Calculating Moment of Inertia for a Thin-Walled Spherical Object

In summary, you need to calculate the moment of inertia of a thin wall, the surface area of a ball of mass m and radii a. The attempt at a solution is to use spherical coordinates. The answer is I=\frac{m}{4\pi a^2}\cdot a^2\cdot d\theta d\phi.
  • #1
Karol
1,380
22

Homework Statement


How to calculat the moment of inertia of a thin wall, the surface area of a ball of mass m and radii a.

Homework Equations


I=mr2

The Attempt at a Solution


Spherical coordinates.

[tex]dm=\frac{m}{4\pi a^2}\cdot a^2\cdot d\theta d\phi[/tex]

See drawing.

[tex]I=\int_{\theta=0}^{2\pi} \int_{\phi=-\frac{\pi}{2}}^\frac{\pi}{2} dm\cdot a^2[/tex]
[tex]I=\int_{\theta=0}^{2\pi} \int_{\phi=-\frac{\pi}{2}}^\frac{\pi}{2} \frac{m}{4\pi a^2}\cdot a^2\cdot a^2 d\theta d\phi [/tex]
[tex]I=\frac{a^2m}{4\pi}\int_{\theta=0}^{2\pi} \int_{\phi=-\frac{\pi}{2}}^\frac{\pi}{2} d\theta d\phi [/tex]
[tex]I=\frac{2\pi m a^2}{4}[/tex]

The answer should be:

[tex]I=\frac{2ma^2}{3}[/tex]
 

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  • #3
I understand about the axis, but Why is the surface element sinφdθdφ, not dθdφ?
 
  • #4
Because the surface element from θ to θ+dθ and from φ to φ+dφ is a tiny rectangle whose sides are rdθ and rsinφdφ …

making an area of r2sinφdθdφ :smile:

(btw, mathematicians usually define θ and φ the other way round, so you'll see usually see sinθdθdφ instead :wink:)
 
  • #5
I understand the rdθ, but why the sinφ in rsinφdφ?
 
  • #6
Karol said:
I understand the rdθ, but why the sinφ in rsinφdφ?

oops! I got the sinφ in the wrong place … it should be rdφ and rsinφdθ. :redface:

(Because, at latitude φ, the circle of latitude has radius rsinφ, so a slight change from θ to θ+dθ only takes you a distance radius time angle = rsinφdθ …

so that tiny rectangle has sides rdφ and rsinφdθ)
 
  • #7
I understand, but then it should be rcosφdθ!
 
  • #8
We usually measure φ from the pole, ie 0 ≤ φ ≤ π, and then it's rsinφdθ … see http://en.wikipedia.org/wiki/Spherical_coordinates" :wink:

Only if you measure φ from the equator, ie -π/2 ≤ φ ≤ π/2 (not recommended), is it rcosφdθ.
 
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  • #9
Thanks a lot, you are a charming man (or woman...)
 
  • #10
no, fish! :biggrin:
 

FAQ: Calculating Moment of Inertia for a Thin-Walled Spherical Object

What is moment of inertia?

Moment of inertia is a measure of an object's resistance to changes in rotational motion. It is calculated based on an object's mass, shape, and distribution of mass around its axis of rotation.

How do you calculate moment of inertia?

Moment of inertia is calculated using the formula I = mr², where I is the moment of inertia, m is the mass of the object, and r is the distance between the object's axis of rotation and the mass. This formula can be applied to different shapes and objects, but may require integration for more complex shapes.

What are the units of moment of inertia?

The units of moment of inertia depend on the units used for mass and distance. In the SI system, moment of inertia is measured in kilograms per meter squared (kg/m²). In the US customary system, it is measured in slug-feet squared (slug*ft²).

How is moment of inertia related to rotational motion?

Moment of inertia is a crucial factor in rotational motion because it determines how much torque, or rotational force, is required to change an object's angular velocity. Objects with a higher moment of inertia will require more torque to change their rotational motion compared to objects with a lower moment of inertia.

How can moment of inertia be used in real-world applications?

Moment of inertia is used in a variety of real-world applications, such as designing structures and machines that require rotational motion, analyzing the stability of objects in motion, and understanding the behavior of rotating objects in physics and engineering. It is also important in sports, such as figure skating and diving, where athletes use their body's moment of inertia to control their rotational movements.

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