Moment of Inertia

Calculating moment of inertia using integration, parallel axis theorem

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Moment of Inertia

Definition

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

where rir_i is perpendicular distance from rotation axis.

For continuous mass distribution: I=r2dmI = \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: λ=M/L\lambda = M/L

dm=λdx=MLdxdm = \lambda \, dx = \frac{M}{L} \, dx

I=0Lx2dm=0Lx2MLdxI = \int_0^L x^2 \, dm = \int_0^L x^2 \frac{M}{L} \, dx

I=MLx330L=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 ρ\rho.

Use cylindrical shells: dm=ρ2πrdrhdm = \rho \cdot 2\pi r \, dr \cdot h

I=0Rr2dm=0Rr2ρ2πrhdrI = \int_0^R r^2 \, dm = \int_0^R r^2 \cdot \rho \cdot 2\pi rh \, dr

I=2πρh0Rr3dr=2πρhR44I = 2\pi\rho h\int_0^R r^3 \, dr = 2\pi\rho h \frac{R^4}{4}

Using M=ρπR2hM = \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=R2z2r = \sqrt{R^2 - z^2}.

dI=12(dm)r2=12ρπr2dzr2=12ρπ(R2z2)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=20R12ρπ(R2z2)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=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.

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