Moment of inertia of a solid sphere

In summary: This is why you cannot simply use the formula for a cylinder, which assumes all points are equidistant from the axis of rotation. Instead, you need to use the formula for a spherical shell, which takes into account the varying distances of the points from the axis. Your final answer should be (2/5)MR^2, as your book states. In summary, the conversation discusses finding the moment of inertia of a solid sphere of uniform mass density about an axis through its center. The attempt at a solution involves using the equation I = ∫rρdV and integrating from 0 to R using spherical shells, but this is incorrect as it assumes all points are equidistant from the axis of rotation. The correct
  • #1
henry3369
194
0

Homework Statement


Find the moment of inertia of a solid sphere of uniform mass density (like a billiard ball) about an axis through its center

Homework Equations


I = ∫rρdV

The Attempt at a Solution


I =ρ ∫r4πr2dr = ρ4π∫r4
Then I integrate this from 0 (the center) to R, so I = (ρ4π)*(R5/5)
And ρ = mv so ρ = M/(4/3)πR3 = 3M/4πR3. Put ρ into the equation for moment of inertia to get I = 3MR2/5.

My book tells me the answer is (2/5)MR^2. Where did I go wrong?
 
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  • #2
henry3369 said:

Homework Statement


Find the moment of inertia of a solid sphere of uniform mass density (like a billiard ball) about an axis through its center

Homework Equations


I = ∫rρdV

The Attempt at a Solution


I =ρ ∫r4πr2dr = ρ4π∫r4
Then I integrate this from 0 (the center) to R, so I = (ρ4π)*(R5/5)
And ρ = mv so ρ = M/(4/3)πR3 = 3M/4πR3. Put ρ into the equation for moment of inertia to get I = 3MR2/5.

My book tells me the answer is (2/5)MR^2. Where did I go wrong?

Can you explain how you are setting up the integral? What shape are you integrating from 0 to R?
 
  • #3
PeroK said:
Can you explain how you are setting up the integral? What shape are you integrating from 0 to R?
The surface area of a spherical shell from the center of the sphere to the outer shell of radius R.
 
  • #4
I figured it out. If I use spherical shells, the points from the shell aren't all the same distance form the axis of rotation. I was thinking about the distance from the center.
 
  • #5
henry3369 said:
I figured it out. If I use spherical shells, the points from the shell aren't all the same distance form the axis of rotation. I was thinking about the distance from the center.
Exactly!
 

Related to Moment of inertia of a solid sphere

1. What is the definition of moment of inertia of a solid sphere?

The moment of inertia of a solid sphere is a physical property that measures the resistance of the object to changes in its rotational motion. It is the sum of the products of each individual mass element in the object with the square of its distance from the axis of rotation.

2. How is moment of inertia of a solid sphere calculated?

The moment of inertia of a solid sphere can be calculated using the formula I = (2/5) * mr^2, where m is the mass of the sphere and r is the radius. This formula assumes that the object has a uniform mass distribution.

3. What are the units of moment of inertia?

The units of moment of inertia depend on the units of mass and distance used in the calculation. In the International System of Units (SI), the units of moment of inertia are kg*m^2.

4. How does the moment of inertia of a solid sphere affect its rotational motion?

The moment of inertia of a solid sphere determines how much torque is required to accelerate the object in rotational motion. A larger moment of inertia means that more torque is needed to produce the same angular acceleration compared to an object with a smaller moment of inertia.

5. Can the moment of inertia of a solid sphere change?

Yes, the moment of inertia of a solid sphere can change depending on the distribution of mass within the object. For example, if the mass is concentrated towards the outer edge of the sphere, the moment of inertia will be larger compared to a sphere with a more uniform mass distribution.

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