- 1

- 0

**1. Homework Statement**

The rods are hengedly connected. O is fixed, C is free to move in the x-direction and AB is parallel to OC. OA turns around O with an angular velocity of ##\omega##.

further given: $$\theta = 30^{\circ}$$ $$|OA|=|BC|=l_1=0.30m$$ $$|AB|=l_2=0.40m$$ $$\omega=15rad/s$$ $$v_c = 6 m/s$$

Question: Find ##v_{B}##.

**2. Homework Equations**

$$v_{B,\text{abs}} = v_{B,\text{sleep}} + v_{B,\text{rel}}$$ $$v_{B,\text{sleep}} = v_{0'} + \omega '\cdot |OB| = v_A = \omega\cdot l_1$$

**3. The Attempt at a Solution**

I define a non-rotating coordinate system o'x'y' that is fixed in point A.

$$v_{B,\text{abs,x}} = -v_{B,\text{tan}}\cos{\beta} + v_c = v_{B,\text{sleep}}\cos{\beta}$$ $$v_{B,\text{abs,y}} = v_{B,\text{tan}}\sin{\beta} = -v_{B,\text{rel}} - v_{B,\text{sleep}}\sin{\beta}$$ $$v_{B,\text{sleep}}=\frac{-\omega l_1\cos{\beta}+v_c}{\cos{\beta}}=7.5m/s$$ I think this is incorrect. $$v_{B,\text{rel}}=-v_{B,\text{tan}}\sin{\beta}-v_{B,\text{sleep}}\sin{\beta}$$ This can't be correct, because this is a negative number and I know from the solution that ##\omega_{BC} = \omega_{OA}## so, I believe, the velocity of B relative to o'x'y' should work out to zero.

What am I doing wrong?

(extra) written version: http://i.imgur.com/WAeCf8L.jpg

Last edited: