Question

The robot arm is elevating and extending simultaneously. At a given instant, θ = 30°, ˙ θ = 10 deg / s = constant θ˙=10 deg/s=constant , l = 0.5 m, ˙ l = 0.2 m / s l˙=0.2 m/s , and ¨ l = − 0.3 m / s 2 l¨=−0.3 m/s2 . Compute the magnitudes of the velocity v and acceleration a of the gripped part P. In addition, express v and a in terms of the unit vectors i and j.

Answer 1

Explanation:

The position vector r:

$\overrightarrow{r(t)}=lcos\theta\hat{i}+lsin\theta\hat{j}$

The velocity vector v:

$\overrightarrow{v(t)}=\overrightarrow{\frac{dr}{dt}}=\dot{l}cos\theta-lsin\theta\dot{\theta}\hat{i}+\dot{l}sin\theta+lcos\theta\dot{\theta}\hat{j}$

The acceleration vector a:

$\overrightarrow{a(t)}}=cos\theta(\ddot{l}-l\dot{\theta}^2)-sin\theta(2\dot{l}\dot{\theta}+l\ddot{\theta})\hat{i}+cos\theta(2\dot{l}\dot{\theta}+l\ddot{\theta})+sin\theta(\ddot{l}-l\dot{\theta}^2)\hat{j}$

$\overrightarrow{v(t)}=0.13\hat{i}+0.18\hat{j}$

$\overrightarrow{a(t)}}=-0.3\hat{i}-0.1\hat{j}$