Measurement & regression

Torsion pendulum

Release a body on a wire, time 30 oscillations and build the table of 15T differences.

Example: A sphere on a steel wire: 30 oscillations, five 15T differences → G from the period, using Probable error.

Uvolněte těleso na drátu, stopkami s mezičasy změřte 30 kmitů a sestavte tabulku postupných rozdílů. Simulace ukazuje celé měření, modul pružnosti ve smyku G spočítáte v kalkulačkách níže.Release the body on the wire, time 30 oscillations with the lap stopwatch and build the table of successive differences. The simulator shows the whole measurement; you work out the shear modulus G in the calculators below.

Apparatusnot to scale

Table · successive differenceslap every 3 oscillations

0.00s

i	i·T / s	i+15	(i+15)·T / s	15T / s

The five 15T readingsbrass line is the true value

Angleθ against time, dots are the counted zero crossings

Presets

Standard valuesshear modulus G and moment of inertia

Materiál drátuWire material	G / GPa	G / 10¹⁰ Pa
Ocel (uhlíková, konstrukční)Steel (carbon, structural)	77 – 7977 – 79	7,7 – 7,97.7 – 7.9
Nerezová ocelStainless steel	7878	7,87.8
LitinaCast iron	4141	4,14.1
MěďCopper	4545	4,54.5
BronzBronze	4545	4,54.5
MosazBrass	36 – 4036 – 40	3,6 – 4,03.6 – 4.0
Hliník (slitiny)Aluminium (alloys)	25,5 – 2725.5 – 27	2,6 – 2,72.6 – 2.7
TitanTitanium	4141	4,14.1
OlovoLead	1313	1,31.3

Orientační tabulkové hodnoty modulu pružnosti ve smyku při pokojové teplotě. Skutečná hodnota závisí na slitině a tepelném zpracování, proto ji u svého drátu ověřte v datovém listu. 1 × 10¹⁰ Pa = 10 GPa. Zdroje: Wikipedia (Shear modulus), Engineers Edge (Modulus of rigidity).Typical shear modulus values at room temperature. The real value depends on the alloy and its heat treatment, so check the datasheet of your own wire. 1 × 10¹⁰ Pa = 10 GPa. Sources: Wikipedia (Shear modulus), Engineers Edge (Modulus of rigidity).

Moment setrvačnosti tělesaMoment of inertia of the body

TělesoBody	c	Moment setrvačnostiMoment of inertia
Koule (osa středem)Sphere (axis through the centre)	2/5	J = 0,4·m·R²J = 0.4·m·R²
Válec (osa válce)Cylinder (about its own axis)	1/2	J = 0,5·m·R²J = 0.5·m·R²
Kužel (osa kužele)Cone (about its own axis)	3/10	J = 0,3·m·R²J = 0.3·m·R²

Calculate your result

Pět rozdílů 15T zpracujte v Pravděpodobné chybě, průměr drátu d a délku L v kalkulačce Více veličin. Z period a rozměrů pak dosaďte do vzorce nahoře a porovnejte G s tabulkou.Work up the five 15T differences in Probable error, and the wire diameter d and length L in the Several quantities calculator. Then put the period and the dimensions into the formula above and compare G with the table.

Probable error Several quantities

Key equations

(135)

G = \frac{128 π L J}{T^{2} d^{4}}

(136)

J = \frac{2}{5} m R^{2}, J = \frac{1}{2} m R^{2}

sphere, cylinder

(137)

J = \frac{3}{10} m R^{2}, J = \frac{3}{10} m \frac{{R_{1}}^{5} - {R_{2}}^{5}}{{R_{1}}^{3} - {R_{2}}^{3}}

cone, truncated cone (R₁ > R₂)

Parameters

SymbolSymbol	VýznamMeaning	JednotkaUnit
$G$	modul pružnosti ve smyku drátushear modulus of the wire	Pa
$L$	délka drátuwire length	m
$d$	průměr drátuwire diameter	m
$T$	perioda torzních kmitůtorsional oscillation period	s
$J$	moment setrvačnosti zavěšeného tělesamoment of inertia of the suspended body	kg·m²
$m$	hmotnost tělesabody mass	kg
$R$	poloměr tělesabody radius	m

Error calculationformulas without values

A · Koule zavěšená na drátuA · Sphere suspended from the wire

G = \frac{256 π L m R^{2}}{5 T^{2} d^{4}}

ϑ (G) = \sqrt{{(\frac{\partial G}{\partial L} ϑ (L))}^{2} + {(\frac{\partial G}{\partial m} ϑ (m))}^{2} + {(\frac{\partial G}{\partial R} ϑ (R))}^{2} + {(\frac{\partial G}{\partial d} ϑ (d))}^{2}}

Partial derivatives

\frac{\partial G}{\partial L} = \frac{256 π m R^{2}}{5 T^{2} d^{4}}

\frac{\partial G}{\partial m} = \frac{256 π L R^{2}}{5 T^{2} d^{4}}

\frac{\partial G}{\partial R} = \frac{256 π L m \cdot 2 R}{5 T^{2} d^{4}}

\frac{\partial G}{\partial d} = \frac{256 π L m R^{2}}{5 T^{2}} d^{- 5} (- 4)

B · Válec zavěšený na drátuB · Cylinder suspended from the wire

G = \frac{64 π L m R^{2}}{T^{2} d^{4}}

ϑ (G) = \sqrt{{(\frac{\partial G}{\partial L} ϑ (L))}^{2} + {(\frac{\partial G}{\partial m} ϑ (m))}^{2} + {(\frac{\partial G}{\partial R} ϑ (R))}^{2} + {(\frac{\partial G}{\partial d} ϑ (d))}^{2}}

Partial derivatives

\frac{\partial G}{\partial L} = \frac{64 π m R^{2}}{T^{2} d^{4}}

\frac{\partial G}{\partial m} = \frac{64 π L R^{2}}{T^{2} d^{4}}

\frac{\partial G}{\partial R} = \frac{64 π L m \cdot 2 R}{T^{2} d^{4}}

\frac{\partial G}{\partial d} = \frac{64 π L m R^{2}}{T^{2}} d^{- 5} (- 4)