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Measurement & regression

Physical pendulum

Time the plate alone and with a roller and measure its dimensions — then work out the moment of inertia in three ways.

Example: A 600 × 104 mm steel plate: 25T and 25T′, ten readings of each dimension → J₀ three ways.

Time a steel plate on an off-centre axis, then with a roller on it, and measure the dimensions. The simulator runs the whole measurement; you work out the moment of inertia J in three ways in the calculators below.

Apparatusnot to scale
Angleθ against time, dots are the counted passes through zero
How J₀ from method 2 depends on T′the other quantities held at their measured values
Presets
Standard valuesdensity and moment of inertia
Plate and roller materialρ / kg·m-3ρ / g·cm-3
Steel (structural)78507.85
Stainless steel79007.9
Cast iron72007.2
Copper89608.96
Brass8400 – 87008.4 – 8.7
Bronze88008.8
Aluminium27002.7
Titanium45004.5
Lead1134011.34

Typical handbook densities. The mass of a plate is M = ρ·b·c·t (t is the thickness); holes lower it.

Moments of inertia and formulas

Body / situationFormula
Rectangular plate b × c, axis perpendicular through the centreJE = M·(b² + c²) / 12
Thin rod of length l, axis perpendicular through the centreJE = M·l² / 12
Cylinder or disc of radius r, about its own axisJE = M·r² / 2
Sphere of radius r, axis through the centreJE = 2·M·r² / 5
Parallel-axis theorem, axis at distance a from the centre of massJ = JE + M·a²
Physical pendulum, small swingsT = 2π√(J / (M·g·a))
Gravitational acceleration used in the simulationg = 9.81 m·s-2
Key equations
(141)
T=2πJMga⇒J=g4π2aMT2
Method 1 — plate alone
(142)
Jp=18md2
added cylinder about its own axis
(143)
Jop=Jp+mL2
Steiner’s theorem
(145)
J=T2T2−T‾2m[gLT‾24π2−d28−L2]
Method 2 — main result, with the added body
(146)
J=112M(b2+c2)+Ma2
Method 3 — direct calculation for the rectangular plate
Parameters
SymbolMeaningUnit
Jmoment of inertia of the plate about pivot O (J0)kg·m2
Mmass of the platekg
mmass of the added cylinderkg
adistance pivot O – centre of massm
b,cplate length and widthm
ddiameter of the added cylinderm
Ldistance pivot – added cylinderm
Tperiod without the added bodys
T‾period with the added bodys
ggravitational accelerationm·s-2
Error calculationformulas without values

Method 1 · From the period of the plate alone

J=g4π2aMT2
ϑ(J)=(∂J∂aϑ(a))2+(∂J∂Mϑ(M))2
Partial derivatives
∂J∂a=g4π2MT2
∂J∂M=g4π2aT2

Method 3 · Direct calculation (rectangular plate)

J=112M(b2+c2)+Ma2
ϑ(J)=(∂J∂aϑ(a))2+(∂J∂bϑ(b))2+(∂J∂cϑ(c))2+(∂J∂Mϑ(M))2
Partial derivatives
∂J∂a=2Ma
∂J∂b=16Mb
∂J∂c=16Mc
∂J∂M=112(b2+c2)+a2

Method 2 · With the added body (main result)

J=T2T2−T‾2m[gLT‾24π2−d28−L2]
ϑ(J)=(∂J∂mϑ(m))2+(∂J∂Lϑ(L))2+(∂J∂dϑ(d))2
Partial derivatives
∂J∂m=T2T2−T‾2[gLT‾24π2−d28−L2]
∂J∂L=T2T2−T‾2m[gT‾24π2−2L]
∂J∂d=T2T2−T‾2m(−2d8)=−T2T2−T‾2md4