Moment of Inertia of a Small Sphere Revolving on a String

AI Thread Summary
The moment of inertia of a small sphere revolving on a string is calculated differently than for a large sphere. The correct formula for a point mass at a distance from an axis is I = mr^2, where r is the distance from the axis, which in this case is 1.2 m. The initial calculation using I = (2/5)mr^2 is inappropriate for this scenario since it applies to a solid sphere rotating about its center. The discussion highlights the importance of distinguishing between the moment of inertia for extended bodies and point masses. Understanding the application of the Parallel Axis Theorem is crucial for accurate calculations in this context.
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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