Test Index

Mechanical Properties of Solids

Show Para  Hide Para 
© examsnet.com
Question : 11 of 21
Marks: +1, -0
A 14.5 kg mass, fastened to the end of a steel wire of unstretchedlength 1.0 m, is whirled in a vertical circle with an angular velocity of2 rev/s at the bottom of the circle. The cross-sectional area of the wire is0.065 cm2^{2}. Calculate the elongation of the wire when the mass is at thelowest point of its path.
Solution:  
Here, m = 14.5 kg;
A=0.065 cm2=0.065×10−4 m2;A=0.065\,\text{cm}^2=0.065\times 10^{-4}\,\text{m}^2;
L=1 mL=1\,\text{m} and v=2 r.p.s.v=2\,\text{r.p.s.}
When the mass is at the lowest point of its circular path, the stretching force on the wire,
F=mg+mrω2F=mg+mr\omega^{2}
=mg+mr×(2πv)2=mg+mr\times(2\pi v)^{2}
=14.5×9.8+14.5×1×(2π×2)2=14.5\times 9.8+14.5\times 1\times(2\pi\times 2)^{2}
=142.1+2,289.75=2,431.85 N=142.1+2,289.75=2,431.85\,\text{N}
Now,Y=FAlL\text{Now},\quad Y=\frac{\frac{F}{A}}{\frac{l}{L}}
or l=FLAYl=\frac{F L}{AY}
=2,431.85×10.065×10−4×2.0×1011=\frac{2,431.85\times 1}{0.065\times 10^{-4}\times 2.0\times 10^{11}}
=1.87×10−3 m=1.87 mm=1.87\times 10^{-3}\,\text{m}=1.87\,\text{mm}
© examsnet.com
Go to Question:
Prev Question
Next Question