Rotational Inertia
Practice
practice problem 1
 the perpendicular bisector of a side
 a side
 a diagonal
 one corner on an axis perpendicular to the plane containing the masses
solution
In the first case, each of the four masses is a distance ½s from the axis. Thus…
I = ∑ r^{2}m I = 4 ⎛
⎜
⎝s ⎞^{2}
⎟
⎠m 2 I = ms^{2} In the second case, two of the masses are on the axis and contribute nothing to the moment of inertia. The other masses are each s away from the axis. Thus…
I = ∑r^{2}m
I = 2ms^{2}In the third case, two masses lie on the axis and two are half a diagonal away ½s√2 from the axis. Thus…
I = ∑ r^{2}m I = 2 ⎛
⎜
⎝s√2 ⎞^{2}
⎟
⎠2 I = ms^{2} In the fourth case, one mass lies on the axis, two masses are a distance s away, and one is a diagonal away s√2 from the axis. Thus…
I = ∑r^{2}m
I = 2ms^{2} + m(s√2)^{2}
I = 4ms^{2}
practice problem 2
solution
Answer it.
practice problem 3
solution
Answer it.
practice problem 4
 ring, hoop, cylindrical shell, thin pipe
 annulus, hollow cylinder, thick pipe
 disk, solid cylinder
 spherical shell
 hollow sphere
 solid sphere
 rod, rectangular plate (perpendicular bisector)
 rod, rectangular plate (axis along edge)
 rectangular plate, solid box (axis perpendicular to face)
 cube (axis perpendicular to face)
 cone (rotated about its central axis)
 cone (rotated about its vertex)
solution

ring, hoop, cylindrical shell, thin pipe
There isn't much of a proof here. Since all the mass is located the same distance R away from the axis of rotation, the moment of inertia is the same as that for a point mass located a distance R from the axis, namely…
I = ⌠
⌡r^{2} dm I = R^{2} ⌠
⌡dm which has a trivial solution…
I = MR^{2}
Note how the height of the hoop is not a factor. This formula would work equally well for a long thin tube or a flat thin ring.

annulus, hollow cylinder, thick pipe
A hollow cylinder is basically a series of infinitesimally thin nested cylindrical shells all added together. The way to write this in calculus is…
I = ⌠
⌡r^{2} dm The mass of each infinitesimal slice (dm) is the overall density (ρ) times the infinitesimal volume (dV) of the slice.
I = ⌠
⌡r^{2}ρ dV The infinitesimal volume is the surface area of a cylindrical shell (2πrh) times its infinitesimal thickness (dr).
I = ⌠
⌡r^{2}ρ2πrh dr The last piece of the puzzle is density, which is mass divided by volume.
I = ⌠
⎮
⌡r^{2} M 2πrh dr V The volume of a hollow cylinder is the volume of the outer cylinder minus the volume of the inner cylinder.
V = πR_{2}^{2}h − πR_{1}^{2}h V = π (R_{2}^{2} − R_{1}^{2}) h Putting it altogether and integrating from the inner radius (R_{1}) to the outer radius (R_{2}) yields…
I = ⌠
⎮
⌡r^{2} M 2πrh dr π (R_{2}^{2} − R_{1}^{2}) h R_{2} I = 2M ⌠
⎮
⌡r^{3} dr R_{2}^{2} − R_{1}^{2} R_{1} I = 2M R_{2}^{4} − R_{1}^{4} R_{2}^{2} − R_{1}^{2} 4 which simplifies to…
I = M(R_{2}^{2} + R_{1}^{2})(R_{2}^{2} − R_{1}^{2}) 2(R_{2}^{2} − R_{1}^{2}) and eventually simplifies to…
I = ^{1}_{2}M(R_{2}^{2} + R_{1}^{2})
Note how height cancelled out of this equation a few steps back. This formula would work for a long, thickwalled pipe or a flat, hollowed out disk (also known as an annulus).

disk, solid cylinder
A solid cylinder is a hollow cylinder with an inner radius of zero, so this proof is similar to the previous one. Start with the definition of the moment of inertia and substitute density times volume (ρ dV) for mass (dm).
I = ⌠
⌡r^{2} dm I = ⌠
⌡r^{2}ρ dV The infinitesimal volume is the surface area of a cylindrical shell (2πrh) times its infinitesimal thickness (dr). The density of a uniform cylinder is its total mass (M) divided by its total volume (πR^{2}h).
I = ⌠
⌡r^{2} ρ dV I = ⌠
⎮
⌡r^{2} M 2πrh dr πR^{2}h Now, integrate all the infinitesimal shells from r = 0 to r = R…
R I = 2M ⌠
⎮
⌡r^{3} dr = 2M R^{4} R^{2} R^{2} 4 0 and simplify…
I = ^{1}_{2}MR^{2}
Once again, height is not a factor affecting the moment of inertia of this shape. This formula would work for a long solid cylinder or a flat solid disk.

spherical shell
This is a tough proof. As always, start with the basic formula.
I = ⌠
⌡r^{2} dm I = ⌠
⌡r^{2}ρ dV Now the hard part. How do we slice this thing up? I recommend rings. Imagine the standard unit circle from trig class. Start on the x axis as is the usual way and walk counterclockwise across the circumference of the circle measuring and angle θ that starts at 0 radians and ends at π radians taking teeny, tiny dθ steps. (I'll use the x axis as the axis of rotation. I hope that's OK.) The radius of each ring is R sin θ, which means its circumference is 2πR sin θ. The width of one of these rings would be R dθ and its thickness would be something small. Something that will hopefully go away in the math we're about to start. Let's call it t. This gives us a volume element dV = (2πR sin θ)(R dθ)(t) and an integral…
I = ⌠
⌡(R sin θ)^{2}ρ(2πR sin θ Rdθ t) We're getting closer. Replace density with mass per volume. The volume of a spherical shell would equal the surface area of the shell (4πR^{2}) times its thickness (t).
I = ⌠
⎮
⌡(R sin θ)^{2} M (2πR sin θ R dθ t) V I = ⌠
⎮
⌡(R sin θ)^{2} M (2πR^{2}t sin θ dθ) 4πR^{2} t Simplify this beast. I beg you.
I = MR^{2} ⌠
⎮
⌡sin^{3} θ dθ 2 Wow! What happened to all the symbols? I'm telling you this algebra stuff is magic. Oops, I forgot the limits of integration. Let's put them in.
π I = MR^{2} ⌠
⎮
⌡sin^{3} θ dθ 2 0 Hmm, I don't quite know how to solve this one. May I suggest looking up the result in an integral table? Or maybe, perhaps, letting a machine do the work for you? If you tell this one to find the integral of (sin x)^{3} it will return something like this expression without the constants in the front or the limits at the end…
π I = MR^{2} ^{1}_{12} ⎡
⎢
⎣cos 3θ − 9 cos θ ⎤
⎥
⎦2 0 The limits of this integral are… well… something. I feel so lazy today after finding all these moments of inertia. Let me use another online source to calculate the upper limit…
and the lower limit…
of the quantity in the square bracket.
I = MR^{2} ^{1}_{12} ⎡
⎢
⎣(+8) − (−8) ⎤
⎥
⎦2 I = ^{16}_{24}MR^{2} I see the final answer approaching.
I = ^{2}_{3}MR^{2}
I am now officially happy.

hollow sphere
What is a hollow sphere but a series of spherical shells piled on top of one another. Do not use the basic formula.
do not use I = ⌠
⌡r^{2} dm do not use Start with something we just dervied a second ago — the moment of inertia of a spherical shell.
I_{spherical shell} = ^{2}_{3}MR^{2}
Break the hollow sphere up into a series of infinitesimal spherical shells and integrate these infinitesimal moments.
R_{2} I = ⌠
⎮
⌡^{2}_{3}r^{2} dm R_{1} Replace dm with ρ dV. Replace density with total mass (M) over total volume (^{4}_{3}π(R_{2}^{3} − R_{1}^{3})). Replace dV with the surface are of a sphere (4πr^{2}) times its infinitesimal thickness (dr).
R_{2} I = ⌠
⎮
⌡^{2}_{3}r^{2} ρ dV R_{1} R_{2} I = ⌠
⎮
⌡^{2}_{3}r^{2} M 4πr^{2} dr ^{4}_{3}π(R_{2}^{3} − R_{1}^{3}) R_{1} This can be simplified to…
R_{2} I = 2M ⌠
⎮
⌡r^{4} dr R_{2}^{3} − R_{1}^{3} R_{1} which certainly is simple to integrate.
R_{2} I = 2M ⎡
⎢
⎣r^{5} ⎤
⎥
⎦R_{2}^{3} − R_{1}^{3} 5 R_{1} Put the limits in…
I = 2M R_{2}^{5} − R_{1}^{5} R_{2}^{3} − R_{1}^{3} 5 and clean it up a bit.
I = ^{2}_{5}M R_{2}^{5} − R_{1}^{5} R_{2}^{3} − R_{1}^{3} This is as simple as I can make it.

solid sphere
You want an easy proof? What is a solid sphere but a hollow sphere with no inner radius. Start with the hollow sphere formula
I_{hollow sphere} = ^{2}_{5}M R_{2}^{5} − R_{1}^{5} R_{2}^{3} − R_{1}^{3} Let R_{2} = R and take the limit as R_{1} → 0
I = ^{2}_{5}M R^{5} R^{3} Simplify and we're done.
I = ^{2}_{5}MR^{2}
You want a harder proof? A solid sphere is built like an onion from layer upon layer of thin spherical shells. Each shell has moment of inertia equal to
I_{spherical shell} = ⌠
⌡^{2}_{3}r^{2}dm I_{spherical shell} = ⌠
⌡^{2}_{3}r^{2} ρ dV Again, density is total mass (M) divided by total volume (^{4}_{3}πR^{3}) and infinitesimal volume (dV) is the surface area of a spherical shell (4πr^{2}) times its infinitesimal thickness (dr). Substitute these values and simplify…
I = ⌠
⎮
⌡^{2}_{3}r^{2} M 4πr^{2} dr ^{4}_{3}πR^{3} I = 2M ⌠
⌡r^{4} dr Yet another simple integral…
R R I = 2M ⌠
⎮
⌡r^{4} dr = 2M ⎡
⎢
⎣r^{5} ⎤
⎥
⎦5 0 0 and it gives us the right answer…
I = ^{2}_{5}MR^{2}
Dare I try another proof? What is a solid sphere but a stack of disks.
I_{disk} = ⌠
⌡^{1}_{2}r^{2} dm = ⌠
⌡^{1}_{2}r^{2} ρ dV Review your analytical geometry. The formula for a circle is…
R^{2} = x^{2} + y^{2}
The disks of our sphere have radii (represented by the symbol y) that vary according to this formula.
y^{2} = R^{2} − x^{2}
Again, density is total mass (M) divided by total volume (^{4}_{3}πR^{3}), but now the infinitesimal volume (dV) is the surface area of a circular disk (πy^{2}) times its infinitesimal thickness (dx). Substitute, simplify, …
I = ⌠
⎮
⌡^{1}_{2}(R^{2} − x^{2}) M π(R^{2} − x^{2}) dx ^{4}_{3}πR^{3} I = 3M ⌠
⎮
⌡(R^{2} − x^{2})^{2} dx 8R and integrate. It's an ugly one. Viewer discretion is advised.
+R I = 3M ⌠
⎮
⌡(R^{2} − x^{2})^{2} dx 8R^{3} −R +R I = 3M ⎡
⎢
⎣x^{5} − 2R^{2}x^{3} + R^{4}x ⎤
⎥
⎦8R^{3} 5 3 −R All of the stuff in square brackets reduces to ^{16}_{15}R^{5}. Trust me. I've checked it several times.
I = 3M 16R^{5} 8R^{3} 15 One last bit of simplification and we're done.
I = ^{2}_{5}MR^{2}

rod, rectangular plate (perpendicular bisector)
Let M and L be the mass and length of the plate respectively. Then…
λ = M L is its linear density. Divide the rectangle up into thin strips that run parallel to the axis of rotation. The width of these strips, dx, times the linear density is the infinitesimal mass of each. Plop this into the moment of inertia formula and integrate from the left edge of the plate (−½L) to the right edge (+½L).
I = ⌠
⌡r^{2} dm I = ⌠
⌡x^{2} λ dx +½L I = ⌠
⎮
⌡x^{2} M dx L −½L +½L I = ⎡
⎢
⎣Mx^{3} ⎤
⎥
⎦3L −½L Stuff cancels, and with a minimal amount of work you end with…
I = ^{1}_{12}ML^{2}

rod, rectangular plate (axis along edge)
Use the same set up as in the previous proof. Integrate from the left edge of the plate to the right edge; that is, from 0 to L.
I = ⌠
⌡r^{2} dm I = ⌠
⌡x^{2} λ dx L I = ⌠
⎮
⌡x^{2} M dx L 0 L I = ⎡
⎢
⎣Mx^{3} ⎤
⎥
⎦3L 0 Easy peasy, here's the answer…
I = ^{1}_{3}ML^{2}
You could also try using the parallel axis theorem.
I = I_{cm} + ML^{2}
The moment of inertia about the center of mass was determined in the previous proof. Just add on a little correction and we're done.
I = ^{1}_{12}ML^{2} + M(^{1}_{2}L)^{2}
I = (^{1}_{12} + ^{1}_{4})ML^{2}This simplifies to the answer…
I = ^{1}_{3}ML^{2}

rectangular plate, solid box (axis perpendicular to face)
Start with the basic formula, but make one sup change. We'll replace the volume density (ρ = M/V) with surface density (σ = M/A) since the thickness of the plate doesn't contribute anything to the moment of inertia about this axis.
I = ⌠
⌡r^{2} dm I = ⌠
⌡r^{2} σ dA I = ⌠
⎮
⌡r^{2} M dA A Now let's dice the plate up into rectangular strips dx long by dy wide and any old height whatsoever.
I = ⌠⌠
⎮⎮
⌡⌡r^{2} M dx dy LW Since I like food preparation analogies, imagine we're slicing the plate up into infinitesimal french fries. Each french fry has coordinates (x, y) relative to the axis, which means their distances from the axis can be found using Pythagorean theorem.
r^{2} = x^{2} + y^{2}
Now, put everything altogether and set the limits of integration. For a plate of length L and width W, the appropriate limits would be ±½L and ±½W.
+½W +½L I = ⌠
⎮
⌡⌠
⎮
⌡(x^{2} + y^{2}) M dx dy LW −½W −½L Integrate first over x while y stays constant…
+½W +½L I = M ⌠
⎮
⌡⎡
⎢
⎣x^{3} + xy^{2} ⎤
⎥
⎦dy LW 3 −½W −½L +½W I = M ⌠
⎮
⌡⎛
⎜
⎝L^{3} + Ly^{2} ⎞
⎟
⎠dy LW 12 −½W then integrate over y…
+½W I = M ⎡
⎢
⎣L^{3}y + Ly^{3} ⎤
⎥
⎦LW 12 3 −½W I = M ⎛
⎜
⎝L^{3}W + LW^{3} ⎞
⎟
⎠LW 12 12 and simplify.
I = ^{1}_{12}M(L^{2} + W^{2})

cube (axis perpendicular to face)
A cube is a plate with length and width equal. Start with the results of the previous proof…
I = ^{1}_{12}M(L^{2} + W^{2})
and set L = W = S.
I = ^{1}_{12}M(S^{2} + S^{2})
C'est finis et voila!
I = ^{1}_{6}MS^{2}

cone (rotated about its central axis)
A cone is an infinite stack of infinitesimally thin disks of varying radius. If we add up the moments of inertia of all these very, very thin slices we'll get the moment of inertia of the whole cone. Adding up a lot of very small pieces to create a whole is called integration.
I = ⌠
⌡I_{slice} dx I = ⌠
⌡^{1}_{2}m_{slice}r^{2} dx Replace mass with density times volume and proceed.
I = ⌠
⌡^{1}_{2}ρA r^{2} dx I = ⌠
⌡^{1}_{2}ρ (πr^{2}) r^{2} dx I = ⌠
⌡^{1}_{2}πρr^{4} dx The "trick" to solving this part of the problem is determining how the radius of the slices vary from the vertex (x = 0) to the base (x = H). We need a function that begins at 0, ends at R, and increases linearly. May I suggest…
r = R x H Make the switch and integrate.
H I = ⌠
⎮
⌡^{1}_{2}πρ ⎛
⎜
⎝R x ⎞^{4}
⎠dx H 0 H I = πρR^{4} ⎡
⎢
⎣x^{5} ⎤
⎥
⎦2H^{4} 5 0 I = πρR^{4}H 10 Recall that the volume of a cone is…
V = ^{1}_{3}πR^{2}H
Do you see the volume hidden inside the moment of inertia? It's in there.
I = πρR^{4}H 10 I = ρ ⎛
⎝^{1}_{3}πR^{2} H ⎞
⎠⎛
⎝^{3}_{10}R^{2} ⎞
⎠I = ρV^{3}_{10}R^{2} Density times volume is mass. Therefore…
I = ^{3}_{10}MR^{2}

cone (rotated about its vertex)
Here's the answer…
I = ^{3}_{5}M(^{1}_{4}R^{2} + H^{2})
I'll leave it to the bold reader to work out the solution. I don't feel like writing solutions anymore.