Moment of Inertia

Calculating moment of inertia using integration, parallel axis theorem

Moment of Inertia

Definition

For discrete point masses: I=โˆ‘imiri2I = \sum_i m_ir_i^2

where rir_i is perpendicular distance from rotation axis.

For continuous mass distribution: I=โˆซr2โ€‰dmI = \int r^2 \, dm

where rr is perpendicular distance from axis.

Common Moments of Inertia

Thin rod about center: I=112ML2I = \frac{1}{12}ML^2

Thin rod about end: I=13ML2I = \frac{1}{3}ML^2

Solid cylinder/disk about axis: I=12MR2I = \frac{1}{2}MR^2

Hollow cylinder about axis: I=MR2I = MR^2

Solid sphere about diameter: I=25MR2I = \frac{2}{5}MR^2

Hollow sphere about diameter: I=23MR2I = \frac{2}{3}MR^2

Rectangular plate about center: I=112M(a2+b2)I = \frac{1}{12}M(a^2 + b^2)

Calculating I by Integration

Example 1: Thin Rod About End

Rod of length LL, mass MM, uniform density.

Linear mass density: ฮป=M/L\lambda = M/L

dm=ฮปโ€‰dx=MLโ€‰dxdm = \lambda \, dx = \frac{M}{L} \, dx

I=โˆซ0Lx2โ€‰dm=โˆซ0Lx2MLโ€‰dxI = \int_0^L x^2 \, dm = \int_0^L x^2 \frac{M}{L} \, dx

I=MLx33โˆฃ0L=ML23I = \frac{M}{L}\frac{x^3}{3}\Big|_0^L = \frac{ML^2}{3}

Example 2: Solid Cylinder About Axis

Radius RR, mass MM, uniform density ฯ\rho.

Use cylindrical shells: dm=ฯโ‹…2ฯ€rโ€‰drโ‹…hdm = \rho \cdot 2\pi r \, dr \cdot h

I=โˆซ0Rr2โ€‰dm=โˆซ0Rr2โ‹…ฯโ‹…2ฯ€rhโ€‰drI = \int_0^R r^2 \, dm = \int_0^R r^2 \cdot \rho \cdot 2\pi rh \, dr

I=2ฯ€ฯhโˆซ0Rr3โ€‰dr=2ฯ€ฯhR44I = 2\pi\rho h\int_0^R r^3 \, dr = 2\pi\rho h \frac{R^4}{4}

Using M=ฯฯ€R2hM = \rho\pi R^2h:

I=12MR2I = \frac{1}{2}MR^2

Example 3: Solid Sphere About Diameter

Sphere of radius RR, mass MM.

Use disk method. At distance zz from center, disk has radius r=R2โˆ’z2r = \sqrt{R^2 - z^2}.

dI=12(dm)r2=12ฯฯ€r2dzโ‹…r2=12ฯฯ€(R2โˆ’z2)2dzdI = \frac{1}{2}(dm)r^2 = \frac{1}{2}\rho\pi r^2 dz \cdot r^2 = \frac{1}{2}\rho\pi(R^2 - z^2)^2 dz

I=2โˆซ0R12ฯฯ€(R2โˆ’z2)2dzI = 2\int_0^R \frac{1}{2}\rho\pi(R^2 - z^2)^2 dz

After integration: I=25MR2I = \frac{2}{5}MR^2

Parallel Axis Theorem

I=Icm+Md2I = I_{cm} + Md^2

where:

  • II = moment about new axis
  • IcmI_{cm} = moment about parallel axis through center of mass
  • dd = distance between the two parallel axes

Example: Rod About End

Icm=112ML2I_{cm} = \frac{1}{12}ML^2 (about center)

d=L/2d = L/2 (distance from center to end)

I=112ML2+M(L2)2=112ML2+14ML2=13ML2I = \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1}{3}ML^2

Perpendicular Axis Theorem

For planar object in xy-plane: Iz=Ix+IyI_z = I_x + I_y

where axes pass through same point.

Example: Thin Disk

About axis perpendicular to disk through center: Iz=12MR2I_z = \frac{1}{2}MR^2

By symmetry: Ix=IyI_x = I_y

Ix=Iy=12Iz=14MR2I_x = I_y = \frac{1}{2}I_z = \frac{1}{4}MR^2

(moment about diameter)

Composite Objects

For object composed of multiple parts: Itotal=โˆ‘iIiI_{total} = \sum_i I_i

Calculate moment of each part (using parallel axis if needed), then sum.

Example: T-Shape

Two identical rods (length LL, mass MM each) forming T-shape.

Vertical rod rotating about its end (where horizontal rod attaches): I1=13ML2I_1 = \frac{1}{3}ML^2

Horizontal rod about its center (perpendicular to length): I2=112ML2I_2 = \frac{1}{12}ML^2

Itotal=13ML2+112ML2=512ML2I_{total} = \frac{1}{3}ML^2 + \frac{1}{12}ML^2 = \frac{5}{12}ML^2

Radius of Gyration

I=Mk2I = Mk^2

where kk is radius of gyration.

kk represents the distance from axis where all mass could be concentrated to give same II.

๐Ÿ“š Practice Problems

1Problem 1medium

โ“ Question:

Calculate the moment of inertia of a thin uniform rod (mass M = 3.0 kg, length L = 2.0 m) about an axis: (a) through the center perpendicular to the rod, (b) through one end perpendicular to the rod, and (c) verify the parallel axis theorem.

๐Ÿ’ก Show Solution

Given:

  • M = 3.0 kg
  • L = 2.0 m

(a) Through center:

Icenter=112ML2=112(3.0)(2.0)2I_{center} = \frac{1}{12}ML^2 = \frac{1}{12}(3.0)(2.0)^2

Icenter=1212I_{center} = \frac{12}{12}

Icenter=1.0ย kg\cdotpm2\boxed{I_{center} = 1.0 \text{ kgยทm}^2}

(b) Through one end:

Iend=13ML2=13(3.0)(2.0)2I_{end} = \frac{1}{3}ML^2 = \frac{1}{3}(3.0)(2.0)^2

Iend=4.0ย kg\cdotpm2\boxed{I_{end} = 4.0 \text{ kgยทm}^2}

(c) Parallel axis theorem verification:

Parallel axis theorem: I=Icm+Md2I = I_{cm} + Md^2

Distance from center to end: d=L/2=1.0d = L/2 = 1.0 m

Iend=Icenter+M(L2)2I_{end} = I_{center} + M\left(\frac{L}{2}\right)^2

Iend=1.0+(3.0)(1.0)2=1.0+3.0I_{end} = 1.0 + (3.0)(1.0)^2 = 1.0 + 3.0

Iend=4.0ย kg\cdotpm2I_{end} = 4.0 \text{ kgยทm}^2 โœ“

Verified!

2Problem 2hard

โ“ Question:

A thin spherical shell (mass M = 2.0 kg, radius R = 0.5 m) and a solid sphere (same M and R) roll down an incline (ฮธ = 30ยฐ) without slipping. Find: (a) the acceleration of each object, (b) which reaches the bottom first, and (c) the ratio of their speeds at the bottom.

๐Ÿ’ก Show Solution

Given:

  • M = 2.0 kg, R = 0.5 m
  • ฮธ = 30ยฐ
  • Rolling without slipping

(a) Acceleration of each:

For rolling without slipping: a=gsinโกฮธ1+I/(MR2)a = \frac{g\sin\theta}{1 + I/(MR^2)}

Thin spherical shell: I=23MR2I = \frac{2}{3}MR^2

ashell=gsinโกฮธ1+2/3=gsinโกฮธ5/3=3gsinโกฮธ5a_{shell} = \frac{g\sin\theta}{1 + 2/3} = \frac{g\sin\theta}{5/3} = \frac{3g\sin\theta}{5}

ashell=3(9.8)sinโก(30ยฐ)5=3(9.8)(0.5)5a_{shell} = \frac{3(9.8)\sin(30ยฐ)}{5} = \frac{3(9.8)(0.5)}{5}

ashell=2.94ย m/s2\boxed{a_{shell} = 2.94 \text{ m/s}^2}

Solid sphere: I=25MR2I = \frac{2}{5}MR^2

asphere=gsinโกฮธ1+2/5=gsinโกฮธ7/5=5gsinโกฮธ7a_{sphere} = \frac{g\sin\theta}{1 + 2/5} = \frac{g\sin\theta}{7/5} = \frac{5g\sin\theta}{7}

asphere=5(9.8)(0.5)7a_{sphere} = \frac{5(9.8)(0.5)}{7}

asphere=3.50ย m/s2\boxed{a_{sphere} = 3.50 \text{ m/s}^2}

(b) Which reaches bottom first?

Since asphere>ashella_{sphere} > a_{shell}:

Solidย sphereย reachesย bottomย first\boxed{\text{Solid sphere reaches bottom first}}

(Less rotational inertia = faster)

(c) Ratio of speeds:

For same distance L down incline: v2=2aLv^2 = 2aL

vspherevshell=asphereashell=3.502.94\frac{v_{sphere}}{v_{shell}} = \sqrt{\frac{a_{sphere}}{a_{shell}}} = \sqrt{\frac{3.50}{2.94}}

vspherevshell=1.09\boxed{\frac{v_{sphere}}{v_{shell}} = 1.09}

Sphere is 9% faster!

3Problem 3hard

โ“ Question:

Using integration, derive the moment of inertia of a solid cylinder (mass M, radius R, height h) about its central axis. Then calculate for M = 4.0 kg, R = 0.2 m.

๐Ÿ’ก Show Solution

Derivation:

Consider cylindrical shells of radius r, thickness dr.

Volume of shell: dV=2ฯ€rhโ€‰drdV = 2\pi r h \, dr

Mass of shell: dm=ฯdV=ฯโ‹…2ฯ€rhโ€‰drdm = \rho dV = \rho \cdot 2\pi r h \, dr

where density ฯ=Mฯ€R2h\rho = \frac{M}{\pi R^2 h}

Moment of inertia contribution: dI=r2โ€‰dm=r2โ‹…ฯโ‹…2ฯ€rhโ€‰dr=2ฯ€ฯhr3โ€‰drdI = r^2 \, dm = r^2 \cdot \rho \cdot 2\pi r h \, dr = 2\pi\rho h r^3 \, dr

Total: I=โˆซ0R2ฯ€ฯhr3โ€‰dr=2ฯ€ฯh[r44]0RI = \int_0^R 2\pi\rho h r^3 \, dr = 2\pi\rho h \left[\frac{r^4}{4}\right]_0^R

I=2ฯ€ฯhโ‹…R44=ฯ€ฯhR42I = 2\pi\rho h \cdot \frac{R^4}{4} = \frac{\pi\rho h R^4}{2}

Substitute ฯ=Mฯ€R2h\rho = \frac{M}{\pi R^2 h}:

I=ฯ€R4h2โ‹…Mฯ€R2h=MR22I = \frac{\pi R^4 h}{2} \cdot \frac{M}{\pi R^2 h} = \frac{MR^2}{2}

I=12MR2\boxed{I = \frac{1}{2}MR^2}

Numerical calculation:

I=12(4.0)(0.2)2=12(4.0)(0.04)I = \frac{1}{2}(4.0)(0.2)^2 = \frac{1}{2}(4.0)(0.04)

I=0.08ย kg\cdotpm2\boxed{I = 0.08 \text{ kgยทm}^2}