Calculate the angular momentum of a star of mass 2 ｘ10³⁰ kg and radius 7 ｘ 10⁵ Km. If it makes one complete rotation about its axis once in 20 days, what is its kinetic energy? (Ans: 14ｘ10⁴² J s,  2.5 ｘ10³⁶ J)



Data Given:


Mass of the star =m = 2 ｘ10³⁰ kg
 
Radius = r = 7 ｘ 10⁵ Km = 7 ｘ 10⁸ m

Time Period = T = 20 days = 20 x 24 x 60 x 60 s = 1728000 s = 1.728 ｘ 10⁶ s

To Find:


Angular Momentum  = `\vec L` = ?

Kinetic Energy  = K.E = ?

Solution:


Using formula for Angular momentum L in terms of angular velocity  ധ and Moment of inertia `\I`

`\vec L` = `\I`ധ  

Where ധ = `\frac {2ㅈ}{T}` and `\I` = `\frac {2MR^2}{5}`, Thus

`\vec L` = `\frac {4ㅈMR^2}{5T}`

By putting values

`\vec L` = `\frac {4 ｘ3.1416ｘ2 ｘ10³⁰ kg ｘ(7 ｘ 10⁸ m)²}{5 ｘ 1.728 ｘ 10⁶ s}`

`\vec L` = `\frac {25.1328 ｘ10³⁰ kg ｘ49 ｘ 10¹⁶ m²}{5 ｘ 1.728 ｘ 10⁶ s}`

`\vec L` = `\frac {1,231.5072 ｘ10⁴⁶ kg m²}{8.64ｘ 10⁶ s}`

`\vec L` = 142.535 ｘ10⁴⁰ kg m² s⁻¹  

or

`\vec L` = 1.425 ｘ10⁴² J s  -----------Ans.

Now To Find 
Kinetic Energy  = K.E = ? we know that 

K.E = `\vec L` = `\frac {1}{2}` `Iധ²`}

Where ധ = `\frac {2ㅈ}{T}` ⇒ ധ² = `\frac {4ㅈ²}{T²}` and `\I` = `\frac {2MR^2}{5}`, Thus

K.E = `\frac {4ㅈ²MR²}{5T²}`

By putting the corresponding values

K.E = `\frac {4(3.1416)²ｘ2 ｘ10³⁰ kg ｘ(7 ｘ 10⁸ m)²}{5 ｘ (1.728 ｘ 10⁶ s)²}`

K.E = `\frac {78.95720448 ｘ10³⁰ kg ｘ49 ｘ 10¹⁶ m²}{5 ｘ 2.985984ｘ 10¹² s²}` 

K.E = `\frac {3,868.90301952 ｘ10⁴⁶ kg m²}{14.92992 ｘ 10¹² s²}`

K.E = 259.137 ｘ10³⁴ kg m² s⁻² 

or

K.E = 2.59.137 ｘ10³⁶ J ------------Ans. (2)



************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.