Video Transcript
Find the moment of inertia of a uniform solid sphere of radius 𝑎 and mass 𝑚 about an axis which lies at a distance of root two 𝑎 over two from the center of the sphere. Is it (a) nine 𝑚𝑎 squared over 10, (b) two 𝑚𝑎 squared over five, (c) seven 𝑚𝑎 squared over five, (d) 13𝑚𝑎 squared over 20, or (e) 𝑚𝑎 squared over two?
Let’s clear some space first, and let’s begin by drawing a simple diagram of the scenario. So we have a uniform solid sphere of radius 𝑎 and mass 𝑚. And it’s rotating about an axis that is root two 𝑎 over two from its center. So we’re finding the moment of inertia of this solid about an axis other than its principal axis, one that does not pass through its center of mass.
This axis is, however, parallel to the principal axis, so we can use the parallel axis theorem. This states that any moment of inertia 𝐼 about an axis parallel to the principal axis is given by the moment of inertia of the object about its principal axis, 𝐼 naught, plus the product of the object’s mass, 𝑚, and the square of the distance between the two axes, 𝑙 squared. So we just need to find the moment of inertia of the object about its principal axis, 𝐼 naught, and the rest is very straightforward.
The moment of inertia of a uniform solid sphere about an axis through its center is a well-known figure, but we’ll derive it from first principles. Let’s center the sphere on the origin of a spherical coordinate system, with the axis of rotation along the 𝑦-axis. A typical point in this coordinate system is a distance 𝑟 from the origin. The angle the line from the origin to the point makes with the 𝑥𝑧-plane is the point’s latitude 𝜙, which can be from zero at the north pole on the positive 𝑦-axis to 𝜋 at the south pole on the negative 𝑦-axis. The angle the point makes with the 𝑥-axis in the 𝑥𝑧-plane is the point’s longitude 𝜃, which varies from zero on the positive 𝑥-axis to 𝜋 on the negative 𝑥-axis and up to two 𝜋 back on the positive 𝑥-axis.
Note that the exact conventions on spherical coordinates may vary. But small differences, such as which axis to find the argument from, will not matter for these calculations. Consider a typical infinitesimal piece of this sphere. The volume element in spherical coordinates d𝑣 looks like this. It has depth d𝑟. And sweeping through a small angle d𝜃 and d𝜙 gives it a cross-sectional area of 𝑟 d𝜙 times 𝑟 sin 𝜙 d𝜃. This gives d𝑣 equals 𝑟 squared sin 𝜙 d𝑟 d𝜃 d𝜙. Again, exact conventions may vary. But as long as this volume element is consistent with the coordinate system you’ve used, this will not matter for the calculations.
The moment of inertia of this small piece of the sphere d𝐼 is given by its mass, let’s call this d𝑚, times the distance from the axis of rotation squared, let’s call this 𝑠 squared. We can reexpress d𝑚 as the density at this point 𝜌 times the volume d𝑣. Substituting this and our expression for d𝑣 into the equation for d𝐼, we get d𝐼 equals 𝑠 squared 𝜌𝑟 squared cos 𝜙 d𝑟 d𝜃 d𝜙. 𝑠 is in fact just the piece’s distance from the 𝑦-axis. This distance is sometimes denoted in spherical coordinates with big 𝑅 and may be given in terms of the standard spherical coordinates by little 𝑟 times sin of the latitude 𝜙. And substituting this into our equation for d𝐼 gives 𝜌𝑟 to the fourth sin cubed 𝜙 d𝑟 d𝜃 d𝜙.
To find the moment of inertia of the whole sphere, we need to integrate d𝐼 between the limits of the sphere’s boundaries. So 𝐼 is equal to the triple integral between 𝜙 one and 𝜙 two, 𝜃 one and 𝜃 two, and 𝑟 one and 𝑟 two of 𝜌𝑟 to the fourth sin cubed 𝜙 d𝑟 d𝜃 d𝜙.
Let’s plug in these limits now as this will greatly simplify the integration. Since the sphere is centered on the origin, the lower and upper limits of 𝑟 will just be zero and the radius the sphere 𝑎, respectively. The lower and upper limits of the longitude 𝜃 will just be zero and two 𝜋, respectively. And finally, the lower and upper limits of the latitude 𝜙 will be zero and 𝜋, respectively.
One last thing before we continue, this 𝜌 in the integration is in fact a constant since the sphere is uniform and is given by the sphere’s total mass 𝑚 divided by its total volume 𝑣. But let’s just call it 𝜌 for now and leave this till later. What this means, however, is that we can take 𝜌 outside of the integration. This gives us 𝐼 is equal to the triple integral between 𝜙 equals zero and 𝜋, 𝜃 equals zero and two 𝜋, and 𝑟 equals zero and 𝑎 of 𝜌𝑟 to the fourth sin cubed 𝜙 d𝑟 d𝜃 d𝜙.
We can go about this integration in any order that we choose, but some orders will be easier than others. As a rule of thumb, it’s generally easier to begin with the easiest integration, in other words, the variable with the least dependence. In this integrand, we can see that we have a dependence on 𝑟 to the fourth and sin cubed 𝜙, but no dependence at all on 𝜃. Let’s therefore integrate with respect to 𝜃 first, and let’s integrate with respect to 𝜙 second because trigonometric integrations over spherical coordinate systems generally lead to nice cancellations.
So this gives us 𝜌 times the integral between zero and 𝑎 d𝑟 times the integral between zero and 𝜋 d𝜙 times the integral between zero and two 𝜋𝑟 to the fourth sin cubed 𝜙 d𝜃. Integrating with respect to 𝜃, we get simply 𝜃 between the limits of zero and two 𝜋. So this evaluates to simply two 𝜋, which we can take outside the rest of the integration. So we now have two 𝜋𝜌 times the integral between zero and 𝑎 d𝑟 times the integral between zero and 𝜋𝑟 to the fourth sin cubed 𝜙 d𝜙.
Let’s clear a little more space before continuing. Now, we need to know how to integrate sin cubed 𝜙 with respect to 𝜙. When integrating any trigonometric function raised to a high power, remember your trigonometric identities. Since sin squared 𝜙 is identically equal to one minus cos squared 𝜙, we can rewrite this integral as sin 𝜙 times one minus cos squared 𝜙 d𝜙. Expanding these parentheses, we get sin 𝜙 minus sin 𝜙 cos squared 𝜙. Integrating sin 𝜙 will be easy. And for the second term, notice that negative sin 𝜙 is the derivative of cos 𝜙.
So by recognition and using the chain rule, we can see that this term is actually the derivative of one-third cos cubed 𝜙. So this gives the integral of sin cubed 𝜙 d𝜙 equal to negative cos 𝜙 plus one-third cos cubed 𝜙 plus 𝐶. So clearing some space and integrating with respect to 𝜙, we get 𝐼 equals two 𝜋𝜌 times the integral between zero and 𝑎 of negative cos 𝜙 plus one-third cos cubed 𝜙 evaluated between zero and 𝜋 times 𝑟 to the fourth d𝑟.
These trigonometric functions will simplify greatly because we are evaluating at zero and 𝜋. Evaluating, we get one minus one-third minus negative one plus one-third. Simplifying this and taking the constant outside the integration, we get eight-thirds 𝜋𝜌 times the integral between zero and 𝑎 of 𝑟 to the fourth d𝑟. And finally, integrating with respect to 𝑟 gives us one-fifth 𝑟 to the fifth.
Let’s clear some more space. And evaluating the integral gives us eight-thirds 𝜋𝜌 times one-fifth 𝑎 to the fifth, which further simplifies to eight over 15 times 𝜋𝜌𝑎 to the fifth. We now need to substitute in an expression for 𝜌 in terms of 𝑚 and 𝑎. Again, since the sphere is uniform, 𝜌 is just equal to the mass of the sphere 𝑚 divided by its volume 𝑣. The volume for the sphere is given by the standard formula for a spherical volume: four-thirds times 𝜋 times 𝑎 cubed. So this simplifies to 𝜌 equals three 𝑚 over four 𝜋𝑎 cubed.
Substituting this into our equation for 𝐼 gives us eight over 15 times 𝜋 times three 𝑚 over four 𝜋𝑎 cubed times 𝑎 to the fifth. Clearing some space, and we will now denote 𝐼 as 𝐼 naught, the moment of inertia about the principal axis, equal to two 𝑚𝑎 squared over five, which is indeed the standard formula for the moment of inertia of uniform solid sphere about its principal axis. Now it’s just a simple matter of using the parallel axis theorem.
So the moment of inertia about the given axis 𝐼 is equal to two 𝑚𝑎 squared over five plus the product of the sphere’s mass 𝑚 and its distance from the principal axis squared, which is root two 𝑎 over two all squared. Expanding the parentheses and simplifying gives us two 𝑚𝑎 squared over five plus 𝑚𝑎 squared over two. And finally, placing over a common denominator gives us 𝐼 equals nine 𝑚𝑎 squared over 10. Comparing this with our choice of answers, the answer is therefore (a) nine 𝑚𝑎 squared over 10.
