Moment of Inertia of a Small Sphere Revolving on a String

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SUMMARY

The moment of inertia of a small sphere revolving on a string can be calculated using the formula I = mr², where m is the mass and r is the distance from the axis of rotation. In this case, a 2.0 kg sphere at the end of a 1.2 m string yields I = 2.9 kgm² when using the distance from the axis. The confusion arises from using the formula for a solid sphere, I = (2/5)mr², which is applicable only when calculating the moment of inertia about the sphere's center of mass. The correct approach for a point mass at a distance from the axis is to apply I = mr².

PREREQUISITES
  • Understanding of moment of inertia concepts
  • Familiarity with the Parallel Axis Theorem
  • Basic knowledge of rotational dynamics
  • Ability to perform calculations involving mass and distance
NEXT STEPS
  • Study the Parallel Axis Theorem in detail
  • Learn about the moment of inertia for various geometric shapes
  • Explore applications of moment of inertia in rotational motion problems
  • Investigate the differences between point masses and extended bodies in physics
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cc2hende
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Homework Statement


A small sphere of mass 2.0kg revolves at the end of a 1.2 m long string in a horizontal plane around a vertical axis. Determine its moment of inertia with respect to that axis.


Homework Equations



Moment of inertia for sphere= (2/5)mr^2

The Attempt at a Solution


I=(2/5)(2.0kg)(1.2m)^2
I=1.2 kgm^2

However, my book says to use the equation: I=mr^2. As a result they got I=2.9kgm^2. Why do they do this?? Isn't that the equation for Moment of inertia of a hollow cylinder?
 
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cc2hende said:

Homework Statement


A small sphere of mass 2.0kg revolves at the end of a 1.2 m long string in a horizontal plane around a vertical axis. Determine its moment of inertia with respect to that axis.


Homework Equations



Moment of inertia for sphere= (2/5)mr^2

The Attempt at a Solution


I=(2/5)(2.0kg)(1.2m)^2
I=1.2 kgm^2

However, my book says to use the equation: I=mr^2. As a result they got I=2.9kgm^2. Why do they do this?? Isn't that the equation for Moment of inertia of a hollow cylinder?

The sphere is small, 1.2 m is the radius of its orbit around an axis. You calculated the moment of inertia of a big sphere of radius 1.2 m.


Think of the definition of moment of inertia with respect an axis: for a very small body, a point mass, it is I=mr^2 where r is the distance of the body from the axis.

The moment of inertia of an extended body is the sum (integral) of the moment of inertia of its parts. Imagine you have a very small sphere with radius r=1 cm, for example, and it revolves around the axis of a circle of radius R=1.2 m. How far are the parts of the sphere from the axis? The closest point is 1.19 m, the farthest is at 1.21 m. So the moment of inertia must be close to mR2.
The moment of inertia of a sphere with respect to an axis going through its own centre is I(CM)=2/5 mr2. The Parallel Axis Theorem establishes the relation between the moment of inertia about an arbitrary axis and that about a parallel axis at distance R, going through the centre of mass of an object: I=I(CM)+mR2.


ehild
 

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