Question

Un corp cu masa de 200 g este suspendat de un fir de lungime egala cu 50 cm.In timpul când corpul descrie o circumferinta in planul orizontal,firul formeaza unghiul de 60° fata de verticala.Determinati forta de tensiune a firului,acceleratia corpului, viteza liniara si viteza lui ungbiulara.

Answer 1

Intai, descompunem tensiunea in componente:

$\vec{T}=\vec{T_x}+\vec{T_y}\implies T_x=T\sin\alpha,\:\: T_y=T\cos\alpha$

Aplicam echilibru pe axele de coordonate:

$Oy: \vec{T_y}+\vec{G}=0\implies T\cos\alpha=mg\implies T=\dfrac{mg}{\cos\alpha}=4N\\ \\ Ox: \vec{T_x}+\vec{F_{cf}}=0\implies T\sin\alpha=\dfrac{mv^2}{l\sin\alpha}\\ \\ \dfrac{g}{\cos\alpha}=\dfrac{v^2}{l\sin^2\alpha}\implies v=\sqrt{\dfrac{gl\sin^2\alpha}{\cos\alpha}}=\sqrt{\dfrac{gl}{\cos\alpha}}\sin\alpha\\\\\text{Numeric, }v=\sqrt{10\cdot 0,5\cdot 2}\cdot \dfrac{\sqrt{3}}{2}m/s\approx 2,738m/s$

$v=\omega R\implies \omega=\dfrac{v}{l\sin\alpha}\\\\\text{Numeric, }\omega\approx 6,32455rad/s$

$a_{cp}=v^2/R=\dfrac{v^2}{l\sin\alpha}\approx 17,3205080757m/s^2$