Physics, asked by Juhishrita3329, 10 months ago

a uniform wire of length l and radius r is twisted by a angle alfa.if modulus of rigidity of wire is X then the elastic potential energy stored in wire is???

Answers

Answered by Anonymous
3

The elastic potential energy stored in wire is 1/2Ux²(πR²l)

Length of the wire = l (Given)

Radius of the wire = r (Given)

Thus,

Shear modulus U = The shear stress / The shear strain

Shear stress = U×shear strain

Work done per unit volume will be = 1/ 2 × stress × strain

work done = 1/2 Ux²

A wire can be considered to the cylinder. Hence

PE = 1/2Ux²(πR²l)

Answered by star8661
0

please see the attached file mathematical symbols based

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