A cylinder of length 1.5m and diameter 4cm is fixed at one end. A tangential force of 4N is applied at the other end. If the rigidity modulus of the cylinder is 6 N/msquare then calculate the twist
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Very long and I didn't understand
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Hey dear,
◆ Answer-
x = 9.946 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 = (5×10^5 × 1.5) / (6×10^10 × 3.142 × 4×10^-4)
x = 9.946×10^-3 m
x = 9.946 mm
Twist produced in the wire is 9.946 mm.
Hope this helps you...
◆ Answer-
x = 9.946 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 = (5×10^5 × 1.5) / (6×10^10 × 3.142 × 4×10^-4)
x = 9.946×10^-3 m
x = 9.946 mm
Twist produced in the wire is 9.946 mm.
Hope this helps you...
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