Question

The bent rod ABCDE rotates about a line joining Points A and E with a constant angular velocity of 7.5 rad's. Knowing that the rotation is clockwise as viewed from E, determine the velocity and acceleration of corner C. (Round the final answer to two decimal places.) 200 min 250 mm B D 1:50 100 mm m/si- m/s The velocity of comer is vo- The acceleration of corner C is ac

Answer 1

Since the axis of rotation is AE, we determine the Unit vector of EA ($\lambda$_EA =
ΕΑ
/EA). Then we calculate the common angular velocity vector ($\vec{\omega}$).

Then we calculate the velocity of corner C using V_C = $\vec{\omega}$ x $\vec{r}$ _C/E , where $\vec{r}$ _C/E = position vector of EC.

Velocity of corner C is " $\vec{V}$ _C = -0.37 $\hat{i}$ - 1$\hat{j}$ +1.25 $\hat{k}$ m/s ".

We know that acceleration of corner C is a_C = r$\alpha$ + r$\omega$², But the rod roates with only angular velocity, angular acceleration($\alpha$) =0.

Now, a_C = $\omega$ x ($\omega$ x $\vec{r}$ _C/E) = $\omega$ x $\vec{V}$ _C. Thus we get the acceleration of corner C.

Acceleration of corner C is " $\vec{a}$ _C = 8.75 $\hat{i}$ + 5.31 $\hat{j}$ + 6.87  $\hat{k}$m/s ".

All the calculations are shown below for clear understanding.

Given, WAE = 7.5 rad/sec Dintance bten ARE EA = 0.4² +0.4 to 2 EA=0.6m vector form EA = - (0.4m)i + (0.4m)j + (0.2m) ER I tonito uj +0.26) EA EA 0.6 *EA = 1/3 E2i + 2) tk) hele = -0.4i tois know that فيه ج دیا AE EA we (1.5) // E2i + zjtk) - si + 5j +2.5k we how that v=WxTCE -S S 25 -oy ois Vai(0-0375) – j (0+1)+ (-0 75+2) velocity of = -0.375 î _ +1.25 mys coher С

Acceleration (ac) exactRelE+ we (wx Teft) (XAC=o - at w x te - ..? ac = 2.5 1-0.375 -1 1.25 ac = i (6-25 +2:3)-j 6625+0.9375) +k(5+1.875) Accleration of comerc a lac=8.75 î+ 5.31 7 + 687 F