- #1
Carla1985
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Im getting stuck on a couple of the questions for my mechanics homework. Il put what I've done so far and hopefully someone can check I am doing it right n push me into the right direction for the rest.
"A grandfather clock has a simple pendulum of length L with a bob of mass m. The effect of air resistance is to produce a force on the bob of magnitude 2cm times its speed, where c > 0 is a positive constant.
(i) Show that, provided the angular displacement θ of the pendulum from the downward vertical is small, θ approximately satisfies the equation,
\[
\ddot{\theta}+2c\dot{\theta}+\frac{g}{L}\theta=0
\]
[Hint: Use polar coordinates and Taylor’s theorem as we did in lectures for the case without air resistance.]
(ii) Suppose now that c = $\sqrt{g/L}$. Show that if $\theta(0) = a$ and $\dot{\theta}(0) = 0$, then
$\theta(t) = a(1 + ct)e^{−ct}$
Will the pendulum ever return to the vertical position in this case? Explain
your answer"
so far I have:
The rod has fixed length r=L so
$\vec{r}=L\vec{e}_r$
as r=L is a constant, $\dot{r}=\ddot{r}=0$
so acceleration is
$\ddot{r}=(\dot{r}-r\dot{\theta}^2)e_r+(2\dot{r}\dot{\theta}+r\ddot{\theta})e_\theta$
$=-L(\dot{\theta}^2e_r+L\ddot{\theta}e_\theta$
so the total force is
$(mgcos\theta-T)e_r-(mgsin\theta+2cm\dot{x})e_\theta$I've completed a very similar question without air resistance without too much problems but not sure if I've added air resistance in right at all. Hope it at least makes some sense :/
Thanks
"A grandfather clock has a simple pendulum of length L with a bob of mass m. The effect of air resistance is to produce a force on the bob of magnitude 2cm times its speed, where c > 0 is a positive constant.
(i) Show that, provided the angular displacement θ of the pendulum from the downward vertical is small, θ approximately satisfies the equation,
\[
\ddot{\theta}+2c\dot{\theta}+\frac{g}{L}\theta=0
\]
[Hint: Use polar coordinates and Taylor’s theorem as we did in lectures for the case without air resistance.]
(ii) Suppose now that c = $\sqrt{g/L}$. Show that if $\theta(0) = a$ and $\dot{\theta}(0) = 0$, then
$\theta(t) = a(1 + ct)e^{−ct}$
Will the pendulum ever return to the vertical position in this case? Explain
your answer"
so far I have:
The rod has fixed length r=L so
$\vec{r}=L\vec{e}_r$
as r=L is a constant, $\dot{r}=\ddot{r}=0$
so acceleration is
$\ddot{r}=(\dot{r}-r\dot{\theta}^2)e_r+(2\dot{r}\dot{\theta}+r\ddot{\theta})e_\theta$
$=-L(\dot{\theta}^2e_r+L\ddot{\theta}e_\theta$
so the total force is
$(mgcos\theta-T)e_r-(mgsin\theta+2cm\dot{x})e_\theta$I've completed a very similar question without air resistance without too much problems but not sure if I've added air resistance in right at all. Hope it at least makes some sense :/
Thanks