Question

A cylinder of length 1.5 m and diameter 4cm is fixed at one end. A tangential gorse of 4x10^5N is applied at the other end. If the rigidity modulus of the cylinder is 6x10^10 Nm^-2 then, calculate the twist produced in the cylinder .

Answer 1

Length of cylinder , L = 1.5m
diameter of base of cylinder , d = 4cm = 0.4m
so, radius of base , r = 0.2m
force , F = 4 × 10^5 N
rigidity modulus of the cylinder, Y = 6 × 10^10 N/m²

use formula, Y = FL/A∆L
∆L = FL/AY

A = πr² = 3.14 × (0.2)² = 3.14 × 4 × 10^-2
= 12.56 × 10^-2 = 0.1256 m²

so, ∆L = 4 × 10^5 × 1.5/(0.1256 × 6 × 10^10)

= 7.96 × 10^-5 m

Answer 2

Hey dear,

◆ Answer-
x = 7.957 mm

◆ Explaination-
# Given-
l = 1.5 m
r = 2 cm = 0.02 m
F = 4×10^5 N
Y = 6×10^10 N/m^2

# Solution-
Youngs modulus is given by-
Y = Stress / Strain
Y = (F/A) / (x/l)
x = Fl / YA
x = Fl / Yπr^2

Substituting values -
x = (4×10^5 × 1.5) / (6×10^10 × 3.142 × 4×10^-4)
x = 7.957×10^-3 m
x = 7.957 mm

Twist produced in the wire is 7.957 mm.

Hope this helps you...