A rod of circular cross section with a diameter of 5cm and length of 1 m possesses a polarmoment of inertia, ?0, and (shear) modulus of rigidity, ? (see table 5.1 for steel andaluminium) and a mass moment of inertia of 0.5 kg m2 of the disc. For the case of a wireor shaft of diameter ?, ?0 = ??4⁄32. The modulus of rigidity has units ? ?2 ⁄ . Thetorsional stiffness is used to describe the vibration problem illustrated in Figure 5.1, wherethe mass of the shaft is ignored. In the figure, ?(?) represents the angular position of theshaft relative to its equilibrium position. The disk of radius ? and rotational moment ofinertia ? will vibrate around the equilibrium position ?(0) with stiffness ??0⁄?.Calculate the natural frequency of oscillation of the torsional system for both steel andaluminium e(0)Figure 5.1A rod of circular cross section with a diameter of 5cm and length of I m possesses a polarmoment of inertia, lg, and (shear) modulus of rigidity, G (see table 5.1 for steel andaluminium) and a mass moment of incrtia of 0.5 kg m of the disc. For the case of a wireor shaft of diameter d, I, = nd*/32. The modulus of rigidity has units N /m2. Thetorsional stiffness is used to describe the vibration problem illustrated in Figure 5.1, wherethe mass of the shaft is ignored. In the figure, 0(t) represents the angular position of theshaft relative to its equilibrium position. The disk of radius r and rotational moment ofinertia / will vibrate around the equilibrium position 8(0) with stiffness Glo/l.Table 5.1 Physical Constants for some common materialsMaterialYoung'smodulus,Density,(kg/m)Shear modulus,G(N/m?)E(N/m?)2.0 x 10"7.1 x 10107.8 x 102.7 x 108.0 x 102.67 x 10SteelAluminiumCalculate the natural frequency of oscillation of the torsional system for both steel andaluminium.

Calculate the torsional stiffness of steel. kθs=GsI0L Here, kθs is the torsional stiffness of steel,…

Answered: 0(0) Figure 5.1 A rod of circular cross…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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A rod of circular cross section with a diameter of 5cm and length of 1 m possesses a polar
moment of inertia, ?0, and (shear) modulus of rigidity, ? (see table 5.1 for steel and
aluminium) and a mass moment of inertia of 0.5 kg m2 of the disc. For the case of a wire
or shaft of diameter ?, ?0 = ??
4⁄32. The modulus of rigidity has units ? ?2 ⁄ . The
torsional stiffness is used to describe the vibration problem illustrated in Figure 5.1, where
the mass of the shaft is ignored. In the figure, ?(?) represents the angular position of the
shaft relative to its equilibrium position. The disk of radius ? and rotational moment of
inertia ? will vibrate around the equilibrium position ?(0) with stiffness ??0⁄?.

Calculate the natural frequency of oscillation of the torsional system for both steel and
aluminium

e(0)
Figure 5.1
A rod of circular cross section with a diameter of 5cm and length of I m possesses a polar
moment of inertia, lg, and (shear) modulus of rigidity, G (see table 5.1 for steel and
aluminium) and a mass moment of incrtia of 0.5 kg m of the disc. For the case of a wire
or shaft of diameter d, I, = nd*/32. The modulus of rigidity has units N /m2. The
torsional stiffness is used to describe the vibration problem illustrated in Figure 5.1, where
the mass of the shaft is ignored. In the figure, 0(t) represents the angular position of the
shaft relative to its equilibrium position. The disk of radius r and rotational moment of
inertia / will vibrate around the equilibrium position 8(0) with stiffness Glo/l.
Table 5.1 Physical Constants for some common materials
Material
Young's
modulus,
Density,
(kg/m)
Shear modulus,
G(N/m?)
E(N/m?)
2.0 x 10"
7.1 x 1010
7.8 x 10
2.7 x 10
8.0 x 10
2.67 x 10
Steel
Aluminium
Calculate the natural frequency of oscillation of the torsional system for both steel and
aluminium.

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