question_answer 11)

A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1 m, is whirled in a vertical circle with an angular velocity of 2 rev is at the bottom of the circle. The cross-sectional area of the wire is \[{{\text{a}}_{\text{1}}}\text{=3 }\!\!\times\!\!\text{ 0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}{{\text{m}}^{\text{2}}}\text{;}\] Calculate the elongation of the wire when the mass is at the lowest point of its path\[{{\text{l}}_{\text{1}}}\text{=4 }\!\!\times\!\!\text{ 7m;}\]

Answer:

Here, m=\[\vartriangle {{l}_{1}}=\vartriangle l;\] rps;                              \[{{F}_{1}}=F\]                 Total pulling force on mass, when it is at the lowest position of the vertical circle is                  \[{{\text{a}}_{\text{2}}}\text{=4 }\!\!\times\!\!\text{ 0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}{{\text{m}}^{\text{2}}}\text{,}\]                 \[{{l}_{2}}=3\cdot 5m,\]                                 \[\vartriangle {{l}_{2}}=\vartriangle l;\]                 \[{{F}_{2}}=F\] or \[{{Y}_{1}},{{Y}_{2}}\]                 \[\therefore \]                 \[{{Y}_{1}}=\frac{{{F}_{1}}}{{{a}_{1}}}\times \frac{{{l}_{1}}}{\vartriangle {{l}_{1}}}=\frac{F}{3\cdot 0\times {{10}^{-5}}}\times \frac{4\cdot 7}{\vartriangle l}\]