛼 = 47.3^0 cos 𝛼 = 0.

𝑇 =^

The definition of the angular momentum of a mass point with respect to a point of reference is

𝐿⃗⃗(𝑂) = 𝑟⃗ × 𝑚𝑣.⃗⃗⃗⃗

From this definition in our case we have got the following relation between 𝐿⃗⃗ and 𝜔⃗⃗⃗:

|𝐿⃗⃗| = √𝛩𝑥𝑧^2 + 𝛩𝑦𝑧^2 + 𝛩𝑧𝑧^2 𝜔𝑧,

where

𝛩𝑥𝑧 = −𝑚𝑥𝑧 𝛩𝑦𝑧 = −𝑚𝑦𝑧 𝛩𝑧𝑧 = 𝑚(𝑥^2 + 𝑦^2 ).

Using these relations:

|𝐿⃗⃗| = √𝛩𝑥𝑧^2 + 𝛩𝑦𝑧^2 + 𝛩𝑧𝑧^2 𝜔𝑧 = 𝑚𝜔𝑧√𝑥^2 𝑧^2 + 𝑦^2 𝑧^2 + (𝑥^2 + 𝑦^2 )^2 =

= 𝑚𝜔𝑧√(𝑥^2 + 𝑦^2 )𝑧^2 + (𝑥^2 + 𝑦^2 )^2 = 𝑚𝜔𝑧√(𝑅 cos 𝛼)^2 (𝑅 sin 𝛼)^2 + ((𝑅 𝑐𝑜𝑠 𝛼)^2 )^2 =

= 𝑚𝜔𝑧𝑅^2 √(cos 𝛼)^2 ((sin 𝛼)^2 + (𝑐𝑜𝑠 𝛼)^2 ) = 𝑚𝜔𝑧𝑅^2 cos 𝛼.

|𝐿⃗⃗| = 𝑚 ∙ 7.27 ∙ 10−5^1 𝑠 ∙ 6.37^2 ∙ 10^12 𝑚^2 ∙ 0.678 = 𝑚 ∙ 2.00 ∙ 10^9 𝑚

2 𝑠.

Be careful! The well known 𝐿(𝑧)𝑧^ = Θ(𝑧)𝜔𝑧 connection applies to the 𝑧-component of the angular momentum vector, here

Θ(𝑧)^ = 𝑚(𝑅 cos 𝛼)^2 ,

𝐿𝑧(𝑧)^ = Θ(𝑧)𝜔𝑧 = 𝑚𝜔𝑧𝑅^2 (cos 𝛼)^2 = 𝑚 ∙ 2.00 ∙ 10^9

𝑚^2

2 𝑠.

→ Θ𝑥𝑧 = −2 ∫ 𝜌𝐴(𝑟 cos 𝛼 cos 𝜑)

𝑙 02 (𝑟 sin 𝛼)𝑑𝑟 = = −2𝜌𝐴 cos 𝛼 cos 𝜑 sin 𝛼 ∙

( 2 𝑙)^3 3 = −^

𝑚 12 𝑙

(^2) cos 𝛼 cos 𝜑 sin 𝛼

𝑙^2 cos 𝛼 sin 𝜑 sin 𝛼

(^2) (cos 𝛼) 2

|𝐿⃗⃗| = √Θxz^2 + Θyz^2 + Θzz^2 𝜔𝑧 = 𝜔𝑧

(^2) √(cos 𝛼) (^2) (sin 𝛼) (^2) + (cos 𝛼) (^4) =

(^2) cos 𝛼.

29.8. Two 10 dkg balls with negligible size are attached to a rod with negligible mass and length of 20 cm. The center of this dumbbell is welded to a vertical axis in a 5 degree angle with respect to the horizontal axis. The 20 cm long vertical axis' endpoints are put in bearings, and than we rotate the whole body with an angular frequency of 400rad/s. What is the force acting on the bearings?

Solution:

𝑚 = 10𝑑𝑘𝑔 = 0.1𝑘𝑔 𝑙 = 20 𝑐𝑚 = 0.2𝑚 𝛼 = 5^0 𝜔 = 400

2 cos 𝛼 cos 𝜑 sin 𝛼

𝛩𝑦𝑧 = −2𝑚𝑦𝑧 = −2𝑚 ( 2 𝑙)

2 cos 𝛼 sin 𝜑 sin 𝛼 𝛩𝑧𝑧 = 2𝑚(𝑥^2 + 𝑦^2 ) = 2𝑚 ( 2 𝑙)

2 (cos 𝛼)^2.

The rate of change of angular momentum on the one hand is

𝑑𝐿⃗⃗ 𝑑𝑡 = 𝜔⃗⃗⃗ × 𝐿⃗⃗,

| = −𝛩𝑦𝑧𝜔𝑧^2 𝑖⃗ + 𝛩𝑥𝑧𝜔𝑧^2 𝑗⃗

on the other hand from the angular momentum theorem

𝑑𝐿⃗⃗ 𝑑𝑡 = −

2 (𝑘⃗⃗ × 𝐹⃗),

(The two bearing-forces are need to be equal in magnitude because of the equilibrium of the center of mass of the system.)

Calculation of Surface Velocity, Angular Momentum, and White Dwarf Angular Frequency, Exercises of Classical Mechanics

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Download Calculation of Surface Velocity, Angular Momentum, and White Dwarf Angular Frequency and more Exercises Classical Mechanics in PDF only on Docsity!

|𝑟⃗ × ∆𝑟⃗| → 𝐴̇ = 𝑑𝐴

2 |𝑟⃗ ×

|𝑟⃗ × 𝑣⃗|,

2 𝑟⃗ × 𝑣⃗.

𝑠 →^ 𝐴̇ = 196 439

𝑘𝑚^2

𝑇 =^

|𝐿⃗⃗| = √𝛩𝑥𝑧^2 + 𝛩𝑦𝑧^2 + 𝛩𝑧𝑧^2 𝜔𝑧,

|𝐿⃗⃗| = √𝛩𝑥𝑧^2 + 𝛩𝑦𝑧^2 + 𝛩𝑧𝑧^2 𝜔𝑧 = 𝑚𝜔𝑧√𝑥^2 𝑧^2 + 𝑦^2 𝑧^2 + (𝑥^2 + 𝑦^2 )^2 =

𝑚^2

𝑚 = 10𝑑𝑘𝑔 = 0.1𝑘𝑔 𝑙 = 20 𝑐𝑚 = 0.2𝑚 𝛼 = 5^0 𝜔 = 400

| = −𝛩𝑦𝑧𝜔𝑧^2 𝑖⃗ + 𝛩𝑥𝑧𝜔𝑧^2 𝑗⃗

2 (𝑘⃗⃗ × 𝐹⃗),

−𝛩𝑥𝑧𝜔𝑧^2

𝑙 𝐹𝑦^ =

−𝛩𝑦𝑧𝜔𝑧^2

|𝐹⃗| = √𝐹𝑥^2 + 𝐹𝑦^2 =

𝜔𝑧^2

𝜔𝑧^2

𝜔𝑧^2 𝑚𝑙

(400)^2

𝑠^2 ∙ 0.1𝑘𝑔 ∙ 0.2𝑚 ∙ 0.078 ∙ 0.997 =

𝜑^ 𝛼