Rotational Inertia

1. In the first case, each of the four masses is a distance ½s from the axis. Thus…

   I = ∑ r2m
   I = 4 ⎛ ⎜ ⎝ s ⎞2 ⎟ ⎠ m 2
   I = ms2

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 = ∑r2m I = 2ms2

3. In the third case, two masses lie on the axis and two are half a diagonal away ½s√2 from the axis. Thus…

   I = ∑ r2m
   I = 2 ⎛ ⎜ ⎝ s√2 ⎞2 ⎟ ⎠ 2
   I = ms2

4. 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 = ∑r2m I = 2ms2 + m(s√2)2 I = 4ms2

Answer it.

Answer it.

1. 

   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 =  ⌠ ⌡ r2 dm

   I =  R2 ⌠ ⌡ dm

   which has a trivial solution…

   I = MR²

   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.

2. 

   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 =

   ⌠ ⌡

   r2 dm

   The mass of each infinitesimal slice (dm) is the overall density (ρ) times the infinitesimal volume (dV) of the slice.

   I =

   ⌠ ⌡

   r2ρ dV

   The infinitesimal volume is the surface area of a cylindrical shell (2πrh) times its infinitesimal thickness (dr).

   I =

   ⌠ ⌡

   r2ρ2πrh dr

   The last piece of the puzzle is density, which is mass divided by volume.

   I =	⌠ ⎮ ⌡	r2	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 = πR22h − πR12h

   V = π (R22 − R12) h

   Putting it altogether and integrating from the inner radius (R₁) to the outer radius (R₂) yields…

   I =  ⌠ ⎮ ⌡ r2  M  2πrh dr π (R22 − R12) h

   R2 I =  2M ⌠ ⎮ ⌡ r3 dr R22 − R12 R1

   I =  2M   R24 − R14 R22 − R12 4

   which simplifies to…

   I =	M(R22 + R12)(R22 − R12)
   	2(R22 − R12)

   and eventually simplifies to…

   I = ¹₂M(R₂² + R₁²)

   Note how height cancelled out of this equation a few steps back. This formula would work for a long, thick-walled pipe or a flat, hollowed out disk (also known as an annulus).

3. 

   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 =  ⌠ ⌡ r2 dm

   I =  ⌠ ⌡ r2ρ 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 (πR²h).

   I =  ⌠ ⌡ r2 ρ dV

   I =  ⌠ ⎮ ⌡ r2  M  2πrh dr πR2h

   Now, integrate all the infinitesimal shells from r = 0 to r = R…

   		R
   I =	2M	⌠ ⎮ ⌡	r3 dr =	2M		R4
   	R2			R2		4
   		0

   and simplify…

   I = ¹₂MR²

   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.

4. 

   spherical shell

   This is a tough proof. As always, start with the basic formula.

   I =  ⌠ ⌡ r2 dm

   I =  ⌠ ⌡ r2ρ 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)2ρ(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 (4πR²) times its thickness (t).

   I =  ⌠ ⎮ ⌡ (R sin θ)2  M  (2πR sin θ R dθ t) V

   I =  ⌠ ⎮ ⌡ (R sin θ)2  M  (2πR2t sin θ dθ) 4πR2 t

   Simplify this beast. I beg you.

   I =	MR2	⌠ ⎮ ⌡	sin3 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 =	MR2	⌠ ⎮ ⌡	sin3 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 =	MR2	112	⎡ ⎢ ⎣	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…

   cos(3·π) − 9 cos(π) = +8

   and the lower limit…

   cos(3·0) − 9 cos(0) = −8

   of the quantity in the square bracket.

   I =  MR2  112 ⎡ ⎢ ⎣ (+8) − (−8) ⎤ ⎥ ⎦ 2

   I = 1624MR2

   I see the final answer approaching.

   I = ²₃MR²

   I am now officially happy.

5. 

   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 =

   ⌠ ⌡

   r2 dm

   do not use

   Start with something we just dervied a second ago — the moment of inertia of a spherical shell.

   I_{spherical shell} = ²₃MR²

   Break the hollow sphere up into a series of infinitesimal spherical shells and integrate these infinitesimal moments.

   	R2
   I =	⌠ ⎮ ⌡	23r2 dm
   	R1

   Replace dm with ρ dV. Replace density with total mass (M) over total volume (⁴₃π(R₂³ − R₁³)). Replace dV with the surface are of a sphere (4πr²) times its infinitesimal thickness (dr).

   R2 I =  ⌠ ⎮ ⌡ 23r2 ρ dV R1

   R2 I =  ⌠ ⎮ ⌡ 23r2  M  4πr2 dr 43π(R23 − R13) R1

   This can be simplified to…

   		R2
   I =	2M	⌠ ⎮ ⌡	r4 dr
   	R23 − R13
   		R1

   which certainly is simple to integrate.

   					R2
   I =	2M		⎡ ⎢ ⎣	r5	⎤ ⎥ ⎦
   	R23 − R13			5
   					R1

   Put the limits in…

   I =	2M		R25 − R15
   	R23 − R13		5

   and clean it up a bit.

   I = 25M  R25 − R15 R23 − R13

   This is as simple as I can make it.

6. 

   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 

   Ihollow sphere = 25M	R25 − R15
   	R23 − R13

   Let R₂ = R and take the limit as R₁ → 0

   I = 25M	R5
   	R3

   Simplify and we're done.

   I = ²₅MR²

   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 

   Ispherical shell =  ⌠ ⌡ 23r2dm

   Ispherical shell =  ⌠ ⌡ 23r2 ρ dV

   Again, density is total mass (M) divided by total volume (⁴₃πR³) and infinitesimal volume (dV) is the surface area of a spherical shell (4πr²) times its infinitesimal thickness (dr). Substitute these values and simplify…

   I =  ⌠ ⎮ ⌡ 23r2  M  4πr2 dr 43πR3

   I = 2M ⌠ ⌡ r4 dr

   Yet another simple integral…

   	R				R
   I = 2M	⌠ ⎮ ⌡	r4 dr = 2M	⎡ ⎢ ⎣	r5	⎤ ⎥ ⎦
   				5
   	0				0

   and it gives us the right answer…

   I = ²₅MR²

   Dare I try another proof? What is a solid sphere but a stack of disks.

   Idisk =

   ⌠ ⌡

   12r2 dm =

   ⌠ ⌡

   12r2 ρ dV

   Review your analytical geometry. The formula for a circle is…

   R² = x² + y²

   The disks of our sphere have radii (represented by the symbol y) that vary according to this formula.

   y² = R² − x²

   Again, density is total mass (M) divided by total volume (⁴₃πR³), but now the infinitesimal volume (dV) is the surface area of a circular disk (πy²) times its infinitesimal thickness (dx). Substitute, simplify, …

   I =  ⌠ ⎮ ⌡ 12(R2 − x2)  M  π(R2 − x2) dx 43πR3

   I =  3M ⌠ ⎮ ⌡ (R2 − x2)2 dx 8R

   and integrate. It's an ugly one. Viewer discretion is advised.

   +R I =  3M ⌠ ⎮ ⌡ (R2 − x2)2 dx 8R3 −R

   +R I =  3M   ⎡ ⎢ ⎣ x5  −  2R2x3  + R4x ⎤ ⎥ ⎦ 8R3 5 3 −R

   All of the stuff in square brackets reduces to ¹⁶₁₅R⁵. Trust me. I've checked it several times.

   I =	3M		16R5
   	8R3		15

   One last bit of simplification and we're done.

   I = ²₅MR²

7. 

   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 =  ⌠ ⌡ r2 dm

   I =  ⌠ ⌡ x2 λ dx

   +½L I =  ⌠ ⎮ ⌡ x2  M  dx L −½L

   +½L I =  ⎡ ⎢ ⎣ Mx3 ⎤ ⎥ ⎦ 3L −½L

   Stuff cancels, and with a minimal amount of work you end with…

   I = ¹₁₂ML²

8. 

   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 =  ⌠ ⌡ r2 dm

   I =  ⌠ ⌡ x2 λ dx

   L I =  ⌠ ⎮ ⌡ x2  M  dx L 0

   L I =  ⎡ ⎢ ⎣ Mx3 ⎤ ⎥ ⎦ 3L 0

   Easy peasy, here's the answer…

   I = ¹₃ML²

   You could also try using the parallel axis theorem.

   I = I_cm + ML²

   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 = 112ML2 + M(12L)2 I = (112 + 14)ML2

   This simplifies to the answer…

   I = ¹₃ML²

9. 

   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 =  ⌠ ⌡ r2 dm

   I =  ⌠ ⌡ r2 σ dA

   I =  ⌠ ⎮ ⌡ r2  M  dA A

   Now let's dice the plate up into rectangular strips dx long by dy wide and any old height whatsoever.

   I =	⌠⌠ ⎮⎮ ⌡⌡	r2	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.

   r² = x² + y²

   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 =	⌠ ⎮ ⌡	⌠ ⎮ ⌡	(x2 + y2)	M	dx dy
   				LW
   	−½W	−½L

   Integrate first over x while y stays constant…

   		+½W				+½L
   I =	M	⌠ ⎮ ⌡	⎡ ⎢ ⎣	x3	+ xy2	⎤ ⎥ ⎦	dy
   	LW			3
   		−½W				−½L

   		+½W
   I =	M	⌠ ⎮ ⌡	⎛ ⎜ ⎝	L3	+ Ly2	⎞ ⎟ ⎠	dy
   	LW			12
   		−½W

   then integrate over y…

   						+½W
   I =	M	⎡ ⎢ ⎣	L3y	+	Ly3	⎤ ⎥ ⎦
   	LW		12		3
   						−½W

   I =	M	⎛ ⎜ ⎝	L3W	+	LW3	⎞ ⎟ ⎠
   	LW		12		12

   and simplify.

   I = ¹₁₂M(L² + W²)

10. 

    cube (axis perpendicular to face)

    A cube is a plate with length and width equal. Start with the results of the previous proof…

    I = ¹₁₂M(L² + W²)

    and set L = W = S.

    I = ¹₁₂M(S² + S²)

    C'est finis et voila!

    I = ¹₆MS²

11. 

    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 =  ⌠ ⌡ Islice dx

    I =  ⌠ ⌡ 12mslicer2 dx

    Replace mass with density times volume and proceed.

    I =  ⌠ ⌡ 12ρA r2 dx

    I =  ⌠ ⌡ 12ρ (πr2) r2 dx

    I =  ⌠ ⌡ 12πρr4 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 =  ⌠ ⎮ ⌡ 12πρ ⎛ ⎜ ⎝ R  x  ⎞4 ⎠ dx H 0

    H I =  πρR4 ⎡ ⎢ ⎣ x5 ⎤ ⎥ ⎦ 2H4 5 0

    I =  πρR4H 10

    Recall that the volume of a cone is…

    V =  ¹₃πR²H

    Do you see the volume hidden inside the moment of inertia? It's in there.

    I =  πρR4H 10

    I = ρ ⎛ ⎝ 13πR2 H ⎞ ⎠ ⎛ ⎝ 310R2 ⎞ ⎠

    I = ρV310R2

    Density times volume is mass. Therefore…

    I = ³₁₀MR²

12. 

    cone (rotated about its vertex)

    Here's the answer…

    I = ³₅M(¹₄R² + H²)

    I'll leave it to the bold reader to work out the solution. I don't feel like writing solutions anymore.