Question

ensino superior ajuda com esses exercicios ?

Answer 1

Questão 1:

$\theta(t)=3-2t^3+t$


Velocidade angular instantânea:

$\omega(t)=\dfrac{d\theta}{dt}\,(t)\\\\\\ \omega(t)=\dfrac{d}{dt}\,(3-2t^3+t)\\\\\\ \omega(t)=-6t^2+1$


Aceleração angular instantânea:

$\alpha(t)=\dfrac{d\omega}{dt}\,(t)\\\\\\ \alpha(t)=\dfrac{d}{dt}\,(-6t^2+1)\\\\\\ \alpha(t)=-12t$


(a) $\theta(1)=3-2\cdot 1^3+1$

$\theta(1)=3-2+1\\\\ \boxed{\begin{array}{c}\theta(1)=2\mathrm{~rad} \end{array}}$


(b) $\omega(1)=-6\cdot 1^2+1$

$\omega(1)=-6\cdot 1^2+1\\\\ \omega(1)=-6+1\\\\ \boxed{\begin{array}{c}\omega(1)=-5\mathrm{~rad/s} \end{array}}$


(c) $\omega(3)=-6\cdot 3^2+1$

$\omega(3)=-54+1\\\\ \boxed{\begin{array}{c}\omega(3)=-53\mathrm{~rad/s} \end{array}}$


(d) $\alpha(6)=-12\cdot 6$

$\boxed{\alpha(6)=\begin{array}{c}-72\mathrm{~rad/s^2} \end{array}}$


A aceleração angular não é constante, pois ela depende do instante $t.$
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Questão 2:

$\theta(t)=2t-2t^2+t^4$


$\omega(t)=\dfrac{d\theta}{dt}\,(t)\\\\\\ \omega(t)=\dfrac{d}{dt}\,(2t-2t^2+t^4)\\\\\\ \omega(t)=2-4t+4t^3$


$\alpha(t)=\dfrac{d\omega}{dt}\,(t)\\\\\\ \alpha(t)=\dfrac{d}{dt}\,(2-4t+4t^3)\\\\\\ \alpha(t)=-4+12t^2$


(a) $\omega(1)=2-4\cdot 1+4\cdot 1^3$

$\omega(1)=2-4+4\\\\ \boxed{\begin{array}{c} \omega(1)=2\mathrm{~rad/s} \end{array}}$


(b) $\omega(3)=2-4\cdot 3+4\cdot 3^3$

$\omega(3)=2-4\cdot 3+4\cdot 3^3\\\\ \omega(3)=2-12+108\\\\ \boxed{\begin{array}{c} \omega(3)=98\mathrm{~rad/s} \end{array}}$


(c) Aceleração angular média no intervalo:

$\overline{\alpha}_{1\to 3}=\dfrac{\omega(3)-\omega(1)}{3-1}\\\\\\ \overline{\alpha}_{1\to 3}=\dfrac{98-2}{2}\\\\\\ \overline{\alpha}_{1\to 3}=\dfrac{96}{2}\\\\\\ \boxed{\begin{array}{c} \overline{\alpha}_{1\to 3}=48\mathrm{~rad/s^2} \end{array}}$


(d) Aceleração angular inicial:

$\alpha(1)=-4+12\cdot 1^2\\\\ \alpha(1)=-4+12\\\\ \boxed{\begin{array}{c}\alpha(1)=8\mathrm{~rad/s^2} \end{array}}$


Aceleração angular final:

$\alpha(3)=-4+12\cdot 3^2\\\\ \alpha(3)=-4+108\\\\ \boxed{\begin{array}{c}\alpha(3)=104\mathrm{~rad/s^2} \end{array}}$

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Questão 3:

Velocidade angular inicial: $\omega(0)=10\mathrm{~rad/s}$

Aceleração angular constante: $\alpha=-2\mathrm{~rad/s^2}$


Velocidade angular em função do tempo:

$\omega(t)=\omega(0)+\alpha t\\\\ \boxed{\begin{array}{c} \omega(t)=10-2t \end{array}}$


(a) Encontrar o instante em que a velocidade angular se anula:

$\omega(t)=0\\\\ 10-2t=0\\\\ 2t=10\\\\ \boxed{\begin{array}{c}t=5\mathrm{~s} \end{array}}$


(b) Deslocamento angular dado pela fórmula de Torricelli:

$\omega^2(t_{f})-\omega^2(t_{i})=2\alpha\cdot \Delta \theta\\\\\\ \Delta \theta=\dfrac{\omega^2(t_{f})-\omega^2(t_{i})}{2\alpha}\\\\\\ \Delta \theta=\dfrac{\omega^2(5)-\omega^2(0)}{2\alpha}\\\\\\ \Delta \theta=\dfrac{0^2-10^2}{2\cdot (-2)}\\\\\\ \Delta \theta=\dfrac{-100}{-4}\\\\\\ \boxed{\begin{array}{c}\Delta \theta=25\mathrm{~rad} \end{array}}$

(este é o deslocamento do tambor até ele parar)